Series and Parallel AC Circuits 1 SERIES RL CIRCUITS (1) Circuit above is a series RL network...

Series and Parallel AC Circuits

1

SERIES RL CIRCUITS (1)

• Circuit above is a series RL network connected to an ac voltage source

• Need to find the phasor form of the total impedance of this combination

• The total impedance of this series combination is

• The magnitude and angle of ZT can be found by converting to polar form:

• |ZT| = √[R2+(ωL)2] and θ = tan-1(ωL/R)

• The plot of ZT:

V 0

ωsin

p

p

E

tEte+

-

00 jRR

ω090ω LjL

R

L

ohms ωω00 LT XjRLjRLjjRZ

Im

Re

θ

R

jωL

|Z T|

ω θZLjRZ TT

RLθ

LRZT

ωtan

ω1

22


2

SERIES RL CIRCUITS (2)

• Example: For a series RL combination circuit, R = 300, L = 0.2H and e(t) = 17sin(2000t) V. Find the total equivalent impedance in polar form and rectangular form. Sketch the impedance in the complex plane.

• We can use Ohm’s law to find the total current supplied by a voltage source: iT = v / ZT

• Mathematical operation must be carried out using phasors since all quantities have both magnitude and angle

• The current iT in a series circuit is the same through every series-connected component

• The ac voltage drop across each component can be found by multiplying each impedance by the current

• v1 = iTZ1, v2 = iTZ2, …

• Example: In a series RL circuit, where e(t) = 300 V, R = 200, XL = j100. Find the total current in the circuit. Find the voltage drops across R and L. Verify KVL around the circuit. Draw a phasor diagram showing e, vR, vL and iT. Sketch the voltage waveforms.


3

SERIES RC CIRCUITS (1)

• Above is a series RC network connected to an ac voltage

• The total impedance of this combination is

• The polar coordinates are

• The phasor diagram of the total impedance is

V 0

ωsin

p

p

E

tEte+

-

00 jRR

ω090ω

1Cj

C

R

C

CT XjRCjRCjjRZ ωω00

RCR

C

CRZT

ω1tanω1

tanθ

ohms ω1

11

22

Im

Re

θ

R

-j/ωC

|ZT |

θ

ω

T

T

Z

CjRZ

RC

CRZ T

ω1tanθ

ω11

22


4

SERIES RC CIRCUITS (2)

• As in the series RL circuit, the total current supplied to the RC network and the voltage drops can be found by using Ohm’s Law:

• iT = e / ZT

• vR = iTR and vC = iTXC

• Example: For a series RC circuit where e(t) = 18sin(240t + 45) V, R = 3.3k and C = 2.2µF:

• a) Find the total current in the circuit in phasor and sinusoidal form

• b) Find the voltage drops across the resistor and capacitor in phasor and sinusoidal form

• c) Verify KVL around the circuit

• d) Draw a phasor diagram showing e, iT, vR and vC

• e) Sketch the waveforms of e, vR and vC versus angle


5

SERIES RLC CIRCUITS (1)

• The total impedance of the RLC circuit is

• In terms of magnitudes it is: ZT = R + j(|XL| - |XC|)

• Inductive and capacitive reactance have opposite signs

• Thus net reactance may be either inductive or capacitive, depending which is larger

• Polar coordinates are

V 0

ωsin

p

p

E

tEte+

-

00 jRR

ω090ω

1Cj

C

R

C

ω090ω LjLL

i(t)

ohms ω1ω CLjRZT

R

XX

R

L

XXRCLRZ

CL

CLT

11

2222

tanωC1ω

tanθ

ω1ω


6


• The phasor diagram of the impedance when inductive reactance is greater than the capacitive reactance, i.e. when |XL|>|XC|

• The phasor diagram of the impedance when capacitive reactance is greater than the inductive reactance, i.e. when |XC|>|XL|

Im

Re

ZT

R

θ

jωL

-j/ωC

ωL – 1/ωCNet reactance

Im

Re

ZT

R

θ

jωL

-j/ωC

ωL – 1/ωCNet reactance


7


• When there is more than one resistor, capacitor and/or inductor in a series circuit, the total impedance has a resistance component equal to the sum of the resistance values and a reactive component equal to the sum of the capacitive reactances subtracted from the sum of the inductive reactances

• Example: In a series RLC circuit, e(t) = 100sinωt, R = 800, ZL = j1250 and ZC = -j450.

• a) Find the current in polar form

• b) Find the voltage drops vR, vL, vC

• c) Verify KVL around the circuit

• d) Draw a phasor diagram showing e, i, vR, vL, vC


8

ADMITTANCE

• The reciprocal of resistance is conductance, with units Siemens

• The reciprocal of impedance is admittance, denoted Y

• Y = 1/Z siemens

• Impedance is a measure of the extent to which a component impedes the flow of ac current through it

• Admittance is a measure of how well it admits the flow of ac current

• The greater the admittance, the smaller the impedance, and vice versa

• Resistance R is one form of impedance Z, conductance G is one form of admittance Y

• Phasor form of conductance is:

S 01

siemens 01

0

1

jR

G

RRG


9

SUSCEPTANCE

• Reactance is another special case of impedance

• The reciprocal of reactance is called susceptance, B

• B = 1/X siemens

• There are two types of susceptance

• Inductive susceptance:

• Capacitive susceptance:

• Z = 1/Y; R = 1/G; XL = 1/BL; XC = 1/BC

• Example: Find the admittance of a 5 resistor; a 5mH inductor at f = 60Hz; a 0.2µF capacitor at ω = 1.25106 rad/s.

• Example: Find the admittance Y corresponding to Z = 30 + