22 01 2014_03_23_31_eee_formula_sheet_final

One stop series for

GATE/IES/PSU-2014

(Formula Sheet for EEE Dept.,)

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Formula Sheet (EEE Department)

Index Page. No.

1. Mathematics --- --- 1 to 9

2. Electromagnetic Fields --- 10 to 16

3. Signals & Systems --- --- 17 to 22

4. Electrical Machines --- --- 23 to 58

5. Power Systems --- --- 59 to 89

6. Control Systems --- --- 90 to 92

7. Measurements --- --- 93 to 114

8. Analog Electronics --- --- 115 to 123

9. Digital Electronics --- --- 124 to 126

10. Microprocessors --- --- 127 to 130

11. Power Electronics --- --- 131 to 143

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Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet MatheMatics

Matrix :- • If |A| = 0 → Singular matrix ; |A| ≠ 0 Non singular matrix • Scalar Matrix is a Diagonal matrix with all diagonal elements are equal • Unitary Matrix is a scalar matrix with Diagonal element as ‘1’ (AQ = (A∗)T = A−1 ) • If the product of 2 matrices are zero matrix then at least one of the matrix has det zero • Orthogonal Matrix if AAT = AT.A = I ⇒ AT = A−1 • A = AT → Symmetric

A = - AT → Skew symmetric Properties :- (if A & B are symmetrical ) • A + B symmetric • KA is symmetric • AB + BA symmetric • AB is symmetric iff AB = BA • For any ‘A’ → A + AT symmetric ; A - AT skew symmetric. • Diagonal elements of skew symmetric matrix are zero • If A skew symmetric A2n → symmetric matrix ; A2n−1 → skew symmetric • If ‘A’ is null matrix then Rank of A = 0.

Consistency of Equations :- • r(A, B) ≠ r(A) is consistent • r(A, B) = r(A) consistent &

if r(A) = no. of unknowns then unique solution r(A) < no. of unknowns then ∞ solutions .

Hermition , Skew Hermition , Unitary & Orthogonal Matrices :- • AT = A∗ → then Hermition • AT = −A∗ → then Hermition • Diagonal elements of Skew Hermition Matrix must be purely imaginary or zero • Diagonal elements of Hermition matrix always real . • A real Hermition matrix is a symmetric matrix. • |KA| = Kn |A| Eigen Values & Vectors :- • Char. Equation |A – λI| = 0. Roots of characteristic equation are called eigen values . Each eigen value corresponds to non zero solution X such that (A – λI)X = 0 . X is called Eigen vector . • Sum of Eigen values is sum of Diagonal elements (trace) • Product of Eigen values equal to Determinent of Matrix . • Eigen values of AT & A are same • λ is Eigen value of A then 1/ λ → A−1 & |A|

λ is Eigen value of adj A.

• λ1, λ2 …… λn are Eigen values of A then

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Pg.No. 1 of 143



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Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet KA → K λ1 , K λ2 ……..K λn Am → λ1

m , λ2m ………….. λn

m . A + KI → λ1 + k , λ2 + k , …….. λn + k (A − KI)2 → (λ1 − k)2 , ……… (λn − k)2

• Eigen values of orthogonal matrix have absolute value of ‘1’ . • Eigen values of symmetric matrix also purely real . • Eigen values of skew symmetric matrix are purely imaginary or zero . • λ1 , λ2 , …… λn distinct eigen values of A then corresponding eigen vectors X1 , X2, .. … Xn for

linearly independent set . • adj (adj A) = |A|n−2 ; | adj (adj A) | = |A|(n−1)2 Complex Algebra :- • Cauchy Rieman equations

𝜕𝜕𝑢𝑢𝜕𝜕𝜕𝜕= 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 ; 𝜕𝜕𝑢𝑢𝜕𝜕𝜕𝜕= − 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕𝜕𝜕𝑢𝑢𝜕𝜕𝜕𝜕= 1r 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕= − 1 r 𝜕𝜕𝑢𝑢

𝜕𝜕𝜕𝜕

� Neccessary & Sufficient Conditions for f(z) to be analytic

• ∫ f(z)/(Z − a)n+1

c dz = 2πin!

[ f n(a) ] if f(z) is analytic in region ‘C’ & Z =a is single point

• f(z) = f(z0) + f ′(z0) (z−z0)1!

+ f ′′(z0) (z−z0)2

2! + …… + f n(z0) (z−z0)n

n! + ………. Taylor Series

⇓ if z0 = 0 then it is called Mclauren Series f(z) = ∑ an (z − z0)n∞

0 ; when an = fn(z0)n !

• If f(z) analytic in closed curve ‘C’ except @ finite no. of poles then

∫ f(z)dzc = 2πi (sum of Residues @ singular points within ‘C’ ) Res f(a) = lim

𝑧𝑧→𝑎𝑎(𝑍𝑍 − 𝑎𝑎 𝑓𝑓(𝑧𝑧)

= Φ(a) / φ′(a) = lim

𝑍𝑍→𝑎𝑎1

(𝑛𝑛−1)! 𝑑𝑑𝑛𝑛−1

𝑑𝑑𝑧𝑧𝑛𝑛−1 ((Z − a)n f(z) )

Calculus :-

Rolle’s theorem :-

If f(x) is

(a) Continuous in [a, b]


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Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet (b) Differentiable in (a, b)

(c) f(a) = f(b) then there exists at least one value C ϵ (a, b) such that f ′(c) = 0 .

Langrange’s Mean Value Theorem :-

If f(x) is continuous in [a, b] and differentiable in (a, b) then there exists atleast one value ‘C’ in (a, b) such that f ′(c) = f(b)−f(a)

b−a

Cauchy’s Mean value theorem :-

If f(x) & g(x) are two function such that

(a) f(x) & g(x) continuous in [a, b]

(b) f(x) & g(x) differentiable in (a, b)

(c) g′(x) ≠ 0 ∀ x in (a, b)

Then there exist atleast one value C in (a, b) such that

f ′(c) / g′(c) = f(b)−f(a)g(b)−g(a)

Properties of Definite integrals :-

• a < c < b ∫ f(x). dxba = ∫ f(x). dxc

a + ∫ f(x). dxbc

• ∫ f(x)dxa0 = ∫ f(a − x)dx a

0

• ∫ f(x). dxa−a = 2 ∫ f(x)dx a

0 f(x) is even

= 0 f(x) is odd

• ∫ f(x). dxa0 = 2 ∫ f(x)dx a

0 if f(x) = f(2a- x)

• = 0 if f(x) = - f(2a – x)

• ∫ f(x). dxna0 = n ∫ f(x)dx a

0 if f(x) = f(x + a)

• ∫ f(x). dxba = ∫ f(a + b − x). dx b

a


Pg.No. 3 of 143





Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet • ∫ x f(x). dxa

0 = a2 ∫ f(x). dx a

0 if f(a - x) = f(x)

• ∫ sinnxπ/20 = ∫ cosnxπ/2

0 = (n−1)(n−3)(n−5)………2n (n−2)(n−4)……….3

if ‘n’ odd

= (n−1)(n−3)……1

n (n−2)(n−4)……….2 . �𝜋𝜋

2� if ‘n’ even

• ∫ sinmxπ/20 . cosnx . dx = {(m−1)(m−3)….(m−5)……(2 or 1)} {(n−1)(n−3)…….(2 or 1)}.K

(m+n) (m+n−2)(m+n−4)………2 or 1

Where K = π / 2 when both m & n are even otherwise k = 1

Maxima & Minima :-

A function f(x) has maximum @ x = a if f ′(a) = 0 and f ′′ (a) < 0

A function f(x) has minimum @ x = a if f ′(a) = 0 and f ′′ (a) > 0

Constrained Maximum or Minimum :-

To find maximum or minimum of u = f(x, y, z) where x, y, z are connected by Φ (x, y, z) = 0

Working Rule :-

(i) Write F(x, y, z) = f(x, y, z) + λ ϕ(x, y, z)

(ii) Obtain Fx = 0, Fy = 0 , Fz = 0

(ii) Solve above equations along with ϕ = 0 to get stationary point .

Laplace Transform :-

• L � 𝑑𝑑𝑛𝑛

𝑑𝑑𝑑𝑑𝑛𝑛 𝑓𝑓(𝑠𝑠)� = sn f(s) - sn−1 f(0) - sn−2 f ′(0) …… f n−1 (0)

• L { tn f(t) } = (−1)n dn

dsn f(s)

• f(t)t

⇔ ∫ f(s)∞s ds

• ∫ f(u)t

0 du ⇔ f(s) / s . Inverse Transforms :-


Pg.No. 4 of 143





Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet • s

(s2+a2)2 = 12a

t sin at

• s2

(s2+a2)2 = 12a

[ sin at + at cos at]

• 1(s2+a2)2 = 1

2a3 [ sin at - at cos at]

• ss2− a2 = Cos hat

• a

s2− a2 = Sin hat

Laplace Transform of periodic function : L { f(t) } = ∫e−st f(t)dtT

01−e−sT

Numerical Methods :-

Bisection Method :-

(1) Take two values of x1 & x2 such that f(x1) is +ve & f(x2) is –ve then x3 = x1+x22

find f(x3) if f(x3) +ve then root lies between x3 & x2 otherwise it lies between x1 & x3 .

Regular falsi method :-

Same as bisection except x2 = x0 - x1−x0f(x1)−f(x0)

f(x0)

Newton Raphson Method :-

xn+1 = xn – f(xn)f′(xn)

Pi cards Method :-

yn+1 = y0 + ∫ f(x, ynx

x0) ← dy

dx = f(x, y)

Taylor Series method :-

dydx

= f(x, y) y = y0 + (x- x0) (y′)0 + (x− x0)2

2! (y)0

′′ + …………. (x− x0)n

n! (y)0

n

Euler’s method :-

y1 = y0 + h f(x0, y0) ← dydx

= f(x, y

y1(1) = y0 + h

2 [f(x0, y0) + f(x0 + h, y1)


Pg.No. 5 of 143





Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet y1

(2) = y0 + h2 [f(x0, y0) + f(x0+h, y1

(1)) ]

:

:

Calculate till two consecutive value of ‘y’ agree

y2 = y1 + h f(x0 + h, y1)

y2(1) = y0 + h

2 [f(x0 + h, y1) + f(x0 + 2h, y2)

………………

Runge’s Method :- k1 = h f(x0, y0)

k2 = h f( x0 + h2, y0 + k1

2 ) finally compute K = 1

6 (K1+ 4K2 + K3)

k′ = h f(x0+h , y0 + k1)

k3 = h ( f (x0+h , y0 + k′ ))

Runge Kutta Method :- k1 = h f(x0, y0)

k2 = h f( x0 + h2, y0 + k1

2 ) finally compute K = 1

6 (K1+ 2K2 + 2K3 + K4)

k3 = h f(x0 + h2

, y0 + k22

) ∴ approximation vale y1 = y0 + K .

k3 = h f (x0+h , y0 + k3 )

Trapezoidal Rule :-

∫ f(x). dxx0+nhx0

= h2 [ ( y0 + yn) + 2 (y1 + y2 + ……. yn−1 )]

f(x) takes values y0, y1 …..

@ x0 , x1 , x2 ……..

Simpson’s one third rule :-


Pg.No. 6 of 143





Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet ∫ f(x). dxx0+nh

x0 = h

3 [ ( y0 + yn) + 4 (y1 + y3 + ……. yn−1 ) + 2 (y2 + y4 + ⋯ … . + yn−2)]

Simpson three eighth rule :-

∫ f(x). dxx0+nhx0

= 3h8

[ ( y0 + yn) + 3 (y1 + y2 + y4 + y5 + ……. yn−1 )+ 2 (y3 + y6 + ⋯ … . + yn−3) ]

Differential Equations :-

Variable & Seperable :-

General form is f(y) dy = ϕ(x) dx

Sol: ∫ f(y) dy = ∫ ϕ(x) dx + C .

Homo generous equations :-

General form dydx

= f(x,y)ϕ(x,y)

f(x, y) & ϕ(x, y) Homogenous of same degree

Sol : Put y = Vx ⇒ dydx

= V + x dvdx

& solve

Reducible to Homogeneous :-

General form dydx

= ax+by+ca′x+b′y+c′

(i) aa′ ≠ b

b′

Sol : Put x = X + h y = Y + k

⇒ dydx

= ax+by+(ah+bk+c)a′x+b′y+(a′h+b′k+c′)

Choose h, k such that dydx

becomes homogenous then solve by Y = VX

(ii) aa′ = b

b′

Sol : Let aa′ = b

b′ = 1m

dydx

= ax+by+cm(ax+by)+c

Put ax + by = t ⇒ dydx

= �𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

− 𝑎𝑎�/b


Pg.No. 7 of 143





Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet Then by variable & seperable solve the equation .

Libnetz Linear equation :-

General form dydx

+py = Q where P & Q are functions of “x”

I.F = e∫ p.dx

Sol : y(I.F) = ∫ Q. (I. F) dx + C .

Exact Differential Equations :-

General form M dx + N dy = 0 M → f (x, y)

N → f(x, y)

If ∂M∂y = ∂N

∂x then

Sol : ∫ M. dx + ∫(terms of N containing x ) dy = C

( y constant )

Rules for finding Particular Integral :-

1f(D)

eax = 1f(a)

eax

= x 1f′(a)

eax if f (a) = 0

= x2 1f′′(a)

eax if f ′(a) = 0

1f(b2)

sin (ax + b) = 1f(−a2)

sin (ax + b) f(- a2) ≠ 0

= x 1f′(−a2)

sin (ax + b) f(- a2) = 0 Same applicable for cos (ax + b)

= x2 1f′′(−a2)

sin (ax + b)

1f(D)

xm = [f(D)]y xm

1f(D)

eax f(x) = eax 1f(D+a)

f(x)

Vector Calculus :-


Pg.No. 8 of 143





Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet Green’s Theorem :-

∫ (ϕ dx + φ dy)C = ∫ ∫ �∂Ψ∂x − ∂ϕ

∂y� dx dy

This theorem converts a line integral around a closed curve into Double integral which is special case of Stokes theorem .

Series expansion :-

Taylor Series :-

f(x) = f(a) + f′(a)1 !

(x-a) + f′′(a)2 !

(x − a)2 + …………+ fn(a)n !

(x − a)n

f(x) = f(0) + f′(0)1 !

x + f′′(0)2 !

x2 + …………+ fn(0)n !

xn + ……. (mc lower series )

(1 + x)n = 1+ nx + n(n−1)2

x2 + …… | nx| < 1

ex = 1 + x + x2

2 ! + ……..

Sin x = x - x3

3 ! + x

5

5! - ……..

Cos x = 1 - x2

2 ! + x

4

4 ! - ……..


Pg.No. 9 of 143





Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet Electromagnetic Fields

Vector Calculus:- → A. (B × C) = C. (A × B) = B. (C × A) → A×(B×C) = B(A.C) – C(A.B) → Bac – Cab rule → Scalar component of A along B is AB = A Cos θAB = A . aB = (A.B)

|B|

→ Vector component of A along B is A�B = A Cos θAB . aB = (A.B) B|B|2

Laplacian of scalars :- • ∮ A. ds = 𝑣𝑣∫(∇.𝐴𝐴)𝑑𝑑𝑑𝑑 → Divergence theorem • L∮ A.dI = 𝑠𝑠∫(∇× 𝐴𝐴)𝑑𝑑𝑑𝑑 → Stokes theorem • ∇2A = ∇ (∇ .A) - ∇ × ∇ × A • ∇ .A = 0 → solenoidal / Divergence loss ; ∇ .A > 0 → source ; ∇ .A < 0 ⇒ sink • ∇ × A = 0 → irrotational / conservative/potential. • ∇2A = 0 → Harmonic .

Electrostatics :- • Force on charge ‘Q’ located @ r F = Q

4πε0 ∑ Qk(r−rk)

|r−rk|3Nk=1 ; F12 = Q1Q2

4πε0R3 . R�12

• E @ point ‘r’ due to charge located @ 𝑟𝑟′ 𝑠𝑠 𝐸𝐸� = 14πε0

∑ (r−rk)|r−rk

3NK=1 Qk

• E due to ∞ line charge @ distance ‘ ρ ‘ E = ρL2πε0 ρ

. aρ (depends on distance)

• E due to surface charge ρs is E = ρs2ε0

an . an → unit normal to surface (independent of distance) • For parallel plate capacitor @ point ‘P’ b/w 2 plates of 2 opposite charges is

E = ρs2ε0

an - � ρs2ε0

� (−𝑎𝑎𝑛𝑛)

• ‘E’ due to volume charge E = Q4πε0 R2 ar .

→ Electric flux density D = ε0 E D → independent of medium Flux Ψ = s∫ D .ds

Gauss Law :- → Total flux coming out of any closed surface is equal to total charge enclosed by surface . Ψ = Qenclosed ⇒ ∫ D . ds = Qenclosed = ∫ ρv. dv ρv = ∇. D → Electric potential VAB = w

Q = - ∫ E. dIB

A (independent of path)

VAB = - ∫ Q4πε0r2

BA ar . dr ar = VB - VA (for point charge )

• Potential @ any point (distance = r), where Q is located same where , whose position is vector @ r′

V = Q4πε0 |r−r′|

→ V(r) = Q4πε0r

+ C . [ if ‘C’ taken as ref potential ] → ∇ × E = 0, E = - ∇V


Pg.No. 10 of 143





Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet → For monopole E ∝ 1

r2 ; Dipole E ∝ 1r3 .

V ∝ 1r ; V ∝ 1

r2 • Electric lines of force/ flux /direction of E always normal to equipotential lines . • Energy Density WE = 1

2 ∑ QkVk

Nk=1 = 1

2 ∫ D. E dv = 1

2 ∫ ε0E2 dv

• Continuity Equation ∇.J = - ∂ρv∂t

. • ρv = ρv0 e−t/Tr where Tr = Relaxation / regeneration time = ε/σ (less for good conductor )

Boundary Conditions :- Et1 = Et2 • Tangential component of ‘E’ are continuous across dielectric-dielectric Boundary . • Tangential Components of ‘D’ are dis continues across Boundary . • Et1 = Et2 ; D1t

D2t = ε1/ ε2 .

• Normal components are of ‘D’ are continues , where as ‘E’ are dis continues. • D1n- D2n = ρs ; E1n = ε2

ε1 E2n ; tan θ1

tan θ2 = ε1

ε2 = εr1

εr2

• H1t = H2t B12= µ1µ2

B2t B1n = B2n H1n= µ2

µ1 H2n

Maxwell’s Equations :- → faraday law Vemf = ∮ E. dI = - d

dt ∫ B. ds

→ Transformer emf = ∮ E. dI = - ∫ ∂B

∂t ds ⇒ ∇ × E = - ∂B

∂t

s → Motional emf = ∇ × Em = ∇ × (μ × B). → ∇ × H = J + ∂D

∂t

Electromagnetic wave propagation :- • ∇ × H = J + �̇�𝐷 D = εE ∇2E = με�̈�𝐸

∇ × E = - �̇�𝐵 B = μH ∇2H = με�̈�𝐻 ∇. D = ρv J = σE ∇.B = 0

• Ey

Hz = - Ez

Hy = �μ/ε ; E.H = 0 E ⊥ H in UPW

For loss less medium ∇2E - ρ2E = 0 ρ = �jωμ(σ + jωϵ) = α + jβ.

α = ω �µϵ2

��1 + � σωϵ

�2

− 1�


Pg.No. 11 of 143





Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet

β = ω �µϵ2

��1 + � σωϵ

�2

+ 1�

• E(z, t) = E0 e−αz cos(ωt – βz) ; H0 = E0 / η .

• η = � jωµσ+jωϵ

|η | < θη

• |η| = �µ/ε

�1+� σωϵ�

2�

1/4 tan 2θη = σ/ωε.

• η= α + jβ α → attenuation constant → Neper /m . | Np| = 20 log10 𝑒𝑒 = 8.686 dB • For loss less medium σ = 0; α = 0. • β → phase shift/length ; μ = ω / β ; λ = 2π/β . • Js

Jd = � 𝜎𝜎𝜎𝜎

𝑗𝑗ωϵE� = σ / ωϵ = tan θ → loss tanjent θ = 2θη

• If tan θ is very small (σ < < ωϵ) → good (lossless) dielectric • If tan θ is very large (σ >> ωϵ) → good conductor • Complex permittivity ϵC = ϵ �1 − 𝑗𝑗𝜎𝜎

𝜔𝜔𝜔𝜔� = ε′ - j ε′′ .

• Tan θ = ε′′

ε′ = σωϵ

. Plane wave in loss less dielectric :- ( σ ≈ 0) • α = 0 ; β = ω√μϵ ; ω = 1

√µϵ ; λ = 2π/β ; η = �μr/εr ∠0.

• E & H are in phase in lossless dielectric Free space :- (σ = 0, μ = μ0 , ε = ε0 ) • α = 0 , β = ω �μ0ε0 ; u = 1/ �μ0ε0 , λ = 2π/β ; η = �μ0/ε0 < 0 = 120π ∠0 Here also E & H in phase . Good Conductor :- σ > > ωϵ σ/ωϵ → ∞ ⇒ σ = ∞ ε = ε0 ; μ = μ0μr

• α = β = �πfμσ ; u = �2ω/μσ ; λ = 2π / β ; η = �Wµσ

∠450

• Skin depth δ = 1/α • η = 1

σδ √2 ejπ/4 = 1+j

σδ

• Skin resistance Rs = 1σδ

= �πfµσ

• Rac = Rs.lw

• Rdc = lσs

. Poynting Vector :- • ∫ (E × H) ds = - ∂

dt ∫ 1

2 [ εE2 + μH2] dv – ∫ σ E2 dv

S v


Pg.No. 12 of 143





Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet • δave(z) = 1

2 E0

2

|η| e−2αz cos θη az

• Total time avge power crossing given area Pavge = ∫ Pave (s) ds S

Direction of propagation :- (𝐚𝐚𝐤𝐤) ak × aE = aH aE × aH = ak → Both E & H are normal to direction of propagation → Means they form EM wave that has no E or H component along direction of propagation . Reflection of plane wave :- (a) Normal incidence Reflection coefficient Γ = Er0

Ei0 = η2− η1

η2+ η1

Txn coefficient Τ = Et0Ei0

= 2η2η2+ η1

Medium-I Dielectric , Medium-2 Conductor :- 𝛈𝛈𝟐𝟐 > 𝛈𝛈𝟏𝟏 :- Γ>0 , there is a standing wave in medium & Txed wave in medium ‘2’. Max values of | E1| occurs Zmax = - nπ/β1 = −nλ1

2 ; n = 0, 1, 2….

Zmin = −(2n+1)π2β1

= −(2n+1)λ14

𝛈𝛈𝟐𝟐 < 𝛈𝛈𝟏𝟏 :- E1max occurs @ β1 Zmax = −(2n+1)π

2 ⇒ Zmax = −(2n+1)π

2β1 = −(2n+1)λ1

4

β1 Zmin = nπ ⇒ Zmin = −nπ

β1 = −nλ1

2

H1 min occurs when there is |t1|max S = |E1|max

|E1|min = |H1|max

|H1|min = 1+|Γ|

1−|Γ| ; | Γ | = s−1

s+1

Since |Γ| < 1 ⇒ 1 ≤ δ ≤ ∞ Transmission Lines :- • Supports only TEM mode • LC = με ; G/C = σ /ε . • d2Vs

dz2 - r2 Vs = 0 ; d2Is

dz2 - r2 Is = 0

• Γ = �(R + jωL)(G + jωC) = α + jβ • V(z, t) = V0

+ e−αz cos (ωt- βz) + V0− eαz cos (ωt + βz)

• Z0 = − V0−

I0− = R+jωL

γ = γ

G+jωC = �R+jωL

G+jωC

Lossless Line : (R = 0 =G; σ = 0)


Pg.No. 13 of 143





Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet → γ = α + jβ = jω√LC ; α = 0, β = w √LC ; λ = 1/ f √LC , u = 1/ √LC Z0 = �L/C Distortion less :(R/L = G/C)

→ α = √RG ; β = ωL�GR

= ωC�RG

= ω √LC

→ Z0 = �RG

= �LC

; λ = 1/f √LC ; u = 1√LC

= Vp ; uz0 = 1/C , u /z0 = 1/L

i/p impedance :- Zin = Z0 �𝑍𝑍𝐿𝐿+ Z0 tan hl

𝑍𝑍0+ ZL tan hl� for lossless line γ = jβ ⇒ tan hjβl = j tan βl

Zin = Z0 �𝑍𝑍𝐿𝐿+ 𝑗𝑗Z0 tan βl𝑍𝑍0+ ZL tan βl

� • VSWR = ΓL = ZL−Z0

ZL+Z0

• CSWR = - ΓL • Transmission coefficient S = 1 + Γ • SWR = Vmax

Vmin = Imax

Imin = 1+| ΓL|

1−|ΓL| = ZL

Z0 = Z0

ZL

(ZL > Z0) (ZL < Z0) • |Zin|max = Vmax

Imin = SZ0

• |Zin|min = VminImax

= Z0/S Shorted line :- ΓL = -1 , S = ∞ Zin = Zsc = jZ0 tan βl • ΓL = -1 , S = ∞ Zin = Zsc = j Z0 tan βl.

• Zin may be inductive or capacitive based on length ‘0’

If l < λ / 4 → inductive (Zin +ve) λ4 < l < λ/2 → capacitive (Zin -ve)

Open circuited line :- Zin = Zoc = -jZ0 cot βl Γl = 1 s = ∞ l < λ / 4 capacitive λ

4 < l < λ/2 inductive

Zsc Zoc = Z02

Matched line : (ZL = Z0) Zin = Z0 Γ = 0 ; s =1 No reflection . Total wave Txed . So, max power transfer possible .


Pg.No. 14 of 143





Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet Behaviour of Transmission Line for Different lengths :- l = λ /4 → 𝑍𝑍𝑠𝑠𝑠𝑠=∞

𝑍𝑍𝑜𝑜𝑠𝑠=0� → impedance inverter @ l = λ /4 l = λ /2 : Zin = Z0 ⇒ 𝑍𝑍𝑠𝑠𝑠𝑠=0

𝑍𝑍𝑜𝑜𝑠𝑠=∞� impedance reflector @ l = λ /2 Wave Guides :- TM modes : (Hz = 0) Ez = E0 sin �𝑚𝑚𝑚𝑚

𝑎𝑎�x sin �𝑛𝑛𝑚𝑚

𝑏𝑏� y e−nz

h2 = kx2 + ky

2 ∴ γ = ��mπa

�2

+ �nπb

�2

− ω2με where k = ω √μϵ m→ no. of half cycle variation in X-direction n→ no. of half cycle variation in Y- direction .

Cut off frequency ωC = 1

√µϵ ��mπ

a�

2+ �nπ

b�

2 γ = 0; α = 0 = β

• k2 < �mπa

�2

+ �nπb

�2 → Evanscent mode ; γ = α ; β = 0

• k2 > �mπa

�2

+ �nπb

�2 → Propegation mode γ = jβ α = 0

β = �k2 − �mπa

�2

− �nπb

�2

• fc = up′

2 ��m

a�

2+ �n

b�

2 up

′ = phase velocity = 1

√µϵ is lossless dielectric medium

• λc = u′/fc = 2

�(ma )2+(n

b)2

• β = β′ �1 − �fcf

�2 β′ = ω/ W β′ = phase constant in dielectric medium.

• up = ω/β λ = 2π/β = up/f → phase velocity & wave length in side wave guide

• ηTM = ExHy

= - Ey

Hx = β

ωϵ = �µ

ε �1 − �fc

f�

2

ηTM = η′ �1 − �fcf

�2 η′ → impedance of UPW in medium

TE Modes :- (𝐄𝐄𝐳𝐳 = 0) → Hz = H0 cos �𝑚𝑚𝑚𝑚𝑚𝑚

𝑎𝑎� cos �𝑛𝑛𝑚𝑚𝑛𝑛

𝑏𝑏� e−nz


Pg.No. 15 of 143





Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet

→ ηTE = wµβ

= η′ / �1 − �fcf

�2

→ ηTE > ηTM → TE10 Dominant mode Antennas :- Hertzian Dipole :- HΦs = jI0β dl

4πr sin θ e−jβγ Eθs = ηHΦs

Half wave Dipole :-

Hϕs = jI0e−jβγ cos�π

2 cos θ�

2πγ sin θ ; Eθs = ηHΦs


Pg.No. 16 of 143





Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet

Signals & Systems → Energy of a signal ∫ |x(t)|2∞

−∞ dt = ∑ |𝑥𝑥[𝑛𝑛]|2∞𝑛𝑛=−∞

→ Power of a signal P = lim

𝑇𝑇→∞1

2𝑇𝑇 ∫ |x(t)|2T

−T dt = lim𝑁𝑁→∞

12𝑁𝑁+1

∑ |x[n]|2Nn=−N

→ x1(t) → P1 ; x2(t) → P2 x1(t) + x2(t) → P1 + P2 iff x1(t) & x2(t) orthogonal → Shifting & Time scaling won’t effect power . Frequency content doesn’t effect power. → if power = ∞ → neither energy nor power signal Power = 0 ⇒ Energy signal Power = K ⇒ power signal → Energy of power signal = ∞ ; Power of energy signal = 0 → Generally Periodic & random signals → Power signals Aperiodic & deterministic → Energy signals Precedence rule for scaling & Shifting : x(at + b) → (1) shift x(t) by ‘b’ → x(t + b) (2) Scale x(t + b) by ‘a’ → x(at + b) x( a ( t + b/a)) → (1) scale x(t) by a → x(at) (2) shift x(at) by b/a → x (a (t+b/a)). → x(at +b) = y(t) ⇒ x(t) = y �𝑡𝑡−𝑏𝑏

𝑎𝑎�

• Step response s(t) = h(t) * u(t) = ∫ h(t)dtt

−∞ S’ (t) = h(t) S[n] = ∑ ℎ[𝑛𝑛]𝑛𝑛

𝑘𝑘=0 h[n] = s[n] – s[n-1] • e−at u(t) * e−bt u(t) = 1

b−a [ e−at - e−bt] u(t) .

• A1 Rect (t / 2T1) * A2 Rect(t / 2T2 ) = 2 A1A2 min (T1, T2) trapezoid (T1, T2) • Rect (t / 2T) * Rect (t / 2T) = 2T tri(t / T) Hilbert Transform Pairs : ∫ e−x2 /2σ2 dx∞

−∞ = σ √2π ; ∫ x2∞−∞ e−x2/2σ2 dx = σ3 √2π σ > 0


Pg.No. 17 of 143





Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet Laplace Transform :- x(t) = 1

2πj ∫ X(s) estσ+j∞

σ−j∞ ds X(s) = ∫ x(t) e−st∞

−∞ ds Initial & Final value Theorems : x(t) = 0 for t < 0 ; x(t) doesn’t contain any impulses /higher order singularities @ t =0 then x( 0+) = lim

𝑠𝑠→∞𝑠𝑠 𝑋𝑋(𝑠𝑠)

x(∞) = lim

𝑠𝑠→0𝑠𝑠 𝑋𝑋(𝑠𝑠)

Properties of ROC :- 1. X(s) ROC has strips parallel to jω axis 2. For rational laplace transform ROC has no poles 3. x(t) → finite duration & absolutely integrable then ROC entire s-plane 4. x(t) → Right sided then ROC right side of right most pole excluding pole s = ∞ 5. x(t) → left sided ROC left side of left most pole excluding s= - ∞ 6. x(t) → two sided ROC is a strip 7. if x(t) causal ROC is right side of right most pole including s = ∞ 8. if x(t) stable ROC includes jω-axis Z-transform :- x[n] = 1

2πj ∮ x(z) zn−1 dz

X(z) = ∑ x[n]∞

n=−∞ z−n Initial Value theorem : If x[n] = 0 for n < 0 then x[0] = lim

𝑧𝑧→∞𝑋𝑋(𝑧𝑧)

Final Value theorem :- lim

→∞𝑥𝑥[𝑛𝑛] = lim

𝑧𝑧→1(𝑧𝑧 − 1) X(z)

Properties of ROC :-


Pg.No. 18 of 143





Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet 1.ROC is a ring or disc centered @ origin 2. DTFT of x[n] converter if and only if ROC includes unit circle 3. ROC cannot contain any poles 4. if x[n] is of finite duration then ROC is enter Z-plane except possibly 0 or ∞ 5. if x[n] right sided then ROC → outside of outermost pole excluding z = 0 6. if x[n] left sided then ROC → inside of innermost pole including z =0 7. if x[n] & sided then ROC is ring 8. ROC must be connected region 9.For causal LTI system ROC is outside of outer most pole including ∞ 10.For Anti Causal system ROC is inside of inner most pole including ‘0’ 11. System said to be stable if ROC includes unit circle . 12. Stable & Causal if all poles inside unit circle 13. Stable & Anti causal if all poles outside unit circle. Phase Delay & Group Delay :- When a modulated signal is fixed through a communication channel , there are two different delays to be considered. (i) Phase delay: Signal fixed @ o/p lags the fixed signal by ∅(ωc) phase τP = - ∅(ωc)

ωc where ∅(ωc) = K H(jω)

↓ Frequency response of channel Group delay τg = − d∅(ω)

dω�

ω= ωc for narrow Band signal

↓ Signal delay / Envelope delay Probability & Random Process:- → P (A/B) = P(A∩B)

P(B)

→ Two events A & B said to be mutually exclusive /Disjoint if P(A ∩ B) =0 → Two events A & B said to be independent if P (A/B) = P(A) ⇒ P(A ∩ B) = P(A) P(B)

→ P(Ai / B) = P(Ai ∩B)P(B)

= P� B

Ai� P(Ai)

∑ P� BAi� P(Ai)n

i=1

CDF :- Cumulative Distribution function Fx(x) = P { X ≤ x } Properties of CDF : • Fx (∞) = P { X ≤ ∞ } = 1 • Fx (- ∞) = 0 • Fx (x1≤ X ≤ x2 ) = Fx (x2) - Fx (x1)


Pg.No. 19 of 143





Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet • Its Non decreasing function • P{ X > x} = 1 – P { X ≤ x} = 1- Fx (x) PDF :- Pdf = fx(x) = d

dx Fx(x)

Pmf = fx(x) = ∑ P{X = xi

∞i=−∞ } δ(x = xi)

Properties:- • fx(x) ≥ 0 • Fx(x) = fx (x) * u(x) = ∫ fx

x−∞ (x) dx

• Fx(∞) = ∫ fx

∞−∞ (x) dx =1 so, area under PDF = 1

• P { x1< X ≤ x2 } = ∫ fx(x)dxx2

x1

Mean & Variance :- Mean µx = E {x} = ∫ x fx

∞−∞ (x) dx

Variance σ2 = E { (X − µx)2 } = E {x2} - µx

2 → E{g(x)} = ∫ g(x) fx

∞−∞ (x) dx

Uniform Random Variables : Random variable X ~ u(a, b) if its pdf of form as shown below

fx(x) = �1

𝑏𝑏−𝑎𝑎 ; 𝑎𝑎 < 𝑥𝑥 ≤ 𝑏𝑏0 , 𝑒𝑒𝑒𝑒𝑠𝑠𝑒𝑒

Fx(x) = �1 ; 𝑥𝑥 > 𝑏𝑏

𝑥𝑥−𝑎𝑎𝑏𝑏−𝑎𝑎

; 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏0 ; 𝑒𝑒𝑒𝑒𝑠𝑠𝑒𝑒

Mean = a+b

2

Variance = (b − a)2 / 12 E{ x2 } = a

2+ ab+ b2

3

Gaussian Random Variable :-


Pg.No. 20 of 143





Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet fx(x) = 1

√2πσ2 e−(x−µ)2/2σ2 X ~ N (µ1σ2) Mean = ∫ x 1

√2πσ2∞

−∞ e−(x−µ)2/2σ2 dx = µ Variance = 1

√2πσ2 ∫ x2 ∞−∞ e−(x−µ)2/2σ2 dx = σ2

Exponential Distribution :- fx (x) = λ e−λx u(x) Fx(x) = ( 1- e−λx ) u(x) Laplacian Distribution :- fx(x) = λ

2 e−λ |x|

Multiple Random Variables :- • FXY (x , y) = P { X ≤ x , Y ≤ y } • FXY (x , ∞) = P { X ≤ x } = Fx (x) ; Fxy (∞ , y) = P { Y < y } = FY(y) • Fxy (-∞, y) = Fxy (x, - ∞) = Fxy (-∞, -∞) = 0 • fx(x) = ∫ fxy

∞−∞ (x, y) dy ; fY (y) = ∫ fxy

∞−∞ (x, y) dx

• FY/X �Y

X≤ x� = P{Y ≤y, X ≤x}

P{X ≤x} = FXY(x,y)

FX(x)

• fY/X(y/x) = fxy(x,y)

fx(x)

Independence :- • X & Y are said to be independent if FXY(x , y) = FX(x) FY(y) ⇒ fXY (x, y) = fX(x) . fX(y) P { X ≤ x, Y ≤ y} = P { X ≤ x} . P{Y ≤ y} Correlation: Corr{ XY} = E {XY} = ∫ ∫ fxy

∞−∞

∞−∞ (x, y). xy. dx dy

If E { XY} = 0 then X & Y are orthogonal . Uncorrelated :- Covariance = Cov {XY} = E { (X - µx) (Y- µy} = E {xy} – E {x} E{y}. If covariance = 0 ⇒ E{xy} = E{x} E{y}


Pg.No. 21 of 143





Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet • Independence → uncorrelated but converse is not true. Random Process:- Take 2 random process X(t) & Y(t) and sampled @ t1, t2 X(t1) , X(t2) , Y(t1) , Y (t2) → random variables → Auto correlation Rx(t1, t2) = E {X(t1) X(t2) } → Auto covariance Cx(t1, t2) = E { X(t1) - µx(t1)) (X(t2) - µx(t2) } = Rx(t1, t2) - µx(t1) µx(t2) → cross correlation Rxy(t1, t2) = E { X(t1) Y(t2) } → cross covariance Cxy(t1, t2) = E{ X(t1) - µx(t1)) (Y(t2) - µy(t2) } = Rxy(t1, t2) - µx(t1) µy(t2) → CXY (t1, t2) = 0 ⇒ Rxy(t1, t2) = µx(t1) µy(t2) → Un correlated → RXY (t1, t2) = 0 ⇒ Orthogonal cross correlation = 0 → FXY (x, y ! t1, t2) = Fx (x! t1) Fy(y ! t2) → independent Properties of Auto correlation :- • Rx(0) = E { x2 } • Rx(τ) = Rx(-τ) → even • | Rx(τ) | ≤ Rx(0)

Cross Correlation • Rxy(τ) = Ryx(-τ) • Rxy

2 (τ) ≤ Rx(0) . Ry(0) • 2 | Rxy(τ)| ≤ Rx(0) + Ry(0) Power spectral Density :- • P.S.D Sx(jω) = ∫ Rx

∞−∞ (τ) e−jωτ dτ

Rx(τ) = 1

2π ∫ 𝑆𝑆𝑥𝑥(jω)ejωτdω∞

−∞ • Sy(jω) = Sx(jω) |H(jω)|2 • Power = Rx(0) = 1

2π ∫ 𝑆𝑆𝑥𝑥(jω) dω∞

−∞ • Rx(τ) = k δ(τ) → white process Properties : • Sx(jω) even • Sx(jω) ≥ 0


Pg.No. 22 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet DC MACHINES : -

Lap Winding Wave Winding (1) Coil Span : Ycs = S

P Ycs = S

P

(2) Back Pitch Yb = U Ycs Yb = U Ycs (3) Commutator Pitch Yc = 1

for Progressive winding Yc = -1 for Retrogressive winding

Yc = 2 (c+1)p

for Progressive winding Yc = -2 (c+1)

p

for Restrogressive winding (YcMust be integer)

(4) Front Pitch Yf =Yb +2 for Progressive winding Yf =Yb -2 for Retrogressive winding

Yf =2Yc - Yb

(5) Parallel Paths A = P A = 2 (6) Conductor Current Ic =Ia

A Ic =Ia

2

(7) No of brushes No of brushes = A = P No of brushes = 2

• S = No of commutator segments • P = No of poles

• U = No of coil sides / No of poles = 2CS

• C = No of coils on the rotor • A = No of armature parallel paths • Ia = Armature current

→ Distribution factor (Kd) = phasor sum coil emfarthematic sum of coil emf

= chordarc

= 2π

→ Pitch factor ( Kp) =elecrrical angle of coil1800 *100%

→ θelectrical0 = P

2 θmechanical

0

→ Armature mmf/Pole (Peak) , ATa = ZIa2AP

→ AT (Compensating Winding) = ZIa2AP

* pole arcpole pitch

→ AT(Inter pole) = ATa + Biµ0

lgi

Where Bi= Flux density in inter pole airgap


Pg.No. 23 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet lgi = length of inter pole airgap , µ0 = 4π ∗ 10−7

→ No of turns in each interpole , Ninterpole = AT(Inter pole) Ia

→ The no of compensating conductor per pole, Ncw/pole = Z2 A P

( pole arcpole pitch

)

→ The Mechanical power that is converted is given by Pconv = Tind ωm Where T = Induced torque

ωm = Angular speed of the machines rotor

→ The resulting electric power produced Pconv = EA IA → The power balance equation of the DC Machine is Tind ωm = EA IA

→ The induced emf in the armature is Ea = ∅ZNP60A

→ Torque developed in Dc machine , Te = PZ2πA

∅ Ia

Where ∅ = Flux\pole , Z = No of armature conductors , P = No of poles , N = Speed in rpm ,

A = No of armature parallel paths, Ia = Armature current

→ The terminal voltage of the DC generator is given by Vt = Ea - Ia Ra → The terminal voltage of the DC motor is given by Vt = Ea + Ia Ra

→ Speed regulation of dc machine is given by ,SR = ωnl− ωfl ωfl

* 100 % = Nnl− Nfl

Nfl * 100 %

→ Voltage regulation , VR = Vnl− Vfl

Vfl * 100 %

Shunt Generator:

→ For a shunt generator with armature induced voltage Ea, armature current Ia and armature resistance Ra, the terminal voltage V is: V = Ea - IaRa

→ The field current I f for a field resistance R f is: I f = V / R f

→ The armature induced voltage Ea and torque T with magnetic flux Φ at angular speed ω are: Ea = k fΦω = kmω T = k fΦIa = kmIa where k f and km are design coefficients of the machine.


Pg.No. 24 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Note that for a shunt generator: - induced voltage is proportional to speed, - torque is proportional to armature current.

→ The airgap power Pe for a shunt generator is: Pe = ωT = EaIa = kmω Ia Series Generator:

→ For a series generator with armature induced voltage Ea, armature current Ia, armature resistance Ra and field resistance R f, the terminal voltage V is: V = Ea - ( IaRa + IaR f )= Ea - Ia(Ra + R f) The field current is equal to the armature current.

→ The armature induced voltage Ea and torque T with magnetic flux Φ at angular speed ω are: Ea = k fΦω Ia = kmω Ia T = k fΦIa2 = kmIa2 where k f and km are design coefficients of the machine.

Note that for a series generator: - induced voltage is proportional to both speed and armature current, - torque is proportional to the square of armature current, - armature current is inversely proportional to speed for a constant Ea

→ The airgap power Pe for a series generator is: Pe = ωT = EaIa = kmω Ia2

→ Cumulatively compounded DC generator : - ( long shunt) (a) Ia = If + IL (b) Vt = Ea - Ia (Ra + Rs )

(c) Isf = VxRf

= shunt field current

(d) The equivalent effective shunt field current for this machine is given by

Isf =Isf + NseNf

Ia - (Armature reaction MMFNf

)

Where Ns e = No of series field turns Nf = = No of shunt field turns


Pg.No. 25 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet → Differentially compounded DC generator : - ( long shunt)

(a) Ia = If + IL (b) Vt = Ea - Ia (Ra + Rs )

(c) Isf = VxRf

= shunt field current

(d) The equivalent effective shunt field current for this machine is given by

Isf =Isf - NseNf

Ia - (Armature reaction MMFNf

)

Where Ns e = No of series field turns

Nf = = No of shunt field turns

Shunt Motor:

→ For a shunt motor with armature induced voltage Ea, armature current Ia and armature resistance Ra, the terminal voltage V is: V = Ea + IaRa The field current I f for a field resistance R f is: I f = V / R f

→ The armature induced voltage Ea and torque T with magnetic flux Φ at angular speed ω are: Ea = k fΦω = kmω T = k fΦIa = kmIa where k f and km are design coefficients of the machine.

Note that for a shunt motor: - induced voltage is proportional to speed, - torque is proportional to armature current.

→ The airgap power Pe for a shunt motor is: Pe = ωT = EaIa = kmω Ia

→ The speed of the shunt motor , ω = VK∅

- Ra(K∅)2T

Where K = PZ2πA

Series Motor :

→ For a series motor with armature induced voltage Ea, armature current Ia, armature resistance Ra and field resistance R f, the terminal voltage V is:


Pg.No. 26 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet V = Ea + IaRa + IaR f = Ea + Ia(Ra + R f) The field current is equal to the armature current.

→ The armature induced voltage Ea and torque T with magnetic flux Φ at angular speed ω are: Ea = k fΦω Ia = kmω Ia T = k fΦIa2 = kmIa2 where k f and km are design coefficients of the machine.

Note that for a series motor: - induced voltage is proportional to both speed and armature current, - torque is proportional to the square of armature current, - armature current is inversely proportional to speed for a constant Ea

→ The airgap power Pe for a series motor is: Pe = ωT = EaIa = kmω Ia2

Losses:

→ constant losses (P k) = Pw f + Pi o

Where, Pio = No of load core loss

→ Pwf = Windage & friction loss → Variable losses (Pv) = Pc + Ps t + Pb

where Pc= Copper losses = Ia2 Ra

Ps t = Stray load loss = α I2

Pb = Brush Contact drop = VbIa , Where Vb = Brush voltage drop

→ The total machine losses , PL = Pk +VbIa+ Kv Ia2

Efficiency

→ The per-unit efficiency η of an electrical machine with input power Pin, output power Pout and power loss Ploss is: η = Pout / Pin = Pout / (Pout + Ploss) = (Pin - Ploss) / Pin

→ Rearranging the efficiency equations:


Pg.No. 27 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Pin = Pout + Ploss = Pout / η = Ploss / (1 - η) Pout = Pin - Ploss = ηPin = ηPloss / (1 - η) Ploss = Pin - Pout = (1 - η)Pin = (1 - η)Pout / η

Temperature Rise:

→ The resistance of copper and aluminium windings increases with temperature, and the relationship is quite linear over the normal range of operating temperatures. For a linear relationship, if the winding resistance is R1 at temperature θ1 and R2 at temperature θ2, then: R1 / (θ1 - θ0) = R2 / (θ2 - θ0) = (R2 - R1) / (θ2 - θ1) where θ0 is the extrapolated temperature for zero resistance.

→ The ratio of resistances R2 and R1 is: R2 / R1 = (θ2 - θ0) / (θ1 - θ0)

→ The average temperature rise ∆θ of a winding under load may be estimated from measured values of the cold winding resistance R1 at temperature θ1 (usually ambient temperature) and the hot winding resistance R2 at temperature θ2, using: ∆θ = θ2 - θ1 = (θ1 - θ0) (R2 - R1) / R1

→ Rearranging for per-unit change in resistance ∆Rpu relative to R1: ∆Rpu = (R2 - R1) / R1 = (θ2 - θ1) / (θ1 - θ0) = ∆θ / (θ1 - θ0)

.Copper Windings:

→ The value of θ0 for copper is - 234.5 °C, so that: ∆θ = θ2 - θ1 = (θ1 + 234.5) (R2 - R1) / R1

→ If θ1 is 20 °C and ∆θ is 1 degC: ∆Rpu = (R2 - R1) / R1 = ∆θ / (θ1 - θ0) = 1 / 254.5 = 0.00393

→ The temperature coefficient of resistance of copper at 20 °C is 0.00393 per degC.

Aluminium Windings:


Pg.No. 28 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet → The value of θ0 for aluminium is - 228 °C, so that:

∆θ = θ2 - θ1 = (θ1 + 228) (R2 - R1) / R1

→ If θ1 is 20 °C and ∆θ is 1 degC: ∆Rpu = (R2 - R1) / R1 = ∆θ / (θ1 - θ0) = 1 / 248 = 0.00403

→ The temperature coefficient of resistance of aluminium at 20 °C is 0.00403 per degC.

Dielectric Dissipation Factor:

→ If an alternating voltage V of frequency f is applied across an insulation system comprising capacitance C and equivalent series loss resistance RS, then the voltage VR across RS and the voltage VC across C due to the resulting current I are: VR = IRS VC = IXC V = (VR2 + VC2)½

→ The dielectric dissipation factor of the insulation system is the tangent of the dielectric loss angle δ between VC and V: tanδ = VR / VC = RS / XC = 2πfCRS RS = XCtanδ = tanδ / 2πfC

→ The dielectric power loss P is related to the capacitive reactive power QC by: P = I2RS = I2XCtanδ = QCtanδ

→ The power factor of the insulation system is the cosine of the phase angle φ between VR and V: cosφ = VR / V so that δ and φ are related by: δ + φ = 90°

→ tanδ and cosφ are related by: tanδ = 1 / tanφ = cosφ / sinφ = cosφ / (1 - cos2φ)½ so that when cosφ is close to zero, tanδ ≈ cosφ

TRANSFORMERS:

→ Gross cross sectional area = Area occupied by magnetic material + Insulation material.


Pg.No. 29 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet → Net cross sectional area = Area occupied by only magnetic material excluding area

of insulation material. → Hence for all calculations, net cross sectional area is taken since ϕ (flux) majorly

flows in magnetic material. ϕ = BAn

→ Specific weight of t/f = Weight oft

f

KVA rating oftf

→ Stacking/iron factor :- (ks) = Net Cross Sectional area

Gross Cross Sectional area

→ ks is always less than 1 → Gross c.s Area = AG = length × breadth → Net c.s Area = An = ks × AG → Utilization factor of transformer core = Effective C.S.Area

Total C.S Area U.F of cruciform core = 0.8 to

0.85 → Flux = mmF

Reluctance = = ϕm sin ωt

→ According to faradays second law e1 = −N1dϕdt

= −N1ddt

�ϕm sin ωt�

→ Transformer emf equations :- E1 = 4.44 N1 Bmax Anf (1) E2 = 4.44 N2 Bmax Anf (2) → Emf per turn in Iry = E1

N1 = 4.44 BmaxAnf

→ Emf per turn in IIry = E2N2

= 4.44 Bmax An f

⟹ Emf per turn on both sides of the transformer is same

E1N1

= E2N2

⟹ E1E2

= N1N2

= 1k

Transformation ratio = K = E2E1

=

Instantaneous value of emf in primary

e1 = N1ϕm ω sin�ωt − π 2� �


Pg.No. 30 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet

N2N1

Turns ratio = 1K

= N1 ∶ N2

→ For an ideal two-winding transformer with primary voltage V1 applied across N1 primary turns and secondary voltage V2 appearing across N2 secondary turns: V1 / V2 = N1 / N2

→ The primary current I1 and secondary current I2 are related by: I1 / I2 = N2 / N1 = V2 / V1

→ For an ideal step-down auto-transformer with primary voltage V1 applied across (N1 + N2) primary turns and secondary voltage V2 appearing across N2 secondary turns: V1 / V2 = (N1 + N2) / N2

→ The primary (input) current I1 and secondary (output) current I2 are related by: I1 / I2 = N2 / (N1 + N2) = V2 / V1.

→ For a single-phase transformer with rated primary voltage V1, rated primary current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere rating S is: S = V1I1 = V2I2

→ For a balanced m-phase transformer with rated primary phase voltage V1, rated

primary current I1, rated secondary phase voltage V2 and rated secondary current I2, the voltampere rating S is: S = mV1I1 = mV2I2

→ The primary circuit impedance Z1 referred to the secondary circuit for an ideal

transformer with N1 primary turns and N2 secondary turns is: Z12 = Z1(N2 / N1)2

→ During operation of transformer :-

Bm ∝ E1

f ∝ V1

f

Bmax = constant ⟹ V1

f = constant

Equivalent ckt of t/f under N.L condition :-


Pg.No. 31 of 143






→ No load current = I0 = Iµ + Iw = I0 �−ϕ0

Iw = I0 cos ϕ0 Iµ = I0 sin ϕ0 → No load power = v1I0 cos ϕ0 = v1Iw = Iron losses.

R0 = v1Iw1

; X0 = v1Iµ

⟹ Iw = No load powerV1

Transferring from 𝐈𝐈𝐈𝐈𝐫𝐫𝐫𝐫 to 𝐈𝐈𝐫𝐫𝐫𝐫:-

I2

2 R2 = I1 2 R2

1

R2 1 = R2 �I2

I1�

2

= R2K2

∴ R2 1 = R2

K2

From 𝐈𝐈𝐫𝐫𝐫𝐫 to 𝐈𝐈𝐈𝐈𝐫𝐫𝐫𝐫 :- I1

2R1 = I2 2. R1

1

R1 1 = I1

2

I2 2 . R1

R1 1 = R1. K2

→ Total resistance ref to primary = R1 + R2 1

R01 = R1 + R2/k2 → Total resistance ref to secondary = R2 + R1

1 R02 = R2 + k2R1

→ Total Cu loss = I1 2R01

Or I2

2 R02

R21 R1

No load /shunt branch.

N1 N2

E2 E1 V1

I0

Iw R0 X0

Iµ


Pg.No. 32 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Per unit resistance drops :- → P.U primary resistance drop = I1R1

E1

→ P.U secondary resistance drop = I2 R2E2

→ Total P.U resistance drop ref to Iry = I1R01E1

→ Total P.U resistance drop ref to IIry = I2 R02E2

→ The P.U resistance drops on both sides of the t/f is same I1 R01

E1= I2R02

E2

Losses present in transformer :-

1. Cu losses in t/f:

Total Cu loss = I1

2R1 + I2 2R1

= I1 2R01

= I2 2 R02

→ Rated current on Iry = VA rating of t/fE1

Similarly current on IIry = VA rating of t/fE2

→ Cu losses ∝ I1 2 or I2

2. Hence there are called as variable losses.

→ P.U Full load Cu loss = FL Cu loss in wattsVA rating of t/f

= I1 2R01E1I1

→ If VA rating of t/f is taken as base then P.U Cu loss ∝ I1 2 as remaining terms are constant.

→ P.U Cu loss at x of FL = x2 × PU FL Cu loss

→ P. U resistance drop ref to Iry

or P. U resistance ref to Iry

� = I1R01E1

× I1I1

= I1 2R01E1I1

1. Copper losses

2. Iron losses

3. Stray load losses

4. Dielectric losses

major losses

minor losses

t/f windings

t/f core cu parts Iron parts

insulating materials.


Pg.No. 33 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet ∴ P.U Resistance drop = P.U FL cu loss

% FL Cu loss = % R = % Resistance drop.

Iron (or) Core losses in t/f :- 1. Hysteresis loss :

Steinmetz formula :-

Where η = stienmetz coefficient Bmax = max. flux density in transformer core. f = frequency of magnetic reversal = supply freq. v = volume of core material x = Hysteresis coeff (or) stienmetz exponent

= 1.6 (Si or CRGo steel) 2. Eddycurrent loss: Eddy current loss ,(We) ∝ Rce × Ie

2 As area decreases in laminated core resistance increases as a result conductivity decreases.

During operation of transformer :- Bm ∝ V1

f

Case (i) :- V1f

= constant, Bmax = const.

we ∝ f 2

we = B f 2

Const.

∴ wi = wh + we

wi = Af + Bf 2

When Bmax = const.

We = K. Bmax 2 f 2. t2

Constant Supply freq

thickness of laminations.

(it is a function of σ )

Area under one hysteresis loop. η Bmax x . f . v Wh =


Pg.No. 34 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Case (ii) :- V1

f ≠ constant, Bm ≠ const.

we ∝ �V1f

�2

. f 2

we ∝ V1 2

wi = wh + we

wi = A V1 1.6

f0.6 + BV1 2

P.U iron loss :- → P.U iron loss = Iron loss in watts

VA rating of t/f

→ As VA rating is choosen as base then the P.U iron loss are also constant at all load conditions. To find out constant losses :- W0 = Losses in t/f under no load condition

= Iron losses + Dielectric loss + no load primary loss (I0 2 R1)

Constant losses = W0 − I0

2R1 Where , R1 = LV winding resistance.

To find out variable losses :- Wsc = Loss in t/f under S.C condition

= F.L Cu loss + stray load losses (Cu and Iron) + Iron losses in both wdgs Variable losses = WSC − Iron losses corresponding to VCC

O.C test :- V1 rated → Wi S.C test :- VSC → (Wi)S.C Wi ∝ V1

2

Wi(Wi)SC

= �V1 ratedVSC

�2

(Wi)S.C = Wi × � VSCV1 rated

�2

∴ Variable losses = WSC − (Wi)SC � VSCV1 rated

�2


Pg.No. 35 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet → Under the assumption that small amount of iron losses corresponds to VSC and stray load

losses are neglected the wattmeter reading in S.C test can be approximately taken as F.L Cu losses in the transformer.

→ Wse ≃ F.L Cu loss ≃ ISC

2 . R01

R01 = WSC

ISC 2

Efficiency :- Efficiency of transformer is given by η = output power

input power

= output poweroutput power+losses

= E2 I2 cos ϕ2 E2 I2 cos ϕ2+ F.L cu losses+Iron losses

ηF.L = E2 I2 cos ϕ2E2I2 cos ϕ2+ I2

2 R02 + Wi

→ Transformer efficiency = KVA × cos ϕ

KVA × cos ϕ + wi + Cu losses

→ Efficiency = η = x × MVA × Pf

x × MVA × Pf + wcu × x2 + wi

→ Total losses in transformer = �1

η− 1� output

→ Condition for maximum effieciency is, Cu losses = Iron losses → Total losses at ηmax = 2Wi

→ %load at which maximum efficiency occurs % x = � WiI2

2 R02 *100 %= � Iron loss

F.L.cu loss*100 %

→ KVA corresponding to ηmax = F.L KVA � Iron lossF.L culoss

ηx of F.L = x (E2 I2) cos ϕ2

x (E2 I2) cos ϕ2 + x2 (I2 2 R02) + Wi

O.C test S.C test


Pg.No. 36 of 143






→ ηall day =Output energy in kwhInput energy in kwh

24hrs

→ Voltage drop in t/f at a Specific load p.f = I2R02 cos ϕ2 ± I2X02 sin ϕ2

→ % Voltage regulation = I2 R02 cos ϕ2 ± I2X02 sin ϕ2

V1 ′ ×100

= �I2 R02

V1 ′ � cos ϕ2 ± �I2 X02

V1 ′ � sin ϕ2

↓ ↓ P.U resistance P.U reactance

% Regulation = �(P. U R) cos ϕ2 + (P. U X) sin ϕ2� × 100 Condition for max. regulation :- % regulation = (% R) cos ϕ2 ± (% X) sin ϕ2

d regnd ϕ2

= 0

Tan ϕ2 = % X% R

= X02R02

ϕ2 = Tan−1 �X02R02

� lagging

At maximum regulation ϕ2 = Tan−1 �XL

RL�

= Tan−1 �X02R02

� XLRL

= X02R02

Value of maximum regulation :- % Regulation = (% R) cos ϕ2 + (% X) sin ϕ2

At max. regulation cos ϕ2 = % R% Z

Sin ϕ2 = % X% Z

max. % regulation = (% R) % R% Z

+ (% X) (% X)% Z

= (% R)2 + (% X)2

% Z

= (% Z)2

% Z

max. % regn = % Z

= % of rated voltage required to produce rated short ckt current .


Pg.No. 37 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Condition for zero regulation :- → If the voltage regulation in the t/f is zero, the t/f voltages are maintained at their nominal

values even under load condition % Regn = (% R) cos ϕ2 ± (% X) sin ϕ2

For zero regulation ⇒ occurs at leading p.f’s (% R) cos ϕ2 − (% X) sin ϕ2 = 0

Tan ϕ2 = % R% X

ϕ2 = Tan−1 �% R% X

� = Tan−1 �R02X02

� leading.

→ At zero regulation condition : ϕ2 = Tan−1 �XC

RL� = Tan−1 �R02

X02�

XCRL

= R02X02

→ Regulation at x of FL = x [% R cosϕ2 ± % X sin ϕ2]

= x × F.L regn

Regulation at U.P.F:- Regulation at UPF = % R

= % F.L Cu loss

Scott Connection:

VAM = 0.866 V1 VAN = �V1

√3� = 0.577 V1

VMN = 0.866 V1 − 0.577 V1 = 0.289 V1

IA2

86.6%

0.289

2 : 1

M

N 0.866 V1 0.577 V1

A

B

C

V1 IBC Vb ib

ia

Va

Ia

IB

IC

�0.866√3

�

IA2�


Pg.No. 38 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet VAN ∶ VMN = 0.577 V1 ∶ 0.289 V1 = 2 : 1 → If a neutral pt is located on 3ϕ side, such that, voltage between any terminal to that neutral

point is 0.577 V1 then such neutral point divides the primary of teaser transformer in the ratio of 2 : 1

→ Location of neutral point from top = 0.866 N1 × 23

→ Location of neutral point from bottom = 0.866 N1 × 13

Operation of Scott Connection with 2ϕ balanced load at UPF :- Teaser t/f :- Ia

ia= N2

0.866 N1

Ia = N2

0.866 N1 × ia

Let N1 ∶ N2 = 1 ∶ 1 IA = 1.15 ia

Main t/f IBC

ib= N2

N1

IBC = N2N1

× ib

Let N1: N2 = 1 ∶ 1 IBC = ib

IB = IBC − IA2� IC = −IBC − IA

2�

→ Capacity of Scott Connection :- (KVA)Scott = √3 VLIL V2 = V1 IL = I1 ↙ ↓ → Vol. rating of 1 – ϕ t/f Current rating of 1 – ϕ t/f (KVA)Scotf = √3 V1 I1

(KVA)Scott = √3 (KVA)1− ϕ

→ Utilization factor = (KVA)Scottavailable KVA

= √3 V1I12V1I1

= 0.866

→ Utilization factor of Scott connection with 2 identical 1 – ϕ t/f’s is 86.6% AUTO TRANSFORMER:


Pg.No. 39 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet → Primary applied voltage, Vab = Secondary voltage V2 referred to primary + primary leakage impedance

drop + secondary leakage impedance drop ref. to primary.

Vab = �N1−N2N2

� V2 + I1(r1 + jx1) + (I2 − I1)(r2 + j x2) �N1− N2N2

�

→ K of auto transformer = LVHV

(KVA)induction = (V1 − V2) I1 I/P KVA = V1 I1 (KVA)induction

i/p KVA= (V1− V2) I1

V1I1

= 1 – LVHV

= 1 – K

∴ (KVA) induction = (1 – K) i/p KVA

(KVA) conduction = I/p KVA – (KVA)ind

(KVA)conduction = K × I/p KVA

→ Wt. of conductor in section AB of auto t/f ∝ (N1 − N2) I1

→ Wt of conductor in section BC of auto t/f ∝ (I2 − I1)N2

∴ Total wt. of conductor in auto t/f is ∝ I1(N1 − N2) + (I2 − I1)N2 ∝ 2 (N1 − N2) I1

→ Total wt. of conductor in 2 wdg transformer ∝ I1N1 + I2 N2 ∝ 2 I1N1

→ wt.of conductor in an auto t/fwt.of conductor in 2 wdg t/f

= 2(N1−N2)I12N1I1

= 1 – N2N1

= 1 – K Wt. of conductor in auto t/f = (1 – K) (wt. of conductor in 2 wdg t/f)

→ Thus saving of conductor material if auto – t/f is used} = K × {conductor wt in 2 wdg transformer. → (% FL losses)Auto t/f = (1 − K)(% FL losses)2 wdg t/f → (% Z)AT = (1 − K) (% Z)2 wdg t/f

→ (KVA)AT = 11−K

(KVA)2 wdg t/f.


Pg.No. 40 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet SYNCHRONOUS MACHINES:

→ Principle of operation :-

Whenever a conductor cuts the magnetic flux, an emf is induced in that conductor”

Faraday’s law of electromagnetic induction.

→ Coil span (β) :- It is the distance between two sides of the coil. It is expressed in terms of degrees, pole pitch, no. of slots / pole etc

→ Pole pitch :- It is the distance between two identical points on two adjacent poles. Pole pitch is always 180° e = slots / pole.

→ θelec = P2 θmech

→ Slot pitch or slot angle :- (T)Slot angle is the angle for each slot.

→ For a machine with ‘P’ poles and ‘s’ no. of slots, the slot angle = γ = P(180°)d

→ Pitch factor or coil span factor or chording factor :- (KP) KP = The emf induced | coil in short pitched winding

The emf induced |coil in full pitched winding

= The vector sum of induced emf | coil

Arithmetic sumof induced emf | coil

KP = 2E cos∝/2

2E

→ Pitch factor for nth harmonic i.e,

→ chording angle to eliminate nth harmonics (α)= 180°n

→ coil spam to eliminate nth harmonics ,(β) = 180 �n−1n

�

→ Distribution factor | spread factor | belt factor | breadth factor(kd) :- Kd = The emf induced when the winding is distributed

The emf induced when the winding is concentrated

Kd = Vector sum of emf induced

Arithmetic sum of emf induced

γ = 180°�s

p�

Kp = cos ∝/2

Kpn = cos n ∝2


Pg.No. 41 of 143






Kd = sinmγ

2m sinγ

2

→ The distribution factor for uniformly distributed winding is

For nth harmonic, kdn = sinmnγ

2m sinnγ

2

→ To eliminate nth harmonics ,phase spread (mγ) = 360°n

→ Generally, KVA rating, power output ∝ kd and Eph(induce emf) ∝ kd . Tph.

∴ KVA60(3− ϕ)KVA120(3− ϕ) = Pout60 (3 ϕ)

Pout120(3ϕ) = kd60kd120

= sin60

2sin120

2 × m120

m60 = sin 30°

sin 60° × 120

60 = 1.15

KVA60(3ϕ)KVA90(2ϕ) = Pout60(3ϕ)

Pout90°(2ϕ) = kd60kd90

= sin60

2sin90

2 × 90

60 = 1.06

KVA60 (3ϕ)KVA180(1ϕ)

= Pout60(3ϕ)Pout180 (1− ϕ) = kd60

kd180=

sin602

sin1802

× 18060

= 1.5

KVA90(2− ϕ)

KVA180(1− ϕ) = Pout90Pout180

= kd90kd180

= sin90 2⁄sin180 2⁄ × 180

90= 1.414

→ Speed of space harmonics of order (6k ± 1) is 1

(6k ±1) . Ns

where Ns = synchronous speed = 120 fp

The order of slot harmonics is �2SP

± 1�

where S = no. of slots , P = no. of poles

→ Slot harmonics can be eliminated by skewing the armature slots and fractional slot winding.

The angle of skew = θs = γ (slot angle)

= 2 harmonic pole pitches = 1 slot pitch.

→ Distribution factor for slot harmonics, kd �2sp

± 1�

kd4 = sim mr

2mr

2 × π180


Pg.No. 42 of 143






Is kd1 = sinm γ

2m sinγ

2 i.e., same that of fundamental

→ Pith factor for slot harmonics, kp �2sp

± 1� = kp1 = cos ∝2�

→ The synchronous speed Ns and synchronous angular speed s of a machine with p pole pairs running on a supply of frequency fs are: ωs = 2πfs / p

→ Slip S = NS− NNS

Where NS = 120 f

p = synchronous speed

→ The magnitude of voltage induced in a given stator phase is Ea= √2 π Nc ∅ f = K∅ω Where K = constant

→ The output power Pm for a load torque Tm is:

Pm = ωsTm

→ The rated load torque TM for a rated output power PM is:

TM = PM / ωs = PM p/ 2πfs = 120PM / 2πNs

Synchronous Generator:

→ For a synchronous generator with stator induced voltage Es, stator current Is and synchronous impedance Zs, the terminal voltage V is:

V = E - IsZs = Es - Is(Rs + jXs)

where Rs is the stator resistance and Xs is the synchronous reactance

E = �(V cos ϕ + Ia Ra)2 + (V sin ϕ ± Ia Xs)2 + ⇒ lag p.f − ⇒ leading p.f.


Pg.No. 43 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Synchronous Motor:

→ For a synchronous motor with stator induced voltage Es, stator current Is and synchronous impedance Zs, the terminal voltage V is:

V = Es + IsZs = Es + Is(Rs + jXs) where Rs is the stator resistance and Xs is the synchronous reactance

Voltage regulation :

→ % regulation = |E|− |V||V| ×100

E – V = IaZs ∴ % regulation = E − V

V

= Ia ZsV

× 100 ∴ regulation ∝ Zs ∴ As Zs increases, voltages regulation increases.

→ Condition for zero | min. voltage regulation is, Cos (θ + ϕ) = − Ia Zs

2V

→ Condition for max. Voltage regulation is, ϕ = θ

→ Short circuit ratio (SCR) = IfmIfa

= 1Zs(adjusted)|unit

= 1Xs(adjusted)|unit

SCR ∝ 1Xa

∝ 1Armature reaction

Voltage regulation ∝ Armature reaction

∴ SCR ∝ 1Voltage regulation

∴ Small value of SCR represent poor regulation. ϕa = armature mmf

reluctance

But reluctance ∝ Air gap ∴ ϕa = armature mmf

airgap

ϕa ∝ 1

Air gap length

Armature reaction ∝ ϕa ∝ 1Airgap length


Pg.No. 44 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet ∴ SCR ∝ 1

Armature reaction ∝ Airgap length

∴ machine size ∝ SCR. Cost ∝ SCR

Power = EVXs

sin δ

⇒ P ∝ 1Xs

∝ SCR

∴ Large value of SCR represent more power output.

→ Synchronizing power coefficient or stability factor Psy is given as

Psy = dpdδ

= ddδ

�EVXs

sin δ�

= EVXs

cos δ

Psy is a measure of stability ∴ stability ∝ Psy

But Psy ∝ 1Xs

∝ SCR

Stability ∝ SCR ∝ Air gap length

→ When the stator mmf is aligned with the d – axis of field poles then flux ϕd perpole is set up

and the effective reactance offered by the alternator is Xd.

Xd = maximum Voltage minimum current

= (Vt)line (at min. Ia )√3 Ia(min )

= Direct axis reactance

→ When the stator mmf is aligned with the q – axis of field poles then flux ϕq per pole is set up

and the effective reactance offered by the alternator is Xq.

Xq = minimum voltge maximum voltage

= Vt line (at maximum Ia )√3 Ia(max )

= Quadrature axis reactance

Air gap length ∝ SCR

Power ∝ SCR

∴ Stability ∝ SCR

∴ Stability ∝ Air gap length


Pg.No. 45 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet → Cylindrical rotor Synchronous machine ,

The per phase power delivered to the infinite bus is given by P = Ef VtXs

sin δ

→ Salient pole synchronous machine ,

The per phase power delivered to the infinite bus is given by

P = Ef VtXd

sin δ + Vt 2

2 � 1

Xq− 1

Xd� sin 2δ

Condition for max. power:-

→ For cylindrical rotor machine :- At constant Vt and Ef, the condition for max. power is obtained by putting dp

dδ = 0

∴ dpdδ

= Ef VtXs

cos δ = 0

Cos δ = 0 δ = 90° Hence maximum power occurs at δ = 90°

→ For salient – pole synchronous machine :- dp

dδ = 0

⇒ Vt EfXd

cos δ + Vt 2 � 1

Xq− 1

Xd� cos 2δ = 0

Cos δ = − Ef Xq

4Vt �Xd− Xq�± �1

2+ � Ef Xq

4Vt�Xd− Xq��

2

The value of load angle is seed to be less than 90°. ∴ max. power occurs at δ < 90°

→ Synchronizing power = Psy. ∆ δ.

= EVXs

cos 𝛿𝛿 . ∆ 𝛿𝛿 .

→ Synchronizing torque = Synchronizing powerω

.

Power flow in Alternator :-

→ Complex power = S = P + jQ = VIa ∗

Where Active power flow (P) = EVZs

cos(θ − δ) − V2

Zscos θ ;


Pg.No. 46 of 143






Reactive power flow (Q) = EVZ2

sin(θ − δ) − V2

Zssin θ ;

→ Condition for max. power output :-

P = EVZs

cos(θ − δ) − V2

Z2cos θ

dpdδ

= 0 for max power condition ie θ – δ = 0

If Ra = 0; θ = δ = 90° ; then max power is given by

SYNCHRONOUS MOTORS: → Speed regulation = NN.L− NF.L

NF.L × 100

= NS− NSNS

× 100 = 0%

⇒ Slip S = NS− NNS

= NS− NSNS

= 0%

NS = 120 fp

→ The speed can be controlled by varying the frequency

↓ϕ ∝ V↑f ↑

v F

ratio control is preferred for rated torque operation Power flow in synchronous motor is given by complex power i/p s = p + jQ = V Ia

∗ where P = Resl power flow , Q = Reactive power flow

: : ⇒ If Ra = 0 ; ZS = XS ; θ 90°

→ Condition for max power :-

θ = δ

Pmax = EVZs

− V2

Zs cos θ

Pin = EVXS

sin δ Q = VXS

[V − E cos δ]

P = V2

Z2cos θ − EV

ZScos(θ + δ) Q = V

2

Z2sin θ − EV

ZSsin(θ + δ)


Pg.No. 47 of 143






Pin = V2

Z2cos θ − EV

ZScos(θ + δ)

d Pind δ

= 0 ⇒ 0 + EVZS

sin(θ + δ) = 0

Sin (θ + δ) = 0 = sin 180°

Expression for mechanical power developed :- → Mechanical power developed = Pm = active component + [E Ia

∗] ⇒

→ Condition for max. mechanical power developed :- d pm

dδ= EV

ZSsin(θ − δ) = 0

Sin (θ – δ) = 0 = sin 0 δ = θ This is the expression for the mechanical power developed interms of load angle and the internal machine angle θ, for constant voltage Vph and constant E i.e., excitation

→ Gross Torque = Pmw

= Pm2π Ns

60

Ns = synchronous speed in r.p.m ∴ Tg = 60

2π . Pm

NS

→ Condition for excitation when motor develops Pmax :- For max power developed is dPm

dE = 0

ddt

�EVZS

cos(θ − δ) − E2

ZScos θ� = 0

→ Condition for excitation when motor develops Pmax is , Eb = VZS2Ra

δ = 180° − 0

Pmax = V2

ZScos θ + EV

ZS

Pm = EVZS

cos(θ − δ) + E2

ZScos θ

Pmmax = EVZS

− E2

ZScos θ

Tg = 9.55 PmNs


Pg.No. 48 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet The corresponding value of max. power is

Pmax = V2

2Ra− V2

4Ra

Power flow in synchronous motors :-

→ For leading p.f

tan δ = Ia �Xq cos ϕ+ Ra sin ϕ�Vt+ Ia �Xq sin ϕ− Ra cos ϕ�

→ The mechanical power developed per phase is given by,

s

Pm = EbVph

Xdsin δ +

Vph 2

2� 1

Xq− 1

Xd� sin 2δ

INDUCTION MACHINES:

→ The power flow diagram of 3 – ϕ induction motor is

Pin √3 VL IL cos ϕ (input)

Stator copper loss 3 Iaph

2 Ra

Mechanical power developed in armature Pm = 3Ebph . Iaph cos (ϕ ± δ) + ve lead –ve for lag

Tsh = Pout 2 π Ns 60. Pout

output

Friction and Iron losses

Tg = Pin 2π Ns 60


Pg.No. 49 of 143






The slip of induction machine is (S) = ns− nrns

= Ns− NrNs

Where Ns is synchronous speed in rpm ns is synchronous speed in rps ⇒ Nr = Ns(1 − s) ⇒ Ns − Nr = SNs ∴ Rotor frequency, f2 = P . SNs

120= S PNs

120= Sf1

For an induction machine with rotor resistance Rr and locked rotor leakage reactance Xr, the rotor impedance Zr at slip s is:Zr = Rr + jsXr The stator circuit equivalent impedance Zrf for a rotor / stator frequency ratio s is: Zrf = Rrs / s + jXrs

For an induction motor with synchronous angular speed ωs running at angular speed ωm and slip s, the airgap transfer power Pt, rotor copper loss Pr and gross output power Pm for a gross output torque Tm are related by: Pt = ωsTm = Pr / s = Pm / (1 - s) Pr = sPt = sPm / (1 - s) Pm = ωmTm = (1 - s)Pt The power ratios are: Pt : Pr : Pm = 1 : s : (1 - s) The gross motor efficiency ηm (neglecting stator and mechanical losses) is: ηm = Pm / Pt = 1 - s

Rotor emf, Current Power :- At stand still, the relative speed between rotating magnetic field and rotor conductors is

synchronous speed Ns; under this condition let the per phase generated emf in rotor circuit be E2.

∴ E2/ph = 4.44 Nphr ϕ1 f1 Kdr Kpr E2/ph = 4.44 Nphr ϕ1 f1 Kwr

Mechanical power developed, Pm Pg

Rotor i/p power = airgap power

Power i/p to stator from mains

Power of rotor shaft

Windage loss

Friction loss at bearings and sliprings of (if any)

Rotor core loss (negligible for small slips)

Rotor I2R loss

Stator core loss

Stator I2R loss


Pg.No. 50 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Kwr = Rotor winding factor → But during running conditions the frequency of the rotor becomes, running with speed Nr

P(Ns− Nr)

120= P SNs

120= Sf1

∴ fr = Sf1 ∴ Emf under running conditions is E = √2 π fr Kw2 Nphr ϕ1 = SE2

→ Rotor leakage reactance = 2π (Rotor frequency) (Rotor leakage Inductance)

∴ Rotor leakage reactance at stand still = 2π f1𝑙𝑙2 = x2 Ω

→ Rotor leakage reactance at any slips = 2π f2𝑙𝑙2

→ Rotor leakage impedance at stand still

= �r2

2 + x22

→ At any slip s, rotor

→ Per phase rotor current at stand still

= E

�r2 2+ x2

2

→ Per phase rotor current at any slip s is given by

I2 =SE2

�r2 2 + (sx2)2

= E2

�(r1/s)2 + x2 2

→ The rotor current I2 lags the rotor voltage E2 by rotor power factor angle θ2 given by

θ2 = tan−1 sx2r2

= sx2 Ω

= �r2 2 + (sx2)2


Pg.No. 51 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet → Per phase power input to rotor is

Pg = F2 I2 cos θ2

cos θ2 = Per phase rotor resistancePer phase rotor impedance

= r2/s

�(r2/s)2+ (x2)2

∴ Pg = E2 I2 × r2/s

�(r2/s)2+ (x2)2

` = E2

�(r2/s)2+ (x2)2 × I2 r2s

= I2

2 r2s

→ Pg is the power transferred from stator to rotor across the air gap. There fore Pg is called air gap power Pg = I2

2 r2s

= I2 2 r2 + I2

2 r2 �1−SS

� Pg = (Rotor ohmic loss) + Internal mechanical power developed in rotor (Pm) = S Pg + (1 − S)Pg ∴ Pm = (1 − S) Pg = I2

2 r2 �1−SS

� Rotor ohmic loss = � S

1−S� Pm = SPg

→ Internal (or gross) torque developed per phase is given by

Te = Internal mechanical power developed inrotor

Rotor speed in mechanical radian per sec

→ Electromagnetic torque Te can also be expressed as

Te = Pg

ωs= 1

ωs × I2

2 r2S

= Rotor ohmic loss(ωs) slip

∴ Te = Rotor ohmic loss(ωs) slip

Te = Pmωr

= (1−S)Pg(1−S)ωs

= Pg

ωs


Pg.No. 52 of 143






→ Power available at the shaft can be obtained from Pg as follows. Output or shaft power, Psh = Pm − Mechanical losses

→ Mechanical losses implies frication and windage losses Psh = Pg − Rotor ohmic loss – Friction and windage losses = Net mechanical power output or net power output Output or shaft torque Tsh = Psh

Rotor speed= Psh

(1−s) ωs

→ If the stator input is known. Then air gap power Pg is given by

Pg = stator power input – stator I2 R loss – stator core loss.

→ Ratio of Rotor input power, rotor copper losses and gross mechanical output is Ir

2 R2/s : Ir2 R2 : Ir

2R2 �1s

− 1�

⇒ ∴ Rotor copper losses = S × Rotor input Gross mechanical output =(1 – S) × Rotor input. Rotor copper losses = (Gross Mechanical output) × S

1−S

Efficiency of the rotor is approximately Equal to ηrotor = Gross mechanical power output

Rotor input

= (1−S) Rotor inputRotor input

= 1 – S = 1 − NS− N

NS

= NNs

Total torque is

Te = mωs

× Ve 2

�Re+ r2s �

2+ (x2+ Xe)2

× r2s

Nm

m is the number of stator phases. Torque equation can be written as Te = m

ωs × I2

2 × r2s

Te = mωs

× rotor input per phase.

1 : S : (1 – S)

ηrotor ≃ NNs


Pg.No. 53 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Thus the slip SmT at which maximum torque occurs is given by SmT = r2

�Re 2+ X2

Substituting the value of maximum slip in the torque equation, gives maximum torque

If stator parameters are neglected then applying maximum transfer theorem to r2/s then r2s

=x2

Slip corresponding to maximum torque is (Breakdown slip) Nm = Ns(1 − Sm) ⇒ Nm = Ns (1 − R2/x2) Nm is the stalling speed at the maximum torque

Starting torque:- At starting, slip S = 1.00, starting torque is given by

Test = m Ve 2

ωs × r2

(Re+ r2)2+ X2

Motor torque in terms of 𝐓𝐓𝐞𝐞𝐞𝐞: → The torque expression of an induction motor can also be expressed in terms of maximum

torque Tem and dimension less ratio SSm,T

. In order to get a simple and approximate

expression, stator resistance r1, or the stator equivalent resistance Re, is neglected.

∴ TeTem

= 2�Re+ �Re

2+ X2 �

�Re+ r2s �

2+ X2

× r2s

→ Since r1 or Re is neglected Te

Tem= 2X

�r2s �

2+ X2

× r2s

→ The slip at which maximum torque occurs is SmT = r2

X ∴ r2 = SmTX

∴ TeTem

= 2X

�SmT Xs �

2+ X2

× SmT Xs

⇒ TeTem

= 2SmT

S + SSmT

Tem = mωs

× Ve 2

2�Re+ �Re 2+ X2

2 �

Sm = r2x2


Pg.No. 54 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Power slip characteristics :- → The total internal mechanical power developed is Pm = m(1 − S)Pg = m I2

2 r2 �1−ss

� w

Pm = mVe 2

�Re+ r2+ r2(1−s)s �

2+ X2

× r2 �1−ss

�

Maximum power transfer theorem is invoked again to obtain maximum value of internal mechanical power developed. Since Pm per phase is the power delivered to r2 �1−s

s�, internal

mechanical power developed is maximum, when rs�1−Smp�

SmP= �(Re + r2)2 + X2

SmP = r2

�(Re+ r2)2+ X2+ r2

In order to get maximum power Pm,substitute r2�1− Smp�Smp

, in place of r2 (1−s)s

in power equation

Pmm = mve 2

2[Re+ r2+ �(Re+ r2)2+ X2

In order to get maximum power output from an induction generator, the rotor must be deiven at a speed given by

ns �1 + r2

�(Re + r2)2 + X2 + r2�

Losses and efficiency :- There are three cases in iron losses. Case (i) : If the ratio of voltage to frequency is constant and flux is also constant then Iron loss = Hysteresis loss + eddy current loss Ph = Kh + Bm

1.6 Pe = Kef 2 Bm 2

Given Vf is constant. As Bm ∝ V

f

⇒ Bm is constant ∴ and Case (ii) : If the ratio of voltage to frequency is not constant and flux is also not constant ⇒ v

f ≠ const ϕ ≠ const

Ph ∝ f Pe ∝ f 2

Te = 2TemSmT

S + SSmT


Pg.No. 55 of 143





Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet Ph = Kh f Bm

1.6 Pe = Ke f 2 Bm2

∴ Case (iii) : If frequency is constant and voltage is variable then Ph = Kh f Bm

1.6 Pe = Khf 2 Bm2

= Kh f �vf�

1.6

→ Short circuit current with normal voltage applied to stator is

I = short circuit current with normal voltage Ibr = short circuit current with voltage Vbr.

→ Power factor on short circuit is found from Pbr = √3 Vbr Ibr cos ϕbr

⇒

→ As Pbr is approximately equal to full load copper losses The blocked rotor impedance is

∴ Blocked rotor reactance = Xbr = �Zbr 2 − Rbr

2

Efficiency of Induction machines :- Generally efficiency = output power

input power

∴ Efficiency of Induction motor = Net mechanical output Electrical power input

∴ Efficiency of Induction generator = Net electrical output

mechanical power input

Squirrel cage rotor: Stator Cu loss = 3 Isc

2 r1 = 3 Isc 2 r1

∴ Rotor Cu loss = Psc − 3 Isc 2 r1

Ph ∝ v1.6 f −0.6 Pe ∝ v2

Ph ∝ v1.6 Pe ∝ v1 2

I = Ibr × VVbr

Cos ϕbr = Pbr

√3 Vbr Ibr

Rbr = PbrIbr

2

Zbr = VbrIbr


Pg.No. 56 of 143






∴ GDGF

= Psc− 3 Isc 2 r1

3 Isc 2 r1

Wound rotor

GDGF

= I2 2 r2

I1 2 r1

= r2r1

�I2I1

�2

Direct – on line (across the line) starting :- → The relation between starting torque and full load torque is

Te = 1ωs

× I2 2 r2

s

∴ Te.stTe.fl

= I2.st

2 r21

I2.ft 2 r2

Sfl

= �I2 stI2 fl

�2

× Sfl

The above equation valids of rotor resistance remains constant. Where Ist

Ifl= (Effective rotor to stator turns ratio) I2 st

(Effective rotor to stator turns ratio) I2 fl

→ Per phase short – circuit current at stand still (or at starting) is,

Where Zsc = (r1 + r2) + j(x1 + x2) Here shunt branch parameters of equivalent circuit are neglected.

→ Therefore, for direct switching,

∴ TestTesf

= �IscIfl

�2

Sfl.

Stator resistor (or reactor) starting :- Since per phase voltage is reduced to xv, the per phase starting current Ist is given by

Ist = xv1Zsc

= xIsc

As be fore Te.stTe.fl

= �IstIfl

�2

× Sfl

= �xIscIfl

�2

× Sfl

→ In an induction motor, torque ∝ (voltage)2 ∴

Te.stTe.fl

= �IstIfl

�2

× Sfl

Isc = V1Zsc

Ist = Isc = V1Zsc

starting torque with reactor startingstarting torque with direct switching

= �xv1v1

�2

= x2


Pg.No. 57 of 143






Auto transformer starting :- → Per phase starting current from the supply mains is Ist = x2Isc

Te.stTe fl

= Per phase starting current in motor windingPer−phase motor full load current

× Sfl

Test with an autotransformerTest with direct switching

= �xv1v1

�2

= x2

Star – delta method of starting :

→ starting torque with star delta starterstarting torque with direct switching in delta

= �VL

√3�2

[VL]2 = 13

∴ star delta starter also reduces the starting torque to one – third of that produced by direct switching in delta.

→ With star – delta starter, a motor behaves as if it were started by an auto transformer starter with x = 1

√3 = 0.58 i.e with 58% tapping.

→ Starting torque with star delta starter Te .stFull load torque with startor winding in delta,Tefld

= 1

ws �Ist .y�

2 r21

1ws

(Ifl d)2 r2Sfl

= � 1

√3 Ist .d�2

(Ifl .d)2 × Sfl = 13

� Isc .dIfl . d

�2

× Sfl

Te.stTe.fl

= Ist IscIfl

2 × Sfl


Pg.No. 58 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet Power systems:

Power Systems Generation and Distribution and Concepts of HVDC: Thermal Power Station :-

Thermal efficiency, ηThermal = Heat equivalent of mech−energy Transmitted toTurbine shatHeat of coal combustion

Thermal efficiency = ηboiler × ηturbine Overall efficiency, ηoverall = Heat equvivalent of electrical o/p

Heat of combustion of coal

Overall efficiency = Thermal efficiency × Electrical efficiency. Energy output = coal consumption × calorific value. = coal consumption × 6500 k.cal η = Output in k.cal

Input in k.cal

Water Power equation:-

Water Head : The difference of water level is called the water head. Gross Head : The total head between the water level at inlet and tail race is called as gross

head Rated Head : Head utilized in doing work on the turbine Net Head : It is the sum of the Rated Head and the loss of head in guide passage and

entrance H = Head of water in metre Q = Quantity of water in m3/sec or lit/sec. W = specific gravity of water = 1 kg/lit when ‘Q’ represented in lit/sec. = 100 kg/m3 when ‘Q’’ represented in m3/sec. η = efficiency of the system.

Effective work done

orOutput of system

� = WQH × η kg- m/sec.

Metric output = WQH × η

75 (H.P)

1H.P = 75 kg-m/sec

Metric output in watt = WQH × η 75

× 735.5

Output = WQH102

× η kw


Pg.No. 59 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet

Volume of water available per annum = catchment area × Annual Rainfall → Electric energy generated = weight × head × overall η.

GAS TURBINE POWER PLANT :

→ The thermal efficiency of gas turbine plant is about 22% to 25% → The air fuel ratio may be of the order of 60 : 1 in this ase.

Engine efficiency ηengines = ηoverall

ηalt

Thermal efficiency ηthey = ηenginemech.η of englnd

Heat produced by fuel per day = coal consumption / day × caloritic value

Terms and Definitions :-

1. Connected load :- It is the sum of ratings in kilo watts of equipment installed in the consumer’s premises 2. Demand :- It is the load or power drawn from the source of supply at the receiving end averaged over a specified period. 3. Maximum Demand :- Maximum demand (M.D) of a power station is the maximum load on the power station in a given period. 4. Average load:- If the number of KWH supplied by a station in one day is divided by 24 hours, then the value so obtained is known as daily average load.

Daily average load = KWH deliverd in one day 24

Monthly average load = KWH delivered in one month30 ×24

Yearly average load = KWH delivered in one year365 ×24

5. Plant capacity :- It is the capacity or power for which a plant or station is designed. It should be slightly more than M.D. it is equal to sum of the ratings of all the generators in a power station. 6. Firm Power :- It is the power which should be always be available even under emergency 7. Prime Power :- It is the maximum power (may be thermal or hydraulic or mechanical) continuously available for conversion into electrical power.


Pg.No. 60 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet 8. Dump power :- This is the term usually used in hydro electric plants and it represents the power in excess of the load requirements. It is made available by surplus water. 9. Spill Power :- Is that power which is produced during floods in a hydro power station. 10. Cold reserve :- Is that reserve generating capacity which is not in operation but can be made available for service. 11. Hot reserve :- It that reserve generating capacity which is in operation but not in service 12. Spinning reserve :- Is that reserve generating capacity which is connected to bus-bars and is ready to take the load.

Load factor :-

It is defined as the ratio of number of units actually generated in a given period to the number of units that could have been generated with maximum demand.

Load factor = Average load or Average DemandMaximum Demand.

= Energy generated in a given period(Maximum Demand) ×(Hours of operaation in the given period)

. → The load factor will be always less than one (< 1)

Demand factor:- It is defined as the ratio of maximum demand on the station to the total connected load to the station.

∴ Demand factor = Maximum Demand on the stationTotal connected load to the station

Its value also will be always less than one (< 1)

Diversity Factor :- Diversity factor may be defined as “the sum of individual maximum demand to the station to the maximum demand on the power station”.

Diversity factor = Sum of individual consumers maximum demandMaximum demand on the station.

Its value will be always greater than one (> 1)

Plant Factor or Plant Use Factor :-

Plant factor = station output in kwhΣ (KW1) H1+ (KW2) H2+ (KW3)H3+ ……

Where KW1, KW2, KW3 etc. are the kilowatt ratings of each generator and H1, H2, H3 etc. are the number of hours for which they have been worked.


Pg.No. 61 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet Capacity Factor or plant capacity factor or capability factor :-

→ It is defined as the ratio of average demand on the station to the maximum installed capacity.

i.e. capacity factor = Average demand on the stationMax.installed capacity of the station

→ Coincidence factor:- It is the reciprocal of diversity factor and is always less than 1

→ Utilization factor = Maximum demand Plant capacity

→ Operation factor = Service hoursTotal duration

→ Use factor = Actual energy produced Plant capacity ×Time (hrs)the plant has been in operation

D.C. Distribution calculations

Uniformly loaded Distributor fed at one end. → Fig (a) shows the single lien diagram of a 2 – wire d.c. distributor AB fed at one end A and

loaded uniformly with i amperes per metre length.

→ Then the current at point c is. = δ𝑙𝑙 − ix amperes = i(𝑙𝑙 – x) amperes. → Total voltage drop is the distributor upto point C is

𝑣𝑣 = ∫ irx0 (𝑙𝑙 − x)dx = ir �𝑙𝑙x − x

2

2�

→ Voltage drop over the distributor AB = 1

2 ir𝑙𝑙2

= 12 IR

Where i𝑙𝑙 = I, the total current entering at point A

Fig. (a)

A

B

𝑙𝑙

i i i i

A

B

𝑙𝑙 C

x dx


Pg.No. 62 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet r𝑙𝑙 = R, the total resistance of the distributor. Uniformly loaded distributor fed at both ends. (i) Distributor fed at both ends with equal voltages

Current supplied from each feeding point = 9𝑙𝑙

2

→ Voltage drop upto point C = ir

2 (𝑙𝑙x − x2).

→ Max. voltage drop = 18 IR

→ Min. voltage = V – IR8

volts (ii) Distributor fed at both ends with unequal voltages :-

The point of minimum potential C is situated at a distance x meters from the feeding point A.

Voltage drop in section AC = irx2

2 volts.

→ x = VA− VB

ir𝑙𝑙+ 𝑙𝑙

2

Comparison of 3 – wire and 2 – wire D.C. distribution

The area of cross – section of neutral is half the cross – section of outers in 3 – wire system.

→ ∴ volume of w for 3−wire systemvolume of w for 2−wire system

= 58

a𝑙𝑙 × 12a𝑙𝑙

= 516

If the neutral has the same cross – section as the outer, then.

A

B

i i i i i

C

VB

x 𝑙𝑙 - x

VA

A

B

𝑙𝑙

i i i i i

C

V

C

dx x


Pg.No. 63 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet volume of w for 3−wire system

volume of w for 2−wire system= 3

4 a𝑙𝑙 × 1

2a𝑙𝑙= 3

8 = 5

= 37.5%

Transmission Lines:

→ The empirical formula for the economical voltage line to line is V = 5.5 � 𝑙𝑙1.6

+ 3P100

where ‘V’ = line pressure in KV, l = distance of transmission in KM, P = estimated max.KW per phase to be delivered over one pole or tower line Performance of Lines

→ By performance of lines is meant the determination of efficiency and regulation of lines.

The efficiency of lines is defined as

→ % efficiency = Power delivered at the receiving endPower sent from sending end

× 100

→ % efficiency = Power delivered at the receiving endPower delivered at the receiving end+losses

× 100

→ % regulation = Vr ′− VrVr

× 100

Where Vr ′ is the receiving end voltage under no load condition and Vr the Receiving end voltage under full load condition.

Effect of Earth on a 3 – 𝛟𝛟 line:- S. No Line Description R L XL C XC

1. Length Increases Increases Increases Increases Increases Decreases 2. Distance of separation

increases No change Increases Increases Decreases Increases


Pg.No. 64 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet 3. Radius of conductor

increases Decreases Decreases Decreases Increases Decreases

4. Symmetrical spacing. Does not depend

Decreases Decreases Increases Decreases

5. Unsymmetrical spacing. Does not depend.

Increases Increases Decreases Increases

6. Effect of earth is taken into account

No change No change No change Increases Decreases

7. Height of the conductor increases

No change No change No change Decreases Increases.

→ The capacitance C of the conductor with reference to grund

C = 2 πε0ln �2h r� �

F/metre

Where h = distance between earth and conductor

Effect of earth on the capacitance of single – phase transmission line :-

Effect of earth on the capacitance of single – phase transmission line is

→ C = π ε0

ln Dabr Daa ′

Dab

F/metre

The effect of earth on capacitance of the system is to increases it. Capacitance of a 1 – 𝛟𝛟 transmission line

a

a′

b

b′

Daa′


Pg.No. 65 of 143






1 – 𝛟𝛟 transmission line

→ C = π ε0hr + �h

2

r2

F/metre.

→ C = ρL

V= π ε0

ln h r� F/metre.

→ The capacitance C of the 3 – ϕ transmission line with reference to grund is

C = PaVa

= 2π εD

ln �DabDbcDca

3

r �Daa′ Dbb′ Dcc′Dab′ Dbc′ Dca′

3 F/metre.

→ Capacitance of a 3 – Phase Unsymmetrically Spaced Transmission Line is

C = PaVa

= 2πε0ln GMD

r

F/metre

Vertical Spacing:-

h

g

f

d

a

b

c

c

a

b

b

c

a

1 2 3

c′

b′

a′

a′

b′

c′

a′

c′

b′

−ρL P

h

r

x

ρL


Pg.No. 66 of 143






→ For Vertical Spacing conductors, The capacitance C per phase of the system is

C = 4 πε0

ln √23 dr �gf�23� F/metre/phase.

→ Capacitance of a hexagonal spacing Double Circuit Line is

C = ρaVa

= 2πε0ln√3D2

F/metre/coloumb

Bundled conductors:-

→ For a two conductor bundle, the equation for maximum gradient at the surface of a sub-conductor is

g = V�1+ 2rs �

2r ln d√rs

where ‘s’ is the seperation between the sub – conductors.

→ Let the equivalent radius or geometric mean radius be P0 then for two conductors P0 = (rd)1 2� = r1 2� d1 2�

→ When there are 3 conductors

P0 = (r d′d′)1 3� = r1 3� d2 3� �34�13�

→ For 4 conductors

P0 = �r d√2

. d√2

d�14� = r1 4� d3 4� �1

2�14� .

→ For six conductors , P0 = r1 6� �d2�56� 61 6� = �6r �d

2�5�16�

Inductance of a double 3 – 𝛟𝛟 line :-


Pg.No. 67 of 143






Transposed double circuit line.

→ Inductance per phase =

2 × 10−7 ln 21 6� �dr1�12� �g

f�13� H/metre/phase

Inductance of composite conductors :-

Inductance of composite conductors – 1 – 𝛟𝛟 transmission line.

→ LA = 2 × 10−7 ln ��D11′ D12′ ⋯ D1n′ � �D21′ D22′ ⋯ D2n′ � ⋯�Dm1

′ Dm2′ Dmn

′ �mn

�(R′ D12 D13 ⋯ D1m) (R′D21D23 ⋯ D2m) ⋯(R′ Dm1 Dm2 ⋯ Dmm)m2

The mnth root of the product of the mn distances between m strands of conductor A and n strands of conductor B is called geometric mean distance (GMD) and is denoted as Dm.

The m2th root of m2 distances i.e., the distance of the various strands from one of the strands and the radius of the same strand, the distances of such m groupings constitute m2 terms in the denominator, is called the geometric mean radius (GMR) of self GMD and is denoted as Ds. → LA = 2 × 10−7 ln Dm

Ds Henry/metre.

A

A

B

1

2

3

m

n

1

2 3

4

h

g

f

d

a

b

c

c

a

b

b

c

a

1 2 3

c′

b′

a′

a′

b′

c′

a′

c′

b′


Pg.No. 68 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet Composite Conductors,

→ Inductance L per unit length

= �µ04π

+ µ0π

ln DR� Henry/metre.

= 4 × 10−7 ln DR′

Henry/metre [where R′ = R. e1 4� ] Inductance of 3 – 𝛟𝛟 Unsymmetrically Spaced Transmission Line

→ L = La + Lb + Lc3

= 2 × 10−7 ln √abc−3

R′ Henry/metre.

Short Transmission Line

→ The equivalent circuit and vector diagram for a short transmission line are shown in fig.

VS = �1 + 2IrR cosϕrVr

+ 2Ir XsinϕrVr

+ Ir 2

Vr 2 (R2 + X2)

Vr

→ In practice the last term under the square root sign is generally negligible; therefore.

VS = Vr �1 + �2IrRVr

cosϕr + 2IrXVr

sinϕr��12�

R + JX

IS

vS vr

Ir

ϕr ϕa

vr IrR

jIrX vS

Ir

3 – ϕ transmission with Unsymmetrical spacing

a

b c

Ia

Ic Ib


Pg.No. 69 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet The terms within the simple brackets is small as compared to unity. Using binomial expansion and limiting only to second term, Vs ≃ Vr + IrRcos ϕr + Ir X sinϕr → The receiving end voltage under no load Vr ′ is the same as the sending end voltage under full

load condition. %regulation = VS− Vr

Vr × 100

= �IrRVr

cosϕr + IrXVr

sinϕr� × 100

Regulation per unit = IrRVr

cosϕr + IrXVr

sinϕr

= Vr cosϕr + Vx sinϕr → Where Vr and Vx are the per unit values of resistance and reactance of the line.

Vs = AVr + BIr Is = CVr + DIr

A = VsVr�

Ir = 0

This means A is the voltage impressed at the sending end per volt at the receiving end when receiving end is open. It is dimensionless. B = Vs

Ir�

Vr = 0

B is the voltage impressed at the sending end to have one ampere at the short circuited receiving end. This is known as transfer impedence in network theory. C = Is

Vr�

Ir = 0

C is the current in amperes into the sending end per volt on the open – circuited receiving end. It has the dimension of admittance. D = Is

Ir�

Vr = 0

D is the current at the sending end for one ampere of current at the short circuited receiving end .

The constants A, B, C, and D are related for a passive network as follows AD – BC = 1

→ The sending end voltage and current can be written from the equivalent network as, Vs = Vr + IrZ Is = Ir

→ The constants for short transmission lines are, A = 1 B = Z C = 0 D = 1

→ % regulation = VS

A�− Vr

Vr ×100


Pg.No. 70 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet → % η = Power received at the receiving end

Power received at the receiving end+losses × 100

Where R is the resistance per phase of the line. Medium Length Lines:- → Transmission lines with lengths between 80 km and 160 km are categorized as medium lines

where the parameters are assumed to be lumped. . → The two configurations are known as nominal – T and nominal – π respectively.

Nominal – T

Vr ′ = |VS|� −jwc�R2+ jX2 − jwc

% of regulation = Vr ′− VrVr

× 100

% η = P

P + 3 R2 �Ir 2+ I5 2� × 100

A, B, C, D constant for nominal – T A = 1 + YZ

2

B = Z �1 + YZ2�

C = Y

D = �1 + YZ2�

Nominal – 𝛑𝛑

jIr X/2

IC

Ir

IS

vr

vc

vS

IS R2

jIS X2

Y = jwc

IC IS

vS vS vr vr Y2 Y

2

Ir Ir I𝑙𝑙 vC R2

+ jX2

R2

+ jX2

R + jX

IC1


Pg.No. 71 of 143






→

Vr ′ = |VS|�−2jωC�

R+jX− jωC

2�

% regulation = VrLVrVr

× 100

% η = PP+3 I𝑙𝑙

2R ×100

A, B, C, D constants for nominal – 𝛑𝛑 A = 1 + Y𝑍𝑍

2

B = Z

C = Y �1 + YZ4�

D = �1 + YZ2�

Long Transmission Lines:- → In case the lines are more than 160 km long

→ Let Z = series impedence per unit length Y = shunt admittance per unit length 𝑙𝑙 = length of line Z = zl = total series impedence Y = yl = total shunt admittance. V = Aerx + Be−rx I = I

ZC(Aerx − Be−rx)

V = Vr+ IrZc2

erx + Vr− Ir ZC2

e−rx

I = 1ZC�Vr+ IrZC

2 erx − Vr− IrZC

2 e−rx�

vS

V + ∆ V1 I + ∆I Z∆X

vr V + ∆V

I + ∆I I

V

∆X X ∆X

C∆X Y∆X

V1I

vS vr Y2

= jwc2

Ir I𝑙𝑙 Z = R + jx

IC1

IS

IC2

Y2

= jwc2

vS

IS

Ir I𝑙𝑙

IC& IC

vr I𝑙𝑙R

jI𝑙𝑙X


Pg.No. 72 of 143






ZC = �zy

= �r+jωLg+jωC

→ The propagation constant r = ∝ + jβ ; the real part is known as attenuation constant and the

quadrature component β the phase constant and is measured in radians per unit length. V = Vr+ Ir ZC

2 e∝x. ejβx + Vr−IrZC

2 e−∝x. e−jβx

Vs = Vr cos hrl + IrZc sin hrl IS = Vr sinhrl

Zc+ Ir cos hrl

A = cosh rl B = Zc sinh rl C = sinh rl

Zc

D = cosh rl

The equivalent Circuit Representation of a Long Line equivalent – 𝛑𝛑 Representation

equivalent – T Representation of Long Line.

Constants for Two networks in Tandem

Y1 = Y sinh r𝑙𝑙r𝑙𝑙

VS Vr

IS Z′

2= Z

2 Tanhr𝑙𝑙/2

r𝑙𝑙/2

VS y1

2 y1

2= y

2 Tanhr𝑙𝑙/2

r𝑙𝑙/2

Z1 = Z sinhr𝑙𝑙r𝑙𝑙


Pg.No. 73 of 143






equivalent �A B

C D� = �A1 B1C1 D1

� �A2 B2C2 D2

�

Constants for networks in parallel

Critical disruptive voltage :-

Critical disruptive voltage is defined as the voltage at which complete disruption of

dielectric occurs

At any other temperature and pressure g0 ′ = g0 . δ

Where δ is the air density correction factor and is given by

δ = 3.92 b273+t

where b is the barometric pressure in cm of Hg and t the temperature in ℃.

∴ The critical disruptive voltage is given by V′ = rg0 δ ln dr kV

A = A1 B2+ A2 B1B1+ B2

B = B1.B2B1+ B2

A = D = A1 B2+ A2 B1B1+ B2

= D1B2+ D2B1B1+ B2

C = C1 + C2 + (A1− A2)(D2− D1)B1+ B2

Equivalent Single Network Parameters

vS vr

IS

IS1

IS2

Ir

A1, B1 C1, D1

A2, B2 C3, D2

vS vr

IS Ir I

V

A1 , B1

C1 , D1

A2 , B2

C2 , D2


Pg.No. 74 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet The final expression for the critical disruptive voltage after taking into account the atmospheric conditions and the surface of the conductor is given by → V′ = rg0 δ m0 ln d

r kV

Where m0 = average value for the ratio of breakdown voltage and smooth conductor

Polished wires − 1 Roughened or weathered wires − 0.98 to 0.93 Seven strand cable − 0.87 to 0.83 Large cables with more then seven strands − 0.90 approx

The distance between gv and g0 is called energy distance. According to peek this distance is equal to (r + 0.301√r) for coaxial conductors.

→ gv = g0δ �1 + 0.3√rδ

� kv/cm for two wires in parallel

→ vv = rg0 δ �1 + 0.3√rδ

� ln dr kV

→ In case the irregularity factor is taken into account,

Vv = 21.1 mv δr �1 + 0.3√rδ

� ln dr kV rms

Where r is the radius in cms. The irregularity factor mv has the following values: mv = 1.0 for polished wires = 0.98 to 0.93 for rough conductor exposed to atmospheric severities = 0.72 for local corona on stranded conductors.

Peek made a number of experiments to study the effect of various parameters on the corona loss and he deduced an empirical relation

P = 241 × 10−5 (f+25)δ

�rd

�Vp − V0�2 kw/km/phase

Where f = frequency of supply δ = The air density correction factor Vp = The operating voltage in kV

V0 = the critical disruptive voltage Sag in Transmission and Distribution Lines

→ When the Supports are at the Same Level , Sag = s = wl2

2T

→ Where w = weight in kg/m run

𝑙𝑙 = half the span length in metre


Pg.No. 75 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet T = Tension in kg

Effect of wind and ice

→ w = weight of conductor kg/m run wi = weight of ice kg/m run ww = wind pressure kg/m run acting horizontally

→ Weight of ice (Wi) = πR (D + R) × 1 × specific wt of ice Specific weight of ice = weight of ice/m3 = 1000 kg

→ Wind pressure = Ww = (D + 2R) × 1 × wind pressure kg/m3 ∴ W = �(w + wi)2 + ww

2

∴ Sag = Wl2

2T m

Factor of Safety

→ factor of safety = Max. Stresspermissible stress

→ Vertical sag = S cos θ When the Supports are at different level

→ Let l = Span length h = difference in levets between two supports x1 = Distance of support at low level from 0 x2 = Distance of support at higher level from 0 T = Tension in the conductor W = weight of the conductor per unit length, then

S1 = Wx1 2

2T ; S2 = Wx2 2

2T

x1 = 𝑙𝑙2− Th

w𝑙𝑙

x2 = 𝑙𝑙2

+ Thw𝑙𝑙

x1 + x2 = 𝑙𝑙 S2 − S1 = h

→ The sag is provided in over headlines so that the safe tension is not exceeded. Most Economical Conductor size in a cable :-

→ gmax = 2V

d loge Dd volts/m


Pg.No. 76 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet → ∴ most economical conductor diameter is

d = D2.718

→ ∴ and the value of gmax under this condition is gmax = 2Vd

volts/m

Power System Operation and Control:

→ incremental fuel rate IFR = Incremental change in inputIncremental change in output

→ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑙𝑙 𝑓𝑓𝑓𝑓𝑖𝑖𝑙𝑙 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = ∆ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖∆ 𝑜𝑜𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖

= d (input)d (output)

= dFdp

→ Incremental efficiency = (output)(input)= dp

dF

→ If Pi and Qi are scheduled electrical generation, PDi and QDi are the respective load demands, Pi − PDi − P𝑙𝑙 = Mi = 0 Qi − QDi − Q𝑙𝑙 = Ni = 0 where

Mi and Ni represent the power residuals at bus i and P𝑙𝑙 and Q𝑙𝑙 the power flow to the neighbouring system given by P𝑙𝑙 = ∑ ViVjYijN

j=1 cos�δij − θij� Q𝑙𝑙 = ∑ ViVj YijN

j=1 �δij − θij� → For proper operation, each generator should have a minimum and maximum permissible output

and the unit production should be constrained to ensure that Pimin ≤ Pi ≤ Pimax, i = 1, 2, … … , NP Qimin ≤ Qi ≤ Qimax, i = 1, 2, … … , Na Np and NQ being total number of real and reactive sources in the system

→ The condition for optimum operation is

dF1dP1

= dF2dP2

= ∙∙ = dFndPn

= λ

→ Here dFndPn

= incremental production cost of plant n in RS. per MWhr.

→ The incremental production cost of a given plant over a limited range is represented by

dFndPn

= Fnn Pn + fn

Where Fnn = slope of incremental production cost curve. fn= intercept of incremental production cost curve.

→ The optimal load dispatch problem including transmission losses is defined as


Pg.No. 77 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet d Fn

d Pn+ λ ∂PL

∂ Pn= λ

→ Here the term ∂ PL∂ Pn

is known as the incremental transmission loss at plant n and λ is

known as the incremental cost of received power in RS per MW The solution of coordination equation requires the calculation of ∂PL/ ∂Pn which is obtained from equation as

∂ PL∂ Pn

= 2 ∑ BmnPmm also d Fnd Pn

= Fnn Pn + fn

The coordination equations can be written as

FnnPn + fn + λ Σ 2Bmn Pm = λ

→ Solving for Pn we obtain

Pn = 1 − fn

λ ∑ 2 Bmn Pmm ≠n

Fnnλ + 2 Bnn

Where Pm and Pn are the source loadings, Bmn the transmission loss co – efficient.

→ The loss formula equation is expressed in terms of generations and is

approximately expressed as

PL = ��Pmnm

BmnPn

Where Pm and Pn are the source loadings, Bmn the transmission loss co – efficient.

→ For two generated plants are there then Power loss, PL = B11 P1 2 + B12 P1 P2 + B22P2

2 Where B11 , B12 , B22 are loss coefficients of transmission line

→ 𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝𝑇𝑇𝑇𝑇𝑖𝑖𝑖𝑖 𝑝𝑝𝑙𝑙𝑖𝑖𝑖𝑖𝑖𝑖𝑝𝑝 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑇𝑇𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏 𝑖𝑖 𝑖𝑖𝑖𝑖𝑖𝑖 – 𝑙𝑙𝑖𝑖𝑖𝑖𝑖𝑖 then the 𝑝𝑝𝑖𝑖𝑖𝑖𝑖𝑖𝑙𝑙𝑖𝑖𝑏𝑏 𝑓𝑓𝑖𝑖𝑖𝑖𝑖𝑖𝑇𝑇𝑖𝑖𝑝𝑝 𝑓𝑓𝑇𝑇𝑖𝑖 𝑝𝑝𝑙𝑙𝑖𝑖𝑖𝑖𝑖𝑖𝑝𝑝 1 𝑖𝑖𝑖𝑖𝑖𝑖 2 𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖𝑝𝑝𝑝𝑝𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑣𝑣𝑖𝑖𝑙𝑙𝑏𝑏

𝐿𝐿1 = 1

1 − 𝜕𝜕 𝑃𝑃𝐿𝐿𝜕𝜕 𝑃𝑃1

& L2 = 1

1 − ∂ PL∂ P2


Pg.No. 78 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet FAULTS:

→ Percentage reactance %𝑋𝑋 = 𝐼𝐼𝐼𝐼𝑉𝑉

× 100 I=full load current

V= phase voltage

X= reactance in ohms per phase

→ Alternatively percentage reactance (%X) (an also be expressed in terms of KVA and KV under

%X= (𝐾𝐾𝑉𝑉𝐾𝐾)𝐼𝐼10(𝐾𝐾𝑉𝑉)2

Where X is the reactance in ohms.

→ If X is the only reactance element in the circuit then short circuit currenr is given by

Isc= 𝑉𝑉𝐼𝐼

= I × (100%𝐼𝐼

)

i.e short circuit current is obtained by multiplying the full load current by 100/%X

Short- circuit KVA=Base KVA × 100%𝐼𝐼

Symmetrical components in terms of phase currents:-

→ The unbalanced phase current in a 3-phase system can be expressed in terms

of symmetrical components as under.

𝐼𝐼𝑅𝑅��⃗ = 𝐼𝐼𝑅𝑅1��⃗ +𝐼𝐼𝑅𝑅2��⃗ +𝐼𝐼𝑅𝑅0��⃗ 𝐼𝐼𝑌𝑌��⃗ = 𝐼𝐼𝑌𝑌1��⃗ +𝐼𝐼𝑌𝑌2��⃗ +𝐼𝐼𝑌𝑌0��⃗ 𝐼𝐼𝐵𝐵��⃗ = 𝐼𝐼𝐵𝐵1��⃗ +𝐼𝐼𝐵𝐵2��⃗ +𝐼𝐼𝐵𝐵0��⃗

Where The positive phase current (𝐼𝐼𝑅𝑅1�� , 𝐼𝐼𝑌𝑌1�� , &𝐼𝐼𝐵𝐵1��)

Negative phase sequence currents (𝐼𝐼𝑅𝑅2�� , 𝐼𝐼𝑌𝑌2�� , &𝐼𝐼𝐵𝐵2��) and


Pg.No. 79 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet Zero phase sequence currents ( 𝐼𝐼𝑅𝑅0�� , 𝐼𝐼𝑌𝑌0�� , &𝐼𝐼𝐵𝐵0��)

→ The operator ‘a’is one, which when multiplied to a vector rotates the vector through 1200 in the anticlockwise direction.

→ A=-0.5+j 0.866 ; a2=-0.5- j 0.866

a3=1

→ Properties of operator ‘a’: 1+a+a2=0 a-a2

=j √3 → Positive sequence current 𝐼𝐼𝐵𝐵1��⃗ in phase B leads 𝐼𝐼𝑅𝑅1��⃗ by 1200 and therefore 𝐼𝐼𝐵𝐵1��⃗ = 𝑖𝑖 𝐼𝐼𝑅𝑅1��⃗

similarly, positive sequence current in phase Y is 2400 ahead of 𝐼𝐼𝑌𝑌1��⃗ = a2𝐼𝐼𝑅𝑅1��⃗ 𝐼𝐼𝑅𝑅��⃗ = 𝐼𝐼𝑅𝑅1��⃗ + 𝐼𝐼𝑅𝑅2��⃗ + 𝐼𝐼𝑅𝑅0��⃗

𝐼𝐼𝑌𝑌��⃗ = 𝐼𝐼𝑌𝑌1��⃗ + 𝐼𝐼𝑌𝑌2��⃗ + 𝐼𝐼𝑌𝑌0��⃗ = a2 𝐼𝐼𝑅𝑅1��⃗ + 𝑖𝑖 𝐼𝐼𝑅𝑅2��⃗ + 𝐼𝐼𝑅𝑅0��⃗

𝐼𝐼𝐵𝐵��⃗ = a 𝐼𝐼𝑅𝑅1��⃗ + a2 𝐼𝐼𝑅𝑅1��⃗ + 𝐼𝐼𝑅𝑅0��⃗ = 𝐼𝐼𝐵𝐵0��⃗ +𝐼𝐼𝐵𝐵1��⃗ +𝐼𝐼𝐵𝐵2��⃗

→ Zero sequence current:

𝐼𝐼𝑅𝑅��⃗ + 𝐼𝐼𝑌𝑌��⃗ + 𝐼𝐼𝐵𝐵��⃗ = 𝐼𝐼𝑅𝑅1��⃗ (1+a+a2) + 𝐼𝐼𝑅𝑅2��⃗ (1+a+a2) + 3𝐼𝐼𝑅𝑅0��⃗ = 3 𝐼𝐼𝑅𝑅0��⃗ ∴ 𝐼𝐼𝑅𝑅0��⃗ = 1

3[ 𝐼𝐼𝑅𝑅��⃗ + 𝐼𝐼𝑅𝑅��⃗ + 𝐼𝐼𝑅𝑅��⃗ ]

→ Positive sequence current : 𝐼𝐼𝑅𝑅��⃗ +a 𝐼𝐼𝑌𝑌��⃗ +a2 𝐼𝐼𝐵𝐵��⃗ = 𝐼𝐼𝑅𝑅1��⃗ (1+a3+a3) + 𝐼𝐼𝑅𝑅2��⃗ (1+a2+a4) + 𝐼𝐼𝑅𝑅0��⃗ (1+a+a2) = 3𝐼𝐼𝑅𝑅1��⃗ ∴ 𝐼𝐼R1 = = 1

3[ 𝐼𝐼𝑅𝑅��⃗ +a 𝐼𝐼𝑌𝑌��⃗ +a2 𝐼𝐼𝐵𝐵��⃗ ]

→ Negative sequence current:- 𝐼𝐼𝑅𝑅��⃗ +a2 𝐼𝐼𝑌𝑌��⃗ +a 𝐼𝐼𝐵𝐵��⃗ = 𝐼𝐼𝑅𝑅1��⃗ (1+a4+a2) + 𝐼𝐼𝑅𝑅2��⃗ (1+a3+a3) + 𝐼𝐼𝑅𝑅0��⃗ (1+a2+a) = 3 𝐼𝐼𝑅𝑅2��⃗ ∴ 𝐼𝐼R2 = 1

3[ 𝐼𝐼𝑅𝑅��⃗ +a2 𝐼𝐼𝑌𝑌��⃗ +a𝐼𝐼𝐵𝐵��⃗ ]


Pg.No. 80 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet Single Line to –Ground Fault:-

→ 𝑉𝑉�⃗ R=0 and 𝐼𝐼B =𝐼𝐼Y=0

The sequence currents in the red phase in terms of line currents shall be :-

→ Fault current:- Fault current, 𝐼𝐼𝑅𝑅��⃗ =3 𝐼𝐼0��⃗ = 3 𝐸𝐸𝑅𝑅��⃗𝑧𝑧0��⃗ +𝑍𝑍1+𝑍𝑍2

��⃗

Phase voltage at fault

Since the generated emf system is of positive sequence only ,the sequence

components of emf in R-phase are:

→

The sequence voltage at the fault for R-phase are: This is ecpected because R-phase is shorted to ground.

∴ The phase voltages at fault are:

Line –To-Line fault:-


Pg.No. 81 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet The condition created by this fault lead to:

→

Again taking R-phase as the reference, we have

→

Expressing in terms of sequence components of red line, we have

→

Also,

Fault current:

→ Phase voltages: - since the generated emf system is of positive phase sequence only,the sequence components of emf in R-phase are:

→ The sequence voltages at the fault for R-phase are:


Pg.No. 82 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet → The phase voltages at the fault are:

→ Double Line- To – Ground Fault:-

The conditions created by this fault lead to :

Also,

→ Fault current:-

→

Phase voltages: - the sequence voltages for phase R are:

→

Now

→

TRANSIENTS IN SIMPLE CIRCUITS:


Pg.No. 83 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet 1. D.C sources

a) Resistance only: - As soon as switch is closed, the current in the circuit will be determined according to ohms law.

𝐼𝐼 = 𝑉𝑉𝑅𝑅

𝑁𝑁𝑇𝑇𝑇𝑇 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑝𝑝𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑝𝑝 𝑇𝑇𝑖𝑖𝑙𝑙𝑙𝑙 𝑏𝑏𝑖𝑖 𝑖𝑖ℎ𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑖𝑖ℎ𝑖𝑖 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑓𝑓𝑖𝑖𝑖𝑖 .

b) Inductance only:- when switch s is closed, the current in the circuit will be given by

𝐼𝐼(𝑝𝑝) =𝑉𝑉(𝑠𝑠)𝑍𝑍(𝑠𝑠)

= 𝑉𝑉𝑆𝑆 . 1𝐿𝐿𝑆𝑆

= 𝑉𝑉𝐿𝐿 . 1𝑆𝑆2

𝑖𝑖(𝑖𝑖) = 𝑉𝑉𝐿𝐿 t

c) Capacitance only:- when switch s is closed, the current in the circuit is given

I(s) = 𝑉𝑉(𝑠𝑠)𝑍𝑍(𝑠𝑠)

= 𝑉𝑉𝑆𝑆 .CS =VC

Which is an impulse of strength (magnitude)VC

d) R-L circuit: when switch s is closed, the current in the circuit is given by


= 𝑉𝑉𝑆𝑆 . 1𝑅𝑅+𝐿𝐿𝑆𝑆

= 𝑉𝑉𝑆𝑆 . 1/𝐿𝐿𝑆𝑆+𝑅𝑅/𝐿𝐿

= 𝑉𝑉𝐿𝐿 �1𝑆𝑆− 1

𝑆𝑆+𝑅𝑅/𝐿𝐿� .𝐿𝐿𝑅𝑅

= 𝑉𝑉𝑅𝑅

�1𝑆𝑆− 1

𝑆𝑆+𝑅𝑅/𝐿𝐿�

𝑖𝑖(𝑖𝑖) = 𝑉𝑉𝑅𝑅

�1 − 𝑖𝑖𝑒𝑒𝑝𝑝 �−𝑅𝑅𝐿𝐿

𝑖𝑖��

e) R-L circuit: After the switch s is closed, current in the circuit is given by


= 𝑉𝑉𝑆𝑆 . 1𝑅𝑅+1/𝐶𝐶𝑆𝑆

= 𝑉𝑉𝑆𝑆 � 1𝑅𝑅𝑅𝑅�𝐶𝐶𝑆𝑆

𝑆𝑆+1/𝑅𝑅𝐶𝐶 = 𝑉𝑉

𝑅𝑅 . 1

𝑆𝑆+1/𝑅𝑅𝐶𝐶

i (t)= 𝑉𝑉𝑅𝑅

.𝑖𝑖−𝑖𝑖/𝐶𝐶𝑅𝑅

→ R-L-C circuit: - After the switch S is closed, the current in the circuit is given by

Type equation here. I(s) = 𝑉𝑉𝑆𝑆

1𝑅𝑅+𝐿𝐿𝑆𝑆+1𝐶𝐶𝑆𝑆

I(s) = 𝑉𝑉𝐿𝐿 . 1

(𝑆𝑆+𝑎𝑎−𝑏𝑏)(𝑆𝑆+𝑎𝑎+𝑏𝑏)


Pg.No. 84 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet i(t) = 𝑉𝑉

2𝑏𝑏𝐿𝐿 �𝑖𝑖−(𝑎𝑎−𝑏𝑏)+ − 𝑖𝑖−(𝑎𝑎+𝑏𝑏)𝑖𝑖�

where 𝑅𝑅2𝐿𝐿

= a and �𝑅𝑅2

4𝐿𝐿2− 1

𝐿𝐿𝐶𝐶 = b; then

→ There are three conditions based on the value of to

∗ If 𝑅𝑅2

4𝐿𝐿2 > 1

𝐿𝐿𝐶𝐶 ,b is real

∗ If 𝑅𝑅2

4𝐿𝐿2 = 1

𝐿𝐿𝐶𝐶 ,b is zero

∗ If 𝑅𝑅2

4𝐿𝐿2 < 1

𝐿𝐿𝐶𝐶 ,b is imaginary

Case I: when b is real

→ i(t) = 𝑉𝑉

2�𝑅𝑅2

4𝐿𝐿2−1𝐿𝐿𝑅𝑅 .𝐿𝐿

�𝑖𝑖𝑒𝑒𝑝𝑝 �−� 𝑅𝑅2𝐿𝐿

+ �𝑅𝑅2

4𝐿𝐿2− 1

𝐿𝐿𝐶𝐶�+� − 𝑖𝑖𝑒𝑒𝑝𝑝 �−� 𝑅𝑅

2𝐿𝐿− �𝑅𝑅2

4𝐿𝐿2− 1

𝐿𝐿𝐶𝐶� 𝑖𝑖� �

Case II: when b= 0

The expression for current becomes

→ i(t)= 𝑉𝑉2𝑏𝑏𝐿𝐿

{𝑖𝑖−𝑎𝑎𝑖𝑖 − 𝑖𝑖−𝑎𝑎𝑖𝑖} which is indeterminate.

→ Now at b=0 i (t) = 𝑉𝑉

𝐿𝐿 𝑖𝑖 𝑖𝑖−𝑎𝑎𝑖𝑖=𝑉𝑉𝑖𝑖𝑒𝑒

𝐿𝐿− (𝑅𝑅/2𝐿𝐿)𝑖𝑖

Case III. When b is imaginary

→ i (t) = 𝑉𝑉 2𝑏𝑏𝐿𝐿

�𝑖𝑖−𝑎𝑎𝑖𝑖. 𝑖𝑖𝑗𝑗𝑗𝑗𝑖𝑖 − 𝑖𝑖−𝑎𝑎𝑖𝑖. 𝑖𝑖−𝑗𝑗𝑗𝑗𝑖𝑖� = 𝑉𝑉2𝑏𝑏𝐿𝐿

𝑖𝑖−𝑎𝑎𝑖𝑖. 2𝑝𝑝𝑖𝑖𝑖𝑖𝑠𝑠𝑖𝑖

= 𝑉𝑉

2𝐿𝐿� 𝑅𝑅2

4𝐿𝐿2−1𝐿𝐿𝑅𝑅

𝑖𝑖−𝑎𝑎𝑖𝑖 .2 sin��−𝑅𝑅2

4𝐿𝐿2+ 1

𝐿𝐿𝐶𝐶� t

A.C source:

→ R-L circuit: when switch is is closed, the current in the circuit is given by


Pg.No. 85 of 143






I (s) = 𝑉𝑉(𝑆𝑆)𝑍𝑍(𝑆𝑆)

= 𝑉𝑉𝑚𝑚 �𝜔𝜔 𝑐𝑐𝑜𝑜𝑠𝑠𝑐𝑐𝑆𝑆2+𝜔𝜔2

+ 𝑆𝑆 𝑠𝑠𝑖𝑖𝑖𝑖𝑐𝑐𝑆𝑆2+𝜔𝜔2

� 1𝑅𝑅+𝐿𝐿𝑆𝑆

= 𝑉𝑉𝑚𝑚𝐿𝐿�𝜔𝜔 𝑐𝑐𝑜𝑜𝑠𝑠𝑐𝑐𝑆𝑆2+𝜔𝜔2

+ 𝑆𝑆 𝑠𝑠𝑖𝑖𝑖𝑖𝑐𝑐𝑆𝑆2+𝜔𝜔2

� 1𝑆𝑆+𝑅𝑅/𝐿𝐿

→ R-L circuit connected to an ac source Let 𝑅𝑅

𝐿𝐿= 𝑖𝑖; 𝑖𝑖ℎ𝑖𝑖𝑖𝑖

I(S) = 𝑉𝑉𝑚𝑚𝐿𝐿� 𝜔𝜔 𝑐𝑐𝑜𝑜𝑠𝑠𝑐𝑐

(𝑠𝑠+𝑎𝑎)(𝑆𝑆2+𝜔𝜔2 )+ 𝑆𝑆 𝑠𝑠𝑖𝑖𝑖𝑖𝑐𝑐

(𝑠𝑠+𝑎𝑎)(𝑆𝑆2+𝜔𝜔2) �

i(t) = 𝑉𝑉𝑚𝑚

�(𝑅𝑅2+𝜔𝜔2𝐿𝐿2)12�

{sin(𝜔𝜔𝑖𝑖 + 𝜑𝜑 − 𝜃𝜃)− sin (𝜑𝜑 − 𝜃𝜃)𝑖𝑖−𝑎𝑎𝑖𝑖}

Where θ= 𝑖𝑖𝑖𝑖𝑖𝑖−1 𝜔𝜔𝐿𝐿𝑅𝑅

Circuit Breaker ratings :

→ The value of resistor required to be connected across the breaker contacts which will

give no transient oscillations, is R= 0.5�𝐿𝐿𝐶𝐶

Where L,C are the inductance and capacitance upto the circuit breaker

→ The average RRRV = 2Vrπ√Lc

→ Maximum value of RRRV = wnEpeak

→ Where wn = 2πfn ,

→ Natural frequency of oscillations, fn = 12π� 1LC

Where L , C are the reactance and capacitance up to the location of circuit breaker

→ Frequency of demand oscillations, f = 12π� 1LC− 1

4R2C2


Pg.No. 86 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet Breaking capacity:

→ Symmetrical breaking current = r.m.s value of a.c component = x

√2

→ Asymmetrical breaking current = r.m.s value of total current.

= �� 𝐼𝐼√2�2

+ 𝑌𝑌2

Where X = maximum value of a.c component Y = d.c component

→ Is the rated service line voltage in volts, then for 3-phae circuit? Breaking capacity = √3 × V × I × 10−6 MVA

→ String efficiency = Voltage across the stringn × voltage acrosss the unit near power conductor

Where, n = no. of insulators Making capacity :-

→ Making capacity = 2.55 × symmetrical breaking capacity. The Universal Relay Torque Equation:-

→ The universal relay torque equation is given as follows T = K1I2 + K2V2 + K3 VI(θ − τ) + K Distance Relays:

impedance relays :

From the universal torque equation putting 𝐾𝐾3 = 0 and giving negative sign to voltage term, it becomes

→ T = K1I2 − K2V2 (Neglecting spring torque) For the operation of the relay the operating toque should be greater than the restraining torque i.e K1I2 > K2V2

→ Here V and I are the voltage and current quantities fed to the relay.

→ V2

I2 < K1

K2�

→ Z < �K1K2�


Pg.No. 87 of 143





Institute of Engineering Studies (IES,Bangalore) Power Systems Formula Sheet → Z < constant (design impedance) This means that the impedance relay will operate only if the impedance seen by the relay is less than a pre-specified value (design impedance). At threshold condition,

Z = �K1K2�

Reactance Relay: The directional element is so designed that its maximum torque angle is 900 i.e. in the universal torque equation. T = K1I2 − K3VI cos(θ − τ) = K1I2 − K3VI cos(θ − 90) = K1I2 − K3VI sinθ For the operation of the relay KI2 > K3 VI sinθ VI

I2 sinθ < K1/K3

Z sinθ < K1K3�

X < K1K3�

The mho relay:-

→ In the relay the operating torque is obtained by the V – I element and restraining torque due tot the voltage elemen

T = K3VI cos(θ − τ) − K2V2 → For relay to operate

K3VI cos (θ − τ) > K2V2

V2

V I< K3

K2� cos (θ − τ)

Z < K3K2� cos (θ − τ)


Pg.No. 88 of 143







Pg.No. 89 of 143





Institute of Engineering Studies (IES,Bangalore) Control Systems Formula Sheet Control Systems

Time Response of 2nd order system :- Step i/P :

• C(t) = 1- e− ζωnt

�1−ζ2 (sin ωn �1 − ζ2 t ± tan−1 ��1−ζ2

ζ� )

• e(t) = e− ζωnt

�1−ζ2 �sin 𝜔𝜔𝑑𝑑𝑡𝑡 ± tan−1 ��1−ζ2

ζ��

• ess = lim𝑡𝑡→∞

e− ζωnt

�1−ζ2 �sin 𝜔𝜔𝑑𝑑𝑡𝑡 ± tan−1 ��1−ζ2

ζ��

→ ζ → Damping ratio ; ζωn → Damping factor ζ < 1(Under damped ) :-

C(t) = 1- = e− ζωnt

�1−ζ2 Sin �𝜔𝜔𝑑𝑑𝑡𝑡 ± tan−1 ��1−ζ2

ζ��

ζ = 0 (un damped) :- c(t) = 1- cos ωnt ζ = 1 (Critically damped ) :- C(t) = 1 - e−ωnt(1 + ωnt) ζ > 1 (over damped) :-

C(t) = 1 - e−�𝛇𝛇− �𝛇𝛇𝟐𝟐−𝟏𝟏� 𝛚𝛚𝐧𝐧𝐭𝐭

2 �𝛇𝛇𝟐𝟐−𝟏𝟏 �𝛇𝛇− �𝛇𝛇𝟐𝟐−𝟏𝟏�

T = 1

�𝛇𝛇− �𝛇𝛇𝟐𝟐−𝟏𝟏�ωn

Tundamped > Toverdamped > Tunderdamped > Tcriticaldamp Time Domain Specifications :-

• Rise time tr = π−∅ωn�1−ζ2 ∅ = tan−1 ��1−ζ2

ζ�

• Peak time tp = nπωd

• Max over shoot % Mp = e−ζωn/�1−ζ2 × 100 • Settling time ts = 3T 5% tolerance


Pg.No. 90 of 143





Institute of Engineering Studies (IES,Bangalore) Control Systems Formula Sheet = 4T 2% tolerance

• Delay time td = 1+0.7ζωn

• Damping factor2 ζ2 = (ln Mp)2

π2+ (ln Mp)2

• Time period of oscillations T = 2πωd

• No of oscillations = ts2π/ωd

= ts×ωd2π

• tr ≈ 1.5 td tr = 2.2 T • Resonant peak Mr = 1

2ζ�1−ζ2 ; ωr = ωn �1 − 2ζ2 𝜔𝜔𝑛𝑛> 𝜔𝜔𝑟𝑟𝜔𝜔𝑏𝑏>𝜔𝜔𝑛𝑛

� ωr < ωn < ωb

• Bandwidth ωb = ωn (1 − 2ζ2 + �4𝜁𝜁4 − 4𝜁𝜁2 + 2)1/2 Static error coefficients :- • Step i/p : ess = lim

t→∞𝑒𝑒(𝑡𝑡) = lim

𝑠𝑠→0𝑠𝑠 𝐸𝐸(𝑠𝑠) = lim

𝑠𝑠→0𝑆𝑆𝑆𝑆(𝑠𝑠)1+𝐺𝐺𝐺𝐺

ess = 1

1+KP (positional error) Kp = lim

𝑠𝑠→0𝐺𝐺(𝑠𝑠) 𝐻𝐻(𝑠𝑠)

• Ramp i/p (t) : ess = 1

Kv Kv = lim

𝑠𝑠→0𝑆𝑆 𝐺𝐺(𝑠𝑠)𝐻𝐻(𝑠𝑠)

• Parabolic i/p (t2/2) : ess = 1/ Ka Ka = lim

𝑠𝑠→0s2 𝐺𝐺(𝑠𝑠)𝐻𝐻(𝑠𝑠)

Type < i/p → ess = ∞ Type = i/p → ess finite Type > i/p → ess = 0

• Sensitivity S = ∂A/A

∂K/K sensitivity of A w.r.to K.

• Sensitivity of over all T/F w.r.t forward path T/F G(s) : Open loop: S =1 Closed loop : S = 1

1+G(s)H(s)

• Minimum ‘S’ value preferable • Sensitivity of over all T/F w.r.t feedback T/F H(s) : S = G(s)H(s)

1+G(s)H(s)

Stability RH Criterion :- • Take characteristic equation 1+ G(s) H(s) = 0 • All coefficients should have same sign • There should not be missing ‘s’ term . Term missed means presence of at least one +ve real part root


Pg.No. 91 of 143





Institute of Engineering Studies (IES,Bangalore) Control Systems Formula Sheet • If char. Equation contains either only odd/even terms indicates roots have no real part & posses only

imag parts there fore sustained oscillations in response. • Row of all zeroes occur if

(a) Equation has at least one pair of real roots with equal image but opposite sign (b) has one or more pair of imaginary roots (c) has pair of complex conjugate roots forming symmetry about origin.


Pg.No. 92 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet Measurements:

Derived Units:

Area m2 L2 MMF A I Volume m3 L3 Frequency Hz T−1 Density Kg/ m3 L−3M Velocity m/sec LT−1 Angular Velocity

rad / sec [L]0T−1 Acceleration m/sec2 LT−2

Angular Acceleration

rad / sec2 [L]0 T−2 Force Kg m/sec2 LMT−2

Pressure, stress

Kg/m/sec2 L−1MT−2 Luminous flux lm(cd Sr)

Luminance cd/m2 Illumination lm/m2 Work , Energy

Joule (Nm)

L2MT−2 Power Watt (J/sec)

L2MT−3

Charge Coulomb TI EMF Volt (W/A) L2MT−3I−1 Electric field strength

V/m LMT−3I−1 Resistance V / A L2MT−3I2

Capacitance (AS/v) L−2M−1T4I2 Magnetic flux Vs L2MT−2I−1 Magnetic flux density

Wb / m2 MT−2I−1 Inductance Vs / A L2MT−2I2

Static error:-

Static error is defined as the difference b/w the measured value and the true value of

the quantity.

δA = ϵ0 = Am − At where Am = measured value of quantity

At = True value of quantity δA = Absolute static error Relative static error (ϵr) = Absolute error

True value

= ∂AAt

= ϵ0At

Percentage static error (% ϵ0) = ϵr × 100 True value At = Am

1+ϵr


Pg.No. 93 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet Static correction :- It is the difference between the true value and the measured value of the quantity

δc = Am − At = - δA

Sensitivity :

It is the ratio of infinitesimal change in output to an infinitesimal change in input.

Static sensitivity = Infinitesimal change in output

Infinitesimal change in input

Inverse sensitivity (or) deflection factor = Infinitesimal change in input

Infinitesimal change in output

Linearity:

A measurement system is considered to be linear if the output is linearly proportional to the input .

Non-linearity = (maximum deviation of output from the idealized straight lineactual reading

× 100

Non-linearity = (maximum deviation of output from the idealized straight line)full scale deflection

× 100

The magnitude of a quantity having a nominal value As and a maximum Aa error or limiting error of ± δA must have a magnitude Aa between the limits As − δA and As ± δA

Actual value of quantity Aa = As + δA Relative limiting error , ϵr = ϵ0 / As

Aa = As [ 1 ± ϵr ]

Magnitude of limiting error = full scale reading × guaranteed accuracy

Mean : Arithmetic Mean = 𝑋𝑋� = ∑ Xini=1

n

The root mean square value is the standard deviation σ .


Pg.No. 94 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet

S.D , σ = �d12 + d2

2 +⋯dn2

n = �∑ di

2ni=1

n

Where d1 = xi − x, d2 = x2 − x, etc the deviation from the mean of the individual readings

If the number of readings is less than 20, then S.D, σ = �∑ di2n

i=1n−1

Variance = σ2

Moving Coil Permanent Magnet Type:

The deflecting torque Td is given by Td= NBI l.b = NBA.I

Where N is the number of turns of the coil, B is air gap flux density , wb/m2. I is current in the coil, ampere. L is active length of conductor , metre and b is width of coil, metre

The controlling torque , TC is given by TC = Sθ Where S is spring constant and θ is the angle of deflection.

When steady deflection is reached , Td= Tc and Hence θ ∝ I.

Moving Coil Dynamometer Type :

The torque developed in a dynamometer type ammeter or voltmeter is given by Td= I2(dM/Dθ)

Where I is the current through fixed coil and moving coil (connected in series) and

M is the mutual inductance between fixed and moving coils.

Moving Iron Instruments :

The deflecting torque is given by Td= (1/2) I2(dL/dθ).

Where I is the current through the coil and

L is the inductance.


Pg.No. 95 of 143






Linearization of scale:

θ.(dL/dθ)=K.

Compensation towards frequency errors can be done by connecting a capacitor across a part of series resistance in MI voltmeter C = 0.41 × (L/R2).

Half wave Rectifier type Instruments:

d.c supply → Im = VdcRm+RL

a.c supply → V = Vrms voltage of source.

Vdc = (Vm/π)

Im= (√2V) / (π(RL + Rm)) = (0.45 V)/(Rm + RL)

= (√2 V)/(π)

In half wave rectifier type instrument , the sensitivity of a.c is 0.45 times that of d.c.

Full wave rectifier type instruments:

d.c. supply → Im= VRm+RL

a.c supply → Im= 2Vmπ(Rm+RL)

= 0.9 VRm+RL

In full wave rectifier type instrument , the sensitivity of a.c is 0.9 times that of d.c.

Thermal instruments :


Pg.No. 96 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet Hot wire type:

The change in the length of the wire due to temperature is given

∆L = [(RLα)/(UA)]I2 (or) ∆L = [(Lα)/(UAR)]V2 Where L in the original length of the wire,

α is the temperature coefficient of linear expansion of wire material,

R is the resistance of wire,

U is the coefficient of heat dissipation,

A is the area of heat dissipating surface.

Thermo couple instruments

The e.m.f generated in thermocouple is given by E = a(∆T) + b(∆T)2

Where ∆T is the temperature difference between hot and cold junctions

‘a’ is a constant (40 to 50 μv/0C )

‘b’ is a constant (1 to 2 μv / 0C)

1. Kelvin’s Double Bridge is used for the measurement of low resistance

Vad = Vam+Vmd =I2R + I2 rp+q+r

P

R= S.PQ

qrp+q+r

�𝑃𝑃𝑄𝑄

− 𝑝𝑝𝑞𝑞

� 2.Measurement of Medium Resistance:

(a)Voltmeter-Ammeter method:

Measured value of resistance, Rml =VR+Va

IR

Where R is the true value of the resistance.

Error = Ra % Error = (Ra/R) In this method, always the measured value of resistance is greater than true value

of resistance.


Pg.No. 97 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet (b)Ammeter-Voltmeter Method:

Measured value of resistance, Rm2 = VRI

Where I= IR = IV= VRR1

+ VRRv

Which gives Rm2 = R

1+ RRv

Error = Rm2 -R = Rm2R

Rv

% error = −Rm2Rv

≅ −RRv

In this method, the measured value of resistance is always less than the true value

of resistance.

This method is suitable for measurement of low resistance among the range.

The resistance where both the methods give same error is obtained by equating the two errors.

RaR

= RRv

,

R = �RaRv Wheatstone Bridge:

Balanced condition → PQ

= RS

QP

= SR

PP+Q

= RR+S

Sensitivity of the galvanometer, Sv = θe

Where θ = deflection of the galvanometer

e = emf across galvanometer

= Vb - Vd

= (E - Vab )- (E - Vad )

= Vad - Vab

= (R+∆R)R+∆R+S

E − E.PP+Q


Pg.No. 98 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet e = E. S∆∆

(R+S)2

∴ Sensitivity of galvanometer, Sv = (R+S)2

E.S∆. .θ

Or θ = Sv E.S∆R(R+S)2 .

Sensitivity of the Bridge = SB = θ∆RR

SB= Sv. ERS+2+S

R

= Sv = EPQ +2+ QP

SBmax = Sv.E4

When PQ

= RS

= 1

Bridge sensitivity is useful for

(a) Selecting the galvanometer with which the given unbalance can be observed. (b) Determining the deflection to be expected for a given unbalance.

Measurement of High Resistance:

(a) Loss of charge method :

At t, voltage across capacitor, v = Ve−t/RC R = −t

2.303C log (V/v)

R = 0.434 t

C log10 (V/v)


Pg.No. 99 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet The general ac bridge circuit

At balance, the Pd across BA = Pd across BC in magnitude and phase

EBA = EBC = I1Z1 = I2 Z2

Z1Z4∠(Q1+Q4) = Z2Z3∠(Q2+Q3)

Equating the magnitudes and angles

Z1Z4=Z2Z3

∠(Q1+Q4) = ∠(Q2+Q3)

Measurement of Self Inductance:-

The following bridges are used for measurement of self inductance.

1. Maxwell’s Inductance Bridge 2. Maxwell’s Inductance- Capacitance Bridge 3. Hay’s Bridge 4. Owen’s Bridge 5. Anderson’s Bridge

1.Maxwell’s Inductance Bridge:-

This bridge circuit measures an inductance by comparison with a variable standard

self-inductance.

At balance condition,

L1 = R3L2/R4 ; R1 = R3(R2+r2)/R4

Where L1 = unknown inductance of resistance R1

L2 = variable inductance of fixed resistance r2

R2 = variable resistance connected in series with inductor L2

R3, R4 = known non-inductive resistance


Pg.No. 100 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet 2.Maxwell’s Inductance-Capacitance Bridge:

In this bridge, an inductance is measured by comparison with a standard

variable capacitance.

At balance conditions,

R1 = R2R3/R4 ; L1= R2R3C4

Q-factor, Q= wL1/R1 =wC4R4

This bridge is used for measurement of low Q coils (1< Q<10).

3.Hay’s Bridge:-

This is a modification of Maxwell’s bridge. It is used for measurement of high Q coils.

L1 = unknown inductance having a resistance

R1, R2, R3, R4 = known non-inductive resistance

C4= standard capacitor


L1= R2R3C41+W2+C4

2+R42 ; R1 = ω2R2R3R4C4

2

1+ω2C42 R4

2

Q=ωL1/R1 = 1/ωC4R4

4.Anderson’s Bridge:-

In this method, the self-inductance is measured in terms of a standard capacitor..



Pg.No. 101 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet R1=(R2R3/R4) − r1 ; L1 = CR3[r(R4 + R2) + R2R4]/R4

5.Owen’s Bridge:-

This bridge may be used for the measurement of an inductance in terms of capacitance.

Under balance conditions,

L1= R2R3C4 R1= R3 C4

C2

Measurement of Capacitance:-

The following bridges are used for the measurement of capacitance:

1. De Sauty’s Bridge 2. Modified De Sauty’s Bridge 3. Schering Bridge.

1.De Sauty’s Bridge :- This bridge is the simplest method of comparing two capacitances.

At balance condition, C1=C2. R4/R3

2.Modified De Sauty’s Bridge:-

At balance condition, C1C2

= R2+r2R1+r1

=R4R3

Dissipation factor, D1= tan δ1= ω c1

D2 = tan δ2 = ω c2r2

D1- D2= ω c2 �𝑅𝑅1𝑅𝑅4𝑅𝑅3

− 𝑅𝑅2� 3.Schering’s Bridge :-

At balance conditions, r1= R3C4/C2 ;


Pg.No. 102 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet C1 = C2(R4/R3)

Dissipation factor, D1 = tan δ1= ω c1r1 = ωc4 r4

Measurement of Mutual Inductance:

The following bridges are used for the measurement of mutual inductance:

(1). Heaviside Mutual Inductance Bridge

(2). Campbell’s modification of Heaviside’s bridge

(3). Heaviside Campbell equal ratio bridge (4). Carry Foster Bridge/ Heydweiler Bridge →

Ammeter Shunts:

These are small resistances in parallel to basic meter to increase current

measuring capacity

Im Rm = Ish Rsh

= Rmm−1

Where m is the multiplying power of ammeter

Series Multipliers:

These are used for increasing the voltage measuring capacity of basic met

VRse+ Rm

= VRm

Rse = Rm (m-1) Where m is the multiplying power of voltmeter.

Multi Range Ammeter: A range of current settings can be obtained using different shunts.


Pg.No. 103 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet Rsh1 = Rm

m1−1 , m1= I1

Im

Rsh2 = Rm

m2−1 , m2= I2

Im

Rsh3 = Rmm3−1

, m3= I3Im

4.Universal Shunt (or) Ayrton shunt:

Im Rm = (I1- Im) R1

R1= Rmm1−1

, m1 = I1Im

R2= Rmm2−1

, m2 = I2Im

R3= Rm

m3−1 , m3 = I3

Im

MEASUREMENT OF POWER & ENERGY:

Power in D.C. Circuits

Power measured = Pm1= VRIR + I2R Ra

True value = Measured power – power loss in ammeter

Power measured = Pm2 = VRIR + (V2R /Rv )


Pg.No. 104 of 143






True power = Measured power- Power loss in voltmeter

Power in A.C. Circuits:

Value of power but in its average value over a cycle. Instantaneous power = VI

Average power = VI cos∅

Where V and I are r.m.s values of voltage and current and cos ∅ is the power factor of the load.

3.Electro Dynamometer Wattmeter:

The deflecting torque in electrodynamometer instruments is given by , Td = i1i2 (dM/dθ)

Where i1 and i2 are instantaneous values of currents in two coils.

Many watt meters are compensated for errors caused by inductance of pressure coil by means of a capacitor connected in parallel with a portion of multiplier.

Capacitance C = (L/r2)

Low Power Factor Wattmeter:

In present case of wattmeter , i1 is the load current and is the current flowing through pressure coil.

Td = iP iC (dM/dθ)

Average deflection torque = IPI cos ∅ (dM/dθ)

= (V/RP).I cos ∅ (dM/dθ)

Td ∝ VI cos ∅ (dM/dθ)

At balance condition, Td = TC K1VI cos ∅ (dM/dθ) = kθ θ ∝ VI cos ∅


Pg.No. 105 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet ∝ power if (dM/dθ) is constant.

Errors in Electro Dynamometer Wattmeter:

In general , pressure coil offers inductance, in addition to resistance. Due to this inductance, wattmeter reads more power on lagging loads and less power in case of leading loads.

Correction is to be made for wattmeter reading to get true value of power.

True power for lagging pf loads = cos ∅cos β ∗cos(∅−β) × actual wattmeter reading

True power for leading pf loads = cos ∅cos β ∗cos(∅+β) × actual wattmeter reading

ERROR = tan ∅ tan β × true power = VI sin ∅ tan β ∅ = pf angle

β = tan−1 (XP/RP)

ERROR = tan ∅ tan β β→ is the angle between PC current and voltage.

Measurement Of Power in Three circuits:

(a)Three watt meter Method:

In the three wattmeter method to determine the power in 3-∅, 4 wire system.

Sum of the instantaneous readings of watt meters = P= P1+P2+P3

P= V1i1 +V2i2+V3i3

Instantaneous power of load = V1i1 +V2i2+V3i3

Hence the summation of readings of three watt meters gives the total power of load.


Pg.No. 106 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet (b)Two wattmeter Method:

In a 3-∅, three wire system we required 3 elements. But if we make the common points of pressure coils coincide with one of the lines, then we will require only (n-1 l =2) elements.

Instantaneous reading of P1 wattmeter (P1) = i1(V1- V3)

Instantaneous reading of P2 wattmeter (P2) = i2(V2- V3)

Sum of instantaneous reading of two watt meters = P1+P2

From Kirchoff’s current law, i1+i2+i3 =0

P1+P2 = i1(V1- V3) + i2(V2- V3)

= V1 i1 - V3 i1 + V2 i2 - V3i2

= V1 i1 + V2 i2 - V3(i1+i2)

= V1 i1 + V2 i2 - V3 (-i3)

= V1 i1 + V2 i2 + V3 i3

Hence, the sum of two watt meter readings is equal to power consumed by load.

Let the load be balanced, V1, V2 , V3 be the rms value of phase voltage and I1 , I2, I3 be the rms values of phase currents.

For star connection, Phase voltages V1 = V2 = V3 = V (say)

Line voltages V13 = V23 = V12 = √3 V


Pg.No. 107 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet Phase currents I1 = I2 = I3 = I (say)

Line currents I1 = I2 = I3 = I

Power factor cos ∅ (lag)

In general , reading of wattmeter is given by

Wattmeter reading = current through wattmeter current coil * voltage across its pressure coil * cos( phase angle between this current and voltage).

∴ Reading of P1 wattmeter = V13 I1 cos(30- ∅) = √3 VI cos(30-∅)

Reading of P2 wattmeter = V23 I2 cos(30+ ∅) = √3 VI cos(30+∅)

Sum of reading of two watt meters = P= P1 +P2 = √3 VI cos (30-∅) + √3 VI cos (30+∅)

= √3 VI cos ∅

This is the total power consumed by load.

Difference of reading of two watt meters = P1 - P2 = √3 VI [cos (30-∅) - cos (30+∅)]

= √3 VI *2* sin 30 * sin ∅

= √3 VI sin ∅

∴ Reactive power consumed by load = √3 (Difference of two wattmeter readings).

= √3 (P1 - P2 )

Power factor cos ∅ = cos tan−1 �√3 (P1− P2)P1+P2

�

7.Effect Of Power Factor On the Readings of Watt Meter:


Pg.No. 108 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet With unity power factor, cos ∅ =1 (or) ∅ = 00 P1 = P2 = √3 VI cos (00) = (3/2)VI

The readings of two watt meters are equal , each watt meter reads half of the total power.

When cos ∅ = 0.5, ∅ =60 P1 = (3/2) VI, P2=0 When power factor is 0.5 , one of the watt meter reads zero and the other reads total power.

When cos ∅ =0 , ∅= 900 P1 = �(3/2) V I , P2 = - �(3/2) V I Therefore , width zero power factor, the readings of two meters are equal but of opposite sign.

8.Measurement of Reactive Power in Three Phase Circuits:

Reading of wattmeter = v23 i1 cos (angle between i1 and v23 ) = v23 i1 cos (90- ∅)

= √3 VI sin ∅

Total reactive power of the circuit = √3 (watt meter reading).

Measurement of Energy:

Energy is the total power delivered or consumed over a time interval, that is. Energy = Power × time

Electrical energy developed as work or dissipated as heat over an interval of time t may be expressed as : W= ∫ VI dtt

0


Pg.No. 109 of 143






Construction and Principle of Single Phase Induction Energy Meter:

Net driving torque is given as, Td= k1 ∅1 ∅2 (f/z) sin β cos α Where β = Phase between fluxes ∅1 and ∅2

α= Phase angle of eddy current paths.

These two fluxes will be produced by two currents which are described earlier.

At steady speed the driving torque must equal to the breaking torque. k4N= k3 VI sin (∆-∅)

If ∆ = 900 , sped, N = K VI sin(90- ∅)

= K VI cos ∅

= k x(power)

Total number of revolutions = ∫ N dt = k ∫ VI sin(∆ − ∅) 𝑑𝑑𝑑𝑑 If ∆= 900, total number of revolutions = k ∫ VI cos ∅ dt

= k * (energy)

Single Phase Electrodynamometer Power Factor Meter

The values of R and L are so adjusted that the two coils carry the same value of current

at normal frequency (i.e) R = ωL

Deflecting torque acting on coil is: TA = KVI Mmax cos ∅ sin θ Where θ= angular deflection from the plane of reference.

Mmax = maximum value of mutual inductance between the two coils.

This torque say acts in the clockwise direction.

Deflecting torque acting on coil B is: TB = KVI Mmax cos (900 -∅) sin (900+θ) = KVI Mmax sin ∅ cos θ

This torque acts in anticlockwise direction.

At equilibrium , TA = TB ⇒ θ = ∅ Therefore the deflection of the instrument is a measure of phase angle of the circuit.


Pg.No. 110 of 143






Q-Meter:

Q Meters are intended to measure the Q (quality factor) of an inductance or capacitor.

Q = ωLR

= 1ωCR

= IXLIR

= 𝐈𝐈𝐗𝐗𝐂𝐂𝐈𝐈𝐈𝐈

= 𝐕𝐕𝐂𝐂 𝐨𝐨𝐨𝐨 𝐕𝐕𝐚𝐚𝐕𝐕𝐚𝐚

So the voltage across the capacitor or coil is Q times the applied voltage

𝑎𝑎) 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑺𝑺𝑪𝑪𝑪𝑪𝑺𝑺𝑪𝑪𝑪𝑪:

In series connection method the unknown value component is connected in series with the resonant circuit. This method is employed for measurement of low value resistors, small coils and large capacitors.

𝐴𝐴𝑑𝑑 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑎𝑎𝑟𝑟𝑟𝑟𝑟𝑟, 𝜔𝜔𝜔𝜔 = 1

𝜔𝜔𝐶𝐶1 𝑎𝑎𝑟𝑟𝑑𝑑 𝑄𝑄1 =

𝜔𝜔𝜔𝜔𝑅𝑅

= 1

𝜔𝜔𝐶𝐶1𝑅𝑅

𝑋𝑋𝑠𝑠 = 𝐶𝐶1 − 𝐶𝐶2

𝜔𝜔𝐶𝐶1𝐶𝐶2

𝑅𝑅𝑠𝑠 = 𝑄𝑄1𝐶𝐶1 − 𝑄𝑄2𝐶𝐶2

𝜔𝜔𝐶𝐶1𝐶𝐶2 𝑄𝑄1𝑄𝑄2

𝑄𝑄𝑥𝑥 = 𝑋𝑋𝑠𝑠

𝑅𝑅𝑠𝑠 =

(𝐶𝐶1 − 𝐶𝐶2)𝑄𝑄1𝑄𝑄2

𝑄𝑄1𝐶𝐶1 − 𝑄𝑄2𝐶𝐶2

b) 𝑷𝑷𝑷𝑷𝑺𝑺𝑷𝑷𝑷𝑷𝑷𝑷𝑺𝑺𝑷𝑷 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑺𝑺𝑪𝑪𝑪𝑪𝑺𝑺𝑪𝑪𝑪𝑪:


Pg.No. 111 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet 𝐼𝐼𝑟𝑟 𝑝𝑝𝑎𝑎𝑟𝑟𝑎𝑎𝑝𝑝𝑝𝑝𝑟𝑟𝑝𝑝 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑑𝑑𝑐𝑐𝑟𝑟𝑟𝑟 , 𝑑𝑑ℎ𝑟𝑟 𝑢𝑢𝑟𝑟𝑢𝑢𝑟𝑟𝑟𝑟𝑢𝑢𝑟𝑟 𝑟𝑟𝑟𝑟𝑐𝑐𝑝𝑝𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑑𝑑 𝑐𝑐𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑑𝑑𝑟𝑟𝑑𝑑 𝑐𝑐𝑟𝑟 𝑝𝑝𝑎𝑎𝑟𝑟𝑎𝑎𝑝𝑝𝑝𝑝𝑟𝑟𝑝𝑝 𝑑𝑑𝑟𝑟 𝑑𝑑ℎ𝑟𝑟 𝑟𝑟𝑎𝑎𝑝𝑝𝑎𝑎𝑟𝑟𝑐𝑐𝑑𝑑𝑟𝑟𝑟𝑟 , 𝑐𝑐𝑟𝑟 𝑑𝑑ℎ𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑐𝑐𝑟𝑟𝑟𝑟 . 𝑇𝑇ℎ𝑐𝑐𝑟𝑟 𝑐𝑐𝑟𝑟𝑑𝑑ℎ𝑟𝑟𝑑𝑑 𝑐𝑐𝑟𝑟 𝑟𝑟𝑐𝑐𝑝𝑝𝑝𝑝𝑟𝑟𝑒𝑒𝑟𝑟𝑑𝑑 𝑓𝑓𝑟𝑟𝑟𝑟 𝑐𝑐𝑟𝑟𝑎𝑎𝑟𝑟𝑢𝑢𝑟𝑟𝑟𝑟𝑐𝑐𝑟𝑟𝑟𝑟𝑑𝑑 𝑟𝑟𝑓𝑓 ℎ𝑐𝑐𝑖𝑖ℎ 𝑣𝑣𝑎𝑎𝑝𝑝𝑢𝑢𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑐𝑐𝑟𝑟𝑑𝑑𝑟𝑟𝑟𝑟𝑟𝑟, 𝑟𝑟𝑟𝑟𝑟𝑟𝑑𝑑𝑎𝑎𝑐𝑐𝑟𝑟 𝑐𝑐𝑟𝑟𝑑𝑑𝑢𝑢𝑟𝑟𝑑𝑑𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑟𝑟𝑑𝑑 𝑟𝑟𝑐𝑐𝑎𝑎𝑝𝑝𝑝𝑝 𝑟𝑟𝑎𝑎𝑝𝑝𝑎𝑎𝑟𝑟𝑐𝑐𝑑𝑑

1RP

= ωC1Q2

= 1RQ1

XP= 1ω(C2−C1)

Qx = Rp

Xp = (C1−C2)Q1Q2

Q1C1−Q2C2

The main error in the measurement of Q is due to distributed or stray capacitance of the circuitry. To check for this, the Q is measured at two frequencies f1 and 2f1. It should be same if not,

Cd = C1−C1C23

Cathode Ray Tube:

For electrostatic deflection, D= LldEd2dEa

Where D-Deflection, m

L – distance from centre of deflection plates to screen, m

ld- effective length of deflection plates, m

Ed- deflection voltage, volts

d- separation between the plates, m

Ea- accelerating voltage, volts

Deflection sensitivity is S = DEd

= Lld2dEa

, m/V


Pg.No. 112 of 143






Deflection factor G is G= 1S = 2dEa

Lld , V/m

Magnetic Deflection D = lLB�Va

� e2m

, m

l- width of magnetic coil, m

L- length from centre of 1 to screen, m

B- magnetic flux density, wb/m2

e, m – charge and mass of electron

Va- acceleration potential

Magnetic Deflection Sensitivity is DB

= IL �e

2mVa , m

Oscilloscope Specifications:

1.Sensitivity:

It means the vertical sensitivity . It refers to smallest deflection factor G = (1/S) and expressed as mv/div. The alternator of the vertical amplifier is calibrated in mv/div.

2.Band width:

It is the range of frequencies between ± 3 dB of centre frequency.

3.Rise Time:

Rise time is the time taken by the pulse to rise from 10% to 90% of its amplitude.

BW = 12πRC

BW = band width in MHz

90% of amplitude is normally reached in 2.2 RC or 2.2 time constants


Pg.No. 113 of 143





Institute of Engineering Studies (IES,Bangalore) Measurements Formula Sheet BW = 1

2πRC = 2.2

2πRC = 0.35

Tr , Tr = rise time in μ seconds.

Synchronization means the frequency of vertical signal input is an integral multiple of the sweep frequency. Fin = n fs


Pg.No. 114 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet EDC & Analog

• Energy gap 𝐸𝐸G/si=1.21− 3.6 × 10−4.T ev𝐸𝐸G/Ge=0.785− 2.23 × 10−4.T ev� Energy gap depending on temperature

• EF = EC - KT ln�𝑁𝑁𝐶𝐶𝑁𝑁𝐷𝐷

� = Ev + KT ln �𝑁𝑁𝑣𝑣𝑁𝑁𝐴𝐴

�

• No. of electrons n = Nc e−(Ec−Ef)/RT (KT in ev) • No. of holes p = Nv e−(Ef−Ev)/RT • Mass action law np = ni

2 = NcNv e−EG/KT • Drift velocity 𝑣𝑣d = μE (for si 𝑣𝑣d ≤ 107 cm/sec) • Hall voltage 𝑣𝑣H = B.I

we . Hall coefficient RH = 1/ρ . ρ → charge density = qN0 = ne …

• Conductivity σ = ρμ ; μ = σRH . • Max value of electric field @ junction E0 = - q

ϵsi Nd. nn0 = - q

ϵsi NA. np0 .

• Charge storage @ junction Q+ = - Q− = qA xn0ND = qA xp0NA

EDC • Diffusion current densities Jp = - q Dp dp

dx Jn = - q Dn dn

dx

• Drift current Densities = q(p μp+ nμn)E • μp , μn decrease with increasing doping concentration . • Dn

µn = Dp

µp = KT/q ≈ 25 mv @ 300 K

• Carrier concentration in N-type silicon nn0 = ND ; pn0 = ni2 / ND

• Carrier concentration in P-type silicon pp0 = NA ; np0 = ni2 / NA

• Junction built in voltage V0 = VT ln �𝑁𝑁𝐴𝐴𝑁𝑁𝐷𝐷𝑛𝑛𝑖𝑖

2 �

• Width of Depletion region Wdep = xp + xn = �2εsq

� 1NA

+ 1ND

� (V0 + VR)

* �2𝜀𝜀𝑓𝑓𝑓𝑓

𝑞𝑞= 12.93𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠𝑠𝑠�

• xnxp

= NAND

• Charge stored in depletion region qJ = q.NA NDNA+ND

. A . Wdep

• Depletion capacitance Cj = εsAWdep

; Cj0 = εsAWdep/ VR=0

Cj = Cj0/�1 + VR

V0�

m

Cj = 2Cj0 (for forward Bias) • Forward current I = Ip + In ; Ip = Aq ni

2 Dp

LpND �𝑒𝑒𝑉𝑉/𝑉𝑉𝑇𝑇 − 1�

In = Aq ni2 Dn

LnNA �𝑒𝑒𝑉𝑉/𝑉𝑉𝑇𝑇 − 1�

• Saturation Current Is = Aq ni2 � Dp

LpND+ Dn

LnNA�

• Minority carrier life time τp = Lp2 / Dp ; τn = Ln

2 / Dn


Pg.No. 115 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet • Minority carrier charge storage Qp = τpIp , Qn = τpIn

Q = Qp + Qn = τTI τT = mean transist time • Diffusion capacitance Cd = � 𝜏𝜏𝑇𝑇

𝜂𝜂𝑉𝑉𝑇𝑇� I = τ.g ⇒ Cd ∝ I.

τ→ carrier life time , g = conductance = I / 𝜂𝜂𝑉𝑉𝑇𝑇 • I02 = 2(T2−T1)/10 I01 • Junction Barrier Voltage Vj = VB = Vr (open condition)

= Vr - V (forward Bias) = Vr + V (Reverse Bias)

• Probability of filled states above ‘E’ f(E) = 11+e(E−Ef)/KT

• Drift velocity of e− 𝑣𝑣d ≤ 107 cm/sec • Poisson equation d

2Vdx2 = −ρv

ϵ = −nq

ϵ ⇒ dv

dx = E = −nqx

ϵ

Transistor :- • IE = IDE + InE • IC = ICo – α IE → Active region • IC = – α IE + ICo (1- eVC/VT ) Common Emitter :- • IC = (1+ β) ICo + βIB β = α

1−α

• ICEO = ICo1−α

→ Collector current when base open • ICBO → Collector current when IE = 0 ICBO > ICo . • VBE,sat or VBC,sat → - 2.5 mv /0 C ; VCE,sat → VBE,sat

10 = - 0.25 mv /0C

• Large signal Current gain β = IC− ICBoIB+ ICBo

• D.C current gain βdc = ICIB

= hFE • (βdc = hFE ) ≈ β when IB > ICBo • Small signal current gain β′ = ∂IC

∂IR�

VCE = hfe = hFE

1−(ICBo+ IB)∂hFE∂IC

• Over drive factor = βactiveβforced→under saturation

∵ IC sat = βforced IB sat Conversion formula :- CC ↔ CE • hic = hie ; hrc = 1 ; hfc = - (1+ hfe) ; hoc = hoe CB ↔ CE • hib = hie

1+hfe ; hib = hie hoe

1+hfe - hre ; hfb = −hfe

1+hfe ; hob = hoe

1+hfe

CE parameters in terms of CB can be obtained by interchanging B & E . Specifications of An amplifier :-


Pg.No. 116 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet • AI = −hf

1+h0ZL Zi = hi + hr AIZL Avs = Av.Zi

Zi+Rs = AI.ZL

Zi+Rs = AIs .ZL

Rs

AV = AI ZL

Zi Y0 = ho - hf hr

hi+ Rs AIs = Av.Rs

Zi+Rs = Avs.Rs

ZL

Choice of Transistor Configuration :- • For intermediate stages CC can’t be used as AV < 1 • CE can be used as intermediate stage • CC can be used as o/p stage as it has low o/p impedance • CC/CB can be used as i/p stage because of i/p considerations.

Stability & Biasing :- ( Should be as min as possible) • For S = ∆IC

∆ICo�

VB0,β S′ = ∆IC

∆VBE�

IC0,β S′′ = ∆IC

∆β�

VBE,ICo

∆IC = S. ∆ICo + S′ ∆VBE + S′′ ∆β

• For fixed bias S = 1+β

1−βdIBdIC

= 1 + β

• Collector to Base bias S = 1+β

1+β RCRC+RB

0 < s < 1+ β = 1+β

1+β� RC+ RERC+ RE+ RB

�

• Self bias S = 1+β

1+β RERE+Rth

≈ 1+ RthRe

βRE > 10 R2

• R1 = Vcc Rth

Vth ; R2 = Vcc Rth

Vcc−Vth

• For thermal stability [ Vcc - 2Ic (RC + RE)] [ 0.07 Ico . S] < 1/θ ; VCE < VCC

2

Hybrid –pi(π)- Model :- gm = |IC | / VT rb′e = hfe / gm rb′b = hie - rb′e rb′c = rb′e / hre gce = hoe - (1+ hfe ) gb′c For CE :- • fβ =

gb′e2π(Ce+ Cc)

= gm hfe2π(Ce+ Cc)


Pg.No. 117 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet • fT = hfe fβ ; fH = 1

2π rb′e C =

gb′e2πC

C = Ce + Cc (1 + gm RL )

fT = S.C current gain Bandwidth product fH = Upper cutoff frequency

For CC :- • fH = 1+gmRL

2πCLRL ≈ gm

2πCL = fT Ce

CL =

gm+ gb′e2π(CL+ Ce)

For CB:- • fα = 1+ hfe

2πrb′e(CC+ Ce) = (1 + hfe) fβ = (1 + β) fβ

• fT = β

1+β fα fα > fT > fβ

Ebress moll model :- IC = - αN IE + ICo (1- eV/VT) IE = - αI IC + IEo (1- eV/VT) αI ICo = αN IEo Multistage Amplifiers :- • fH* = fH √21/n − 1 ; fL

∗ = fL

�21/n−1

• Rise time tr = 0.35fH

= 0.35B.W

• tr∗ = 1.1 �tr1

2 + tr22 + ⋯

• fL∗ = 1.1 �fL1

2 + fL22 + ⋯

• 1fH

∗ = 1.1 �1

fH12 + 1

fH22 + ⋯

Differential Amplifier :- • Zi = hie + (1 + hfe) 2Re = 2 hfe Re ≈ 2βRe • gm = α0|IEE|

4VT = IC

4VT = gm of BJT/4 α0 → DC value of α

• CMRR = hfeRe

Rs+hie ; Re ↑ , → Zi ↑ , Ad ↑ & CMRR ↑

Darlington Pair :-


Pg.No. 118 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet • AI = (1 + β1) (1 + β2) ; Av ≈ 1 ( < 1) • Zi = (1+hfe)2 Re2

1+hfe hoc Re2 Ω [ if Q1 & Q2 have same type ] = AI Re2

• Ro = Rs

(1+hfe)2 + 2 hie1+hfe

• gm = (1 + β2 ) gm1

Tuned Amplifiers : (Parallel Resonant ckts used ) : • f0 = 1

2π√LC Q → ‘Q’ factor of resonant ckt which is very high

• B.W = f0 /Q

• fL = f0 - ∆BW

2

• fH = f0 + ∆BW2

• For double tuned amplifier 2 tank circuits with same f0 used . f0 = �fL fH . MOSFET (Enhancement) [ Channel will be induced by applying voltage] • NMOSFET formed in p-substrate • If VGS ≥ Vt channel will be induced & iD (Drain → source ) • Vt → +ve for NMOS • iD ∝ (VGS - Vt ) for small VDS

• VDS ↑ → channel width @ drain reduces .

VDS = VGS - Vt channel width ≈ 0 → pinch off further increase no effect

• For every VGS > Vt there will be VDS,sat • iD = Kn

′ [ (VGS - Vt ) VDS - 12 VDS

2 ] �𝑊𝑊𝐿𝐿

� → triode region ( VDS < VGS - Vt )

Kn′ = μn Cox

• iD = 1

2 Kn

′ �𝑊𝑊𝐿𝐿

� [ VDS2 ] → saturation

• rDS = 1

Kn′ �𝑊𝑊

𝐿𝐿 �(VGS − Vt ) → Drain to source resistance in triode region

PMOS :-


Pg.No. 119 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet • Device operates in similar manner except VGS , VDS, Vt are –ve • iD enters @ source terminal & leaves through Drain .

VGS ≤ Vt → induced channel VDS ≥ VGS - Vt → Continuous channel iD = Kp


� [(VGS − Vt )2 - 12 VDS

2 ] Kp′ = μp Cox

VDS ≤ VGS - Vt → Pinched off channel .

• NMOS Devices can be made smaller & thus operate faster . Require low power supply . • Saturation region → Amplifier • For switching operation Cutoff & triode regions are used • NMOS PMOS

VGS ≥ Vt VGS ≤ Vt → induced channel VGS - VDS > Vt VGS - VDS < Vt → Continuous channel(Triode region) VDS ≥ VGS - Vt VDS ≤ VGS - Vt → Pinchoff (Saturation)

Depletion Type MOSFET :- [ channel is physically implanted . i0 flows with VGS = 0 ] • For n-channel VGS → +ve → enhances channel .

→ -ve → depletes channel • iD - VDS characteristics are same except that Vt is –ve for n-channel

• Value of Drain current obtained in saturation when VGS = 0 ⇒ IDSS .

∴ IDSS = 1

2 Kn


� Vt2 .

MOSFET as Amplifier :- • For saturation VD > VGS - Vt • To reduce non linear distortion 𝑣𝑣gs < < 2(VGS - Vt ) • id = Kn


� (VGS − Vt ) 𝑣𝑣gs ⇒ gm = Kn′ �𝑊𝑊

𝐿𝐿� (VGS − Vt )

• 𝑣𝑣d

𝑣𝑣gs = - gm RD

• Unity gain frequency fT = gm

2π(Cgs+Cgd)

JFET :- • VGS ≤ Vp ⇒ iD = 0 → Cut off


Pg.No. 120 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet • Vp ≤ VGS ≤ 0, VDS ≤ VGS - Vp

iD = IDSS �2 �1 − 𝑉𝑉𝐺𝐺𝐺𝐺𝑉𝑉𝑝𝑝

� �𝑉𝑉𝐷𝐷𝐺𝐺−𝑉𝑉𝑝𝑝

� − �𝑉𝑉𝐷𝐷𝐺𝐺𝑉𝑉𝑝𝑝

�2

� � → Triode

• Vp ≤ VGS ≤ 0 , VDS ≥ VGS - Vp

iD = IDSS �1−𝑉𝑉𝐺𝐺𝐺𝐺

𝑉𝑉𝑝𝑝�

2 ⇒ VGS = Vp �1−�

I𝐷𝐷IDSS

�

gm = 2IDSS|Vp| �1−𝑉𝑉𝐺𝐺𝐺𝐺

𝑉𝑉𝑝𝑝� = 2IDSS

|Vp| �I𝐷𝐷

IDSS

� → Saturation

Zener Regulators :- • For satisfactory operation Vi− Vz

Rs ≥ IZmin + ILmax

• RSmax = Vsmin− Vz0− IZmin rz

IZmin+ ILmax

• Load regulation = - (rz || Rs )

• Line Regulation = rz

Rs+rz .

• For finding min RL take Vs min & Vzk , Izk (knee values (min)) calculate according to that .

Operational Amplifier:- (VCVS) • Fabricated with VLSI by using epitaxial method • High i/p impedance , Low o/p impedance , High gain , Bandwidth , slew rate . • FET is having high i/p impedance compared to op-amp . • Gain Bandwidth product is constant . • Closed loop voltage gain ACL = AOL

1± β AOL β → feed back factor

• ⇒ V0 = −1

RC ∫ Vi dt → LPF acts as integrator ;

• ⇒ V0 = −R

L ∫ 𝑉𝑉i dt ; V0 = −L

R dvi

dt (HPF)

• For Op-amp integrator V0 = −1

τ ∫ 𝑉𝑉i dt ; Differentiator V0 = - τ dvi

dt

• Slew rate SR = ∆V0

∆t = ∆V0

∆t . ∆Vi

∆t = A. ∆Vi

∆t

• Max operating frequency fmax = slew rate

2π . ∆V0 = slew rate

2π × ∆Vi ×A .


Pg.No. 121 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet

• In voltage follower Voltage series feedback

• In non inverting mode voltage series feedback

• In inverting mode voltage shunt feed back • V0 = -η VT ln � 𝑉𝑉𝑖𝑖

𝑅𝑅I0�

• V0 = - VBE

= - η VT ln � 𝑉𝑉𝑠𝑠

𝑅𝑅I𝐶𝐶0�

• Error in differential % error = 1

CMRR � 𝑉𝑉𝑐𝑐

𝑉𝑉𝑑𝑑�× 100 %

Power Amplifiers :-

• Fundamental power delivered to load P1 = �B1

√2�

2 RL = B1

2

2 RL

• Total Harmonic power delivered to load PT = �𝐵𝐵12

2+ 𝐵𝐵2

2

2+ ⋯ . . � 𝑅𝑅𝐿𝐿

= P1 �1 + �𝐵𝐵2𝐵𝐵1

�2

+ �𝐵𝐵3𝐵𝐵1

�2

+ … … �

= [ 1+ D2] P1 Where D = �+D2

2 + ⋯ . . +Dn2 Dn = Bn

B1

D = total harmonic Distortion . Class A operation :- • o/p IC flows for entire 3600 • ‘Q’ point located @ centre of DC load line i.e., Vce = Vcc / 2 ; η = 25 % • Min Distortion , min noise interference , eliminates thermal run way • Lowest power conversion efficiency & introduce power drain • PT = IC VCE - ic Vce if ic = 0, it will consume more power • PT is dissipated in single transistors only (single ended) Class B:- • IC flows for 1800 ; ‘Q’ located @ cutoff ; η = 78.5% ; eliminates power drain • Higher Distortion , more noise interference , introduce cross over distortion • Double ended . i.e ., 2 transistors . IC = 0 [ transistors are connected in that way ] PT = ic Vce • PT = ic Vce = 0.4 P0 PT → power dissipated by 2 transistors . Class AB operation :-


Pg.No. 122 of 143





Institute of Engineering Studies (IES,Bangalore) EDC & Analog Electronics Formula Sheet • IC flows for more than 1800 & less than 3600 • ‘Q’ located in active region but near to cutoff ; η = 60% • Distortion & Noise interference less compared to class ‘B’ but more in compared to class ‘A’ • Eliminates cross over Distortion Class ‘C’ operation :- • IC flows for < 180 ; ‘Q’ located just below cutoff ; η = 87.5% • Very rich in Distortion ; noise interference is high . Oscillators :- • For RC-phase shift oscillator f = 1

2πRC √6+4K hfe ≥ 4k + 23 +29

k where k = Rc/R

f = 1

2πRC√6 μ > 29

• For op-amp RC oscillator f = 1

2πRC√6 | Af| ≥ 29 ⇒ Rf ≥ 29 R1

Wein Bridge Oscillator :- f = 1

2π√R′ R′′C′C′′ hfe ≥ 3 μ ≥ 3 A ≥ 3 ⇒ Rf ≥ 2 R1 Hartley Oscillator :-

f = 12π�(L1+L2)C

|hfe| ≥ L2L1

| μ | ≥ L2L1

|A| ≥ L2L1

↓ Rf

R1

Colpits Oscillator :- f = 1

2π�L C1C2C1+C2

|hfe| ≥ C1C2

| μ | ≥ C1C2

| A | ≥ C1

C2


Pg.No. 123 of 143





Institute of Engineering Studies (IES,Bangalore) Digital Electronics Formula Sheet

Digital Electronics

• Fan out of a logic gate = IOHIIH

or IOLIIL

• Noise margin : VOH - VIH or VOL - VIL • Power Dissipation PD = Vcc Icc = Vcc �

I𝐶𝐶𝐶𝐶𝐶𝐶+I𝐶𝐶𝐶𝐶𝐶𝐶2

� I𝐶𝐶𝐶𝐶𝐶𝐶 → Ic when o/p low I𝐶𝐶𝐶𝐶𝐶𝐶 → Ic when o/p high .

• TTL , ECL & CMOS are used for MSI or SSI • Logic swing : VOH - VOL • RTL , DTL , TTL → saturated logic ECL → Un saturated logic • Advantages of Active pullup ; increased speed of operation , less power consumption . • For TTL floating i/p considered as logic “1” & for ECL it is logic “0” . • “MOS” mainly used for LSI & VLSI . fan out is too high • ECL is fastest gate & consumes more power . • CMOS is slowest gate & less power consumption • NMOS is faster than CMOS . • Gates with open collector o/p can be used for wired AND operation (TTL) • Gates with open emitter o/p can be used for wired OR operation (ECL) • ROM is nothing but combination of encoder & decoder . This is non volatile memory . • SRAM : stores binary information interms of voltage uses FF. • DRAM : infor stored in terms of charge on capacitor . Used Transistors & Capacitors . • SRAM consumes more power & faster than DRAM . • CCD , RAM are volatile memories . • 1024 × 8 memory can be obtained by using 1024 × 2 memories • No. of memory ICs of capacity 1k × 4 required to construct memory of capacity 8k × 8 are “16”

DAC ADC

• FSV = VR �1 − 12𝑛𝑛� * LSB = Voltage range / 2n

• Resolution = step sizeFSV

= VR/2n

VR �1− 12n�

= 12n−1

× 100% * Resolution = FSV2n−1

• Accuracy = ± 12 LSB = ± 1

2n+1 * Quantisation error = VR

2n %

• Analog o/p = K. digital o/p

PROM , PLA & PAL :-

AND OR

Fixed Programmable

Programmable fixed

Programmable Programmable

PROM

PAL

PLA


Pg.No. 124 of 143





Institute of Engineering Studies (IES,Bangalore) Digital Electronics Formula Sheet • Flash Type ADC : 2n−1 → comparators

2n → resistors 2n × n → Encoder

Fastest ADC :-

• Successive approximation ADC : n clk pulses • Counter type ADC : 2n - 1 clk pulses • Dual slope integrating type : 2n+1 clock pulses .

Flip Flops :-

• a(n+1) = S + R′ Q = D = JQ′ + K′Q = TQ′ + T′ Q

Excitation tables :-

• For ring counter total no.of states = n • For twisted Ring counter = “2n” (Johnson counter / switch tail Ring counter ) . • To eliminate race around condition tpd clock < < tpd FF . • In Master slave master is level triggered & slave is edge triggered

Combinational Circuits :-

Multiplexer :-

0

0

000

1

x

1 1

0

1

R

1 0

1

S

x

0

1

0

00 0

1 01

0

1

D

0

1

01

1 01

011 01

0

1

0

J K

0

1x 1

0x

xx 1

0

1

T

0

1

10


Pg.No. 125 of 143





Institute of Engineering Studies (IES,Bangalore) Digital Electronics Formula Sheet • 2n i/ps ; 1 o/p & ‘n’ select lines. • It can be used to implement Boolean function by selecting select lines as Boolean variables • For implementing ‘n’ variable Boolean function 2n × 1 MUX is enough . • For implementing “n + 1” variable Boolean 2n × 1 MUX + NOT gate is required . • For implementing “n + 2” variable Boolean function 2n × 1 MUX + Combinational Ckt is

required • If you want to design 2m × 1 MUX using 2n × 1 MUX . You need 2m−n 2n × 1 MUXes

Decoder :-

• n i/p & 2n o/p’s • used to implement the Boolean function . It will generate required min terms @ o/p & those terms

should be “OR” ed to get the result . • Suppose it consists of more min terms then connect the max terms to NOR gate then it will give the

same o/p with less no. of gates . • If you want to Design m × 2m Decoder using n × 2n Decoder . Then no. of n × 2n Decoder

required = 2m

2n .

• In Parallel (“n” bit ) total time delay = 2n tpd . • For carry look ahead adder delay = 2 tpd .


Pg.No. 126 of 143





Institute of Engineering Studies (IES,Bangalore) Microprocessors Formula Sheet

Microprocessors

• Clock frequency = 12 crystal frequency

• Hardware interrupts TRAP (RST 4.5) 0024H both edge level RST 7.5 → Edge triggered 003CH RST 6.5 0034 H RST 5.5 level triggered 002C INTR Non vectored

• Software interrupts RST 0 0000H

RST 1 0008H 2 0010H Vectored : 0018H : 7 0038H

•

• HOLD & HLDA used for Direct Memory Access . Which has highest priority over all interrupts .

Flag Registers :-

• Sign flag :- After arthematic operation MSB is resolved for sign flag . S = 1 → -ve result • If Z = 1 ⇒ Result = 0 • AC : Carry from one stage to other stage is there then AC = 1 • P : P =1 ⇒ even no. of one’s in result . • CY : if arthematic operation Results in carry then CY = 1 • For INX & DCX no flags effected • In memory mapped I/O ; I/O Devices are treated as memory locations . You can connect max of

65536 devices in this technique . • In I/O mapped I/O , I/O devices are identified by separate 8-bit address . same address can be used

to identify i/p & o/p device . • Max of 256 i/p & 256 o/p devices can be connected .

1

Haltwrite

Readfetch

S 01 S

0

1

0

10

1

0

ACS X P CYXZ X


Pg.No. 127 of 143





Institute of Engineering Studies (IES,Bangalore) Microprocessors Formula Sheet

Programmable Interfacing Devices :-

• 8155 → programmable peripheral Interface with 256 bytes RAM & 16-bit counter • 8255 → Programmable Interface adaptor • 8253 → Programmable Interval timer • 8251 → programmable Communication interfacing Device (USART) • 8257 → Programmable DMA controller (4 channel) • 8259 → Programmable Interrupt controller • 8272 → Programmable floppy Disk controller • CRT controller • Key board & Display interfacing Device

RLC :- Each bit shifted to adjacent left position . D7 becomes D0 .

CY flag modified according to D7

RAL :- Each bit shifted to adjacent left position . D7 becomes CY & CY becomes D0 .

ROC :-CY flag modified according D0

RAR :- D0 becomes CY & CY becomes D7

CALL & RET Vs PUSH & POP :-

CALL & RET PUSH & POP

• When CALL executes , μp automatically stores * Programmer use PUSH to save the contents 16 bit address of instruction next to CALL on the rp on stack Stack

• CALL executed , SP decremented by 2 * PUSH executes “SP” decremented by “2” . • RET transfers contents of top 2 of SP to PC * same here but to specific “rp” . • RET executes “SP” incremented by 2 * same here

Some Instruction Set information :-

CALL Instruction

CALL → 18T states SRRWW

CC → Call on carry 9 – 18 states

CM → Call on minus 9-18

CNC → Call on no carry


Pg.No. 128 of 143





Institute of Engineering Studies (IES,Bangalore) Microprocessors Formula Sheet CZ → Call on Zero ; CNZ call on non zero

CP → Call on +ve

CPE → Call on even parity

CPO → Call on odd parity

RET : - 10 T

RC : - 6/ 12 ‘T’ states

Jump Instructions :-

JMP → 10 T

JC → Jump on Carry 7/10 T states

JNC → Jump on no carry

JZ → Jump on zero

JNZ → Jump on non zero

JP → Jump on Positive

JM → Jump on Minus

JPE → Jump on even parity

JPO → Jump on odd parity .

• PCHL : Move HL to PC 6T • PUSH : 12 T ; POP : 10 T • SHLD : address : store HL directly to address 16 T • SPHL : Move HL to SP 6T • STAX : Rp store A in memory 7T • STC : set carry 4T • XCHG : exchange DE with HL “4T”

XTHL :- Exchange stack with HL 16 T

• For “AND “ operation “AY” flag will be set & “CY” Reset • For “CMP” if A < Reg/mem : CY → 1 & Z → 0 (Nothing but A-B)

A > Reg/mem : CY → 0 & Z → 0

A = Reg/mem : Z → 1 & CY → 0 .


Pg.No. 129 of 143





Institute of Engineering Studies (IES,Bangalore) Microprocessors Formula Sheet • “DAD” Add HL + RP (10T) → fetching , busidle , busidle • DCX , INX won’t effect any flags . (6T) • DCR, INR effects all flags except carry flag . “Cy” wont be modified • “LHLD” load “HL” pair directly • “ RST “ → 12T states • SPHL , RZ, RNZ …., PUSH, PCHL, INX , DCX, CALL → fetching has 6T states

• PUSH – 12 T ; POP – 10T


Pg.No. 130 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet Power electronics: 1.Turn on time of scr = td + tr +ts Where td = delay time tr = rise time tS = settling time 2.Device turn off time, tq= trr + tgr

Where trr = Reverse recovery time and tgr = gate recovery time

3.SERIES OPERATION: → SCR,s are connected in series to increase the voltage rating.

String efficiency ηs= 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑟𝑟𝑣𝑣𝑣𝑣𝑟𝑟𝑟𝑟𝑣𝑣 𝑣𝑣𝑜𝑜 𝑠𝑠𝑣𝑣𝑟𝑟𝑟𝑟𝑟𝑟𝑣𝑣

𝑁𝑁𝑣𝑣.𝑣𝑣𝑜𝑜 𝑆𝑆𝑆𝑆𝑆𝑆,𝑠𝑠 𝑟𝑟𝑟𝑟 𝑣𝑣ℎ𝑒𝑒 𝑠𝑠𝑣𝑣𝑟𝑟𝑟𝑟𝑟𝑟𝑣𝑣 𝑿𝑿 voltage rating of each SCR = V1+ V2

2 × V1

Derating factor =1 – string efficiency

→ Static equalizing Resistance Rs = 𝑟𝑟𝑉𝑉𝑏𝑏𝑏𝑏−𝑉𝑉𝑠𝑠

(𝑟𝑟−1)∆ 𝐼𝐼𝐼𝐼

→ Dynamic equalizing capacitance C = (n−1)∆ Q

n Vbm−Vs

4. Parallel operation is applied for SCR, s with higher current ratings String efficiency ηS = 𝑐𝑐𝑐𝑐𝑟𝑟𝑟𝑟𝑒𝑒𝑟𝑟𝑣𝑣 𝑟𝑟𝑣𝑣𝑣𝑣𝑟𝑟𝑟𝑟𝑣𝑣 𝑣𝑣𝑜𝑜 𝑠𝑠𝑣𝑣𝑟𝑟𝑟𝑟𝑟𝑟𝑣𝑣

𝑁𝑁𝑣𝑣.𝑣𝑣𝑜𝑜 𝑆𝑆𝑆𝑆𝑆𝑆,𝑠𝑠 𝑟𝑟𝑟𝑟 𝑣𝑣ℎ𝑒𝑒 𝑠𝑠𝑣𝑣𝑟𝑟𝑟𝑟𝑟𝑟𝑣𝑣 𝑿𝑿 current rating of each SCR= I1 + I2

2 I1

1. Single Phase Half wave rectifier R load:

Circuit turn off time = tc = π/ω

Average output voltage V0 = Vm /2π[1+ cos α]

r. m. s output voltage Vor = Vm / (2√π ) [π − α + (Sin 2α) /2]1/2

T1

C

C

T2


Pg.No. 131 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet

Average output current, I0 = V0𝑆𝑆

R .m .s output current, Ior / R

Power factor of input supply = Power supplied to loadSource volt ampere

p f = Vor2 / R

Vs Is p.f = Vor

Vs

2. Single Phase Half wave rectifier R-L load:

Circuit turn off time = tc = (2π- β)/ ω

Average output voltage V0 = (Vm /2π) [cos α – cos β]

r.m.s output voltage Vor = Vm /(2√π ) � β − α + ½ �sin 2α – sin 2β��12

Average output current, I0 = V0 /R

R.m.s value of output current, Ior = Vor / R

Power factor of supply, pf = Vor

Vs

3. Single Phase Half wave rectifier R.L load and free wheeling Diode:

Circuit turn off time = tc = πω

Average output voltageV0 = �Vm

2π� [ 1 + 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶]

r.m.s output voltageVor = Vm / (2 √π ) �π − α + 12

(sin 2α)�12

Average output current, , I0 = V0 /R

R.m.s value of output current, , Ior = Vor / R

Power factor of supply, Pf = VorVs

4. Single Phase Half wave rectifier RLE load:

Circuit turn off time, tc = 2π +θ1−β ω

, θ1 = Sin−1 (𝐸𝐸/𝑉𝑉𝑚𝑚) and θ2 = 1800 - θ1


Pg.No. 132 of 143






Average Output voltage, V0 =E+I0 R

Average Output currentI0 = � 12𝜋𝜋𝑆𝑆

� [𝑉𝑉𝑚𝑚(𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 − 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶) − 𝐸𝐸(𝐶𝐶 − 𝐶𝐶)]

Supply power factor, pf = I2

orR + I0 EVs Ior

R.m.s value of output current

Ior = � 12πR2 ��VS

2 + E2� (β − α) − VS2

2(sin 2β − 2 sin 2α) − 2VmE(cosα − cosβ)�

12

5. Single Phase Full wave rectifier – Mid point converter type:

Circuit turn off time tc = π−αω

Average output voltage , V0 = 2Vm cos απ

6.Single Phase Full wave Bridge type Rectifier:

Circuit turn off time , tc = π−αω

Average output voltage , V0 = 2Vm cos απ

r.m.s output voltageVor = Vs

The inductance of source results in an lesser value of voltage ∴ V0 = 2Vm

π cos α - ωLs

π I0

7. Single Phase semi converter:

Circuit turn off time tc = π−αω

Average output voltage V0 = Vmπ

[1+cos α]


Pg.No. 133 of 143






r.m.s output voltageVor = Vs / (√π ) �π − α + 12

(sin 2α)�12

8.Three Phase controlled half wave rectifier with R load:

Average output voltage V0 = 32π

Vml cos α= 3√32π

Vmp cos α

Average Output current I0 = V0 /R

r.m.s output voltageVor = VMLR

�16

+ √38π

(cos 2α)�12

R.m.s value of output current, Ior = Vor / R

9. Three Phase full converter:

voltage V0 Average output = 3π Vml cos α

r.m.s output voltageVor = VML� 32π

�π3

+ √32

(cos 2α)�12

𝐢𝐢𝐬𝐬= 𝐢𝐢𝐨𝐨 �𝟑𝟑𝟐𝟐

10. Three Phase full converter:

voltage V0 Average output = 32π

Vml (1+cos α)

r.m.s output voltageVor = VML2

�3π

�2π3

+ √32

(1 + cos 2α)�12

11. For a 3-∅ converter , The inductance of source results in an lesser value of voltage

V0 = 3√6π

Vph cos α - 3ωLsπ

I0


Pg.No. 134 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet CHOPPERS: a.STEP UP CHOPPER:

Duty cycle, = �Ton (Ton + Toff)

� � = (Ton / T)

Average output voltage across load ,V0= VS1−α

Energy supplied by inductor = Wout= (𝑉𝑉0 − 𝑉𝑉𝑠𝑠) (Imin +Imax)2

x ToFF b.STEP DOWN CHOPPER :

Duty cycle,=�Ton (Ton + Toff)

� � = (Ton / T)

Average output voltage across load ,V0= VS (Ton / T) = f Ton VS = α Vs Where V =supply voltage

Average output current through load,𝐼𝐼0 =𝛂𝛂 VS

R

Rms value of output voltage = √α Vs

Rms value of thyristor current = √α Vs

R

Effective input resistence of chopper= R

α

The minimum and maximum values of load current is given by

Imax = Vs/R [1−e−Ton

Ta ]

[1−e− T

Ta] - (E/R)

Imn = Vs/R [e−Ton

Ta −1]

[e− T

Ta−1] - (E/R)

Ripple ∆I = VsR

��1−e

−TonTa ��1−e

−ToffTa �

[1−e− T

Ta]�


Pg.No. 135 of 143






Per unit Ripple = �1−e

−αTTa ��1−e

−(1−α)TTa �

[1−e− T

Ta]

Maximum value of ripple current is given by (∆I)max = Vs / 4fL

Ripple factor = A.C.ripple voltageD.C voltage

= ��1α

� − 1

Voltage Commutated Chopper:

Minimum turn on time of chopper → π√LC seconds

Minimum duty cycle of voltage commutated chopper αmn = πf √LC

The output current , I0 = Cvs−(−vs)2tc

where C = I0 tcvs

, L ≥ �VSI0

�2

C

t1= π√LC


Pg.No. 136 of 143






Current Commutated Chopper:

The values of L & C are given by

L = vstc

xI0�π−2 Sin−1 �1x� �

C = x I0tc

vs� π−2 sin−1 �1x� �

Where x = ICp / I0 tc = [ π-2 sin−1(1/x) ] √LC tc = ICp sin ωt At ωt = θ1 , ic = I0= ICp sin θ1 , θ1 = sin−1 [I0/ICp ] = sin−1 �1

x�

Peak capacitor voltage = vs +I0�L/C INVERTERS: Fourier Analysis Of Single Phase Inverter Output Voltage : 1-phase Half bridge Inverter:

V0 = ∑ 2Vs nπ

sin nωt∞n=1,3

i0 = ∑ 2Vs

nπZn Sin(nωt − ∅n)∞

n=1,3

1-Phase Full Bridge Inverter: V0 = ∑ 4Vs

nπ∞n=1,3 sin nωt

i0 = ∑ 4Vs

nπZn

∞n=1,3 sin (nωt - ∅n)

Zn is the impedance offered to nth harmonic

Zn = �R2 + �nωL − 1nωc

�2


Pg.No. 137 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet ∅n = tan−1 �nωL−1/nωc

R�

Vor = Vs - full bridge inverter

Vor = Vs/2 – Half bridge inverte

Harmonic factor of nth harmonic , H.Fn = Vn / V1

Where Vn = r.m.s value of nth harmonic component V1 = fundamental component r.m.s value.

→ Total harmonic distortion (T.H.D) → it is a measure of closeness in shape between a wave form and its fundamental component.

THD = 1/V1 �∑ 𝑉𝑉𝑟𝑟2∞

𝑟𝑟=2,3 �1/2

= �Vor

2 −V12

V1

Distortion factor of nth harmonic is defined as VnV1.n2

Three Phase Bridge Inverter :

Vab = ∑ 4Vsnπ

∞n=1,3 Cos nπ

6 Sin n(ωt + π/6)

Vbc = ∑ 4Vs

nπ∞n=1,3 Cos nπ

6 Sin n(ωt - π/2)

Vca = ∑ 4Vs

nπ∞n=1,3 Cos nπ

6 Sin n(ωt + 5π/6)

All triple n harmonics are absent from the line voltages.

Vl = �1π ∫ Vs

22π/30 d(ωt) = Vs �2/3 = 0.8165 Vs

Vph = 0.8165 Vs

√3 = 0.4714 Vs

Va0 = ∑ 2Vs

nπ∞n=1,3 Cos nπ

6 Sin n(ωt+π

6 )

Vb0 = ∑ 2Vs

nπ∞n=1,3 Cos nπ

6 Sin n(ωt - π

2 )

Vc0 = ∑ 2Vs

nπ∞n=1,3 Cos nπ

6 Sin n(ωt+5π

6 )

�Vph = 1π

∫ �Vs2

�22π/3

0 d(ωt) = Vs �1/6


Pg.No. 138 of 143






Vph = 0.4082 Vs , Vl = 0.4082 Vs × √3 = 0.7071 Vs Voltage Control In 1-∅ Inverter: 1.Single Pulse Modulation:

Vor = Vs�2d/π

V0 = ∑ (4Vsnπ

∞n=1,3 Sin nπ/2. sin nd) sin (nωt)

↓ Maximum value of nth harmonic

To eliminate nth harmonic , nd = π i.e., width of the pulse = 2d = 2π/n

2.Multiple Pulse Modulation:

Vor = √2 d/π V0 = ∑ 8Vs

nπ∞n=1,3 sin nγ sin nd

2 sin (nωt)

Number of pulses per half cycle = Length of half cycle of reference wave form

Width of one cycle of triangle carrier wave

N = 1/2fr1/fc

= fc2fr

= ωc2ωr

N= ωc /2ωr per half cycle

Where fc is carrier wave frequency and fr is reference wave frequency

AC VOLTAGE CONTROLLERS: Single Phase Voltage controller with R Load:

Circuit turn off time = π/ω sec.

VC = ∑ Anωn=1,3,5 sin nωt + ∑ Bn

ωn=1,3,5 cos nωt d(ωt)

Where An= Vm

π�sin (𝑟𝑟+1)𝛼𝛼

(𝑟𝑟+1)− sin (𝑟𝑟−1)𝛼𝛼

(𝑟𝑟−1)�

Bn = Vmπ

�cos(𝑟𝑟+1)𝛼𝛼−1(𝑟𝑟+1)

− sin(𝑟𝑟+1)𝛼𝛼−1(𝑟𝑟−1)

�

Vnm = �An2 + Bn

2 ∅n = tan−1 Bn

An


Pg.No. 139 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet For n=1, Vor of output voltage is given by

Vor = Vm

√2π �(𝜋𝜋 − 𝐶𝐶) + 1

2sin 2𝐶𝐶�

1/2

Ior = Vor

R

P = Ior

2 R = Vor2

R = Vm

2

2πR (π-α) + ½ sin 2α

= Vs

2

πR�(𝜋𝜋 − 𝐶𝐶) + 1

2sin 2𝐶𝐶�

Power factor = Real power

Apparent power = Vor

2 /RVsVor/R

= VorVs

= �1𝜋𝜋

�(𝜋𝜋 − 𝐶𝐶) + 12

sin 2𝐶𝐶��1/2

PHASE VOLTAGE CONTROLLER WITH RL LOAD →

i0 = Vm2

sin (ωt - ∅) - Vm2

sin (α - ∅). exp[𝑅𝑅𝐶𝐶 / 𝜔𝜔𝜔𝜔]. e−RT/L

i0 = Vm2

sin (ωt - ∅) - Vm2

sin (α - ∅). exp �−𝑆𝑆𝜔𝜔𝜔𝜔

(𝜔𝜔𝜔𝜔 − 𝐶𝐶)� DC & AC DRIVES DC Motor equations :

Ea = Z∅N60

𝐏𝐏𝐀𝐀

N → rpm

= Z∅n PA

n → rps

If ωm = 2πn Ea = Z- ∅ ωm

2π 𝐏𝐏𝐀𝐀

= � 𝒁𝒁𝟐𝟐𝟐𝟐

𝑷𝑷𝑨𝑨

� ∅ 𝝎𝝎𝒎𝒎

Ea = ka ∅ωm ka = motor constant = ((z/2π) (P/A)) volts/webers rad/sec

Torque T = 12π

Z∅ Ia PA

= � Z2π

PA

� ∅ Ia


Pg.No. 140 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet T = Ka ∅ Ia ka = Newton meter / wb amper

For a dc separately excited motor:

Flux , ∅ is constant ∴ Ea = km ωm km → motor constant volts/rad/sec

Torque = ka ∅Ia Te = kmIa km → Newton meter/Ampere

For a dc series motor:

∅ ∝ Ia , ∅ = c Ia Ea = kac Ia ωm Ea = k1 Ia ωm k1 → motor constant volt sec

rad.amp

Te = ka∅Ia

= kac Ia2

Te = k1 Ia2 k1 → Nm /Amp2

1-∅ half wave converter drive:

Ia is assumed to be constant for speed control of the drive. Vt = V0 = Vm

2π (1+cos α1)

Vf = Vm

π (1+cos α2 )

Is rms = � 12π ∫ Ia

2πα d(ωt)

= Ia � 12π

(π − α)

Is rms = Ia [(π − α)/2π]1/2 IF.D.R = Ia [(π + α)/2π]1/2

Supply power factor = power delivered to load

Source VA

= EaIa+Ia2ra

VsIsr

= Ia[Ea+Iara]

VsIsr

Input supply pf = (IaVt)/(VsIsr)


Pg.No. 141 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet 1-∅ half controlled converter drive →

v0= vt = Vmπ

[1+ cos α1]

Vf = Vmπ

[1+cos α2] Isr= Ia �(π − α)/π

Ifdr = Ia �2α/2π = Ia �α/π ITr = Ia �(π − α)/2π = I0r Input supply pf = VtIa

VsIsr

Single Phase full wave Converter Drive:

V0 = Vt = 2Vmπ

cos α1

Vf = 2Vmπ

cos α2 Isr = Ia ITr = Ia�π/2π = Ia/√2 Pf = Vt Ia/ Vs Isr

3-∅ half wave converter drive:

V0 = Vt = 3√62π

Vph cos α1 Isr = Ia �120/360 = Ia �1/3 ITr = Ia �120/360 = Ia �1/3

3-∅ full wave converter drive:

V0 = Vt = 3√6π

Vph cos α

Isr= Ia �240/360 = Ia �2/3 ITr = Ia �120/360 = Ia �1/3

3-∅ semi converter drive:

V0 = Vt = 3√62π

Vph (1 + cos α)

For α1 < 600, Isr = Ia �2/3 , ITr = Ia �1/3 For α1 > 600 , Isr = Ia �180 − α1/180 , ITr = Ia �(180 − α1)/360


Pg.No. 142 of 143





Institute of Engineering Studies (IES,Bangalore) Power Electronics Formula Sheet Static Rotor Resistance Control:

Reff = R × Toff𝑇𝑇

= R × (T-Ton)/T

Reff = R(1-α) α → duty cycle of the chopper


Pg.No. 143 of 143





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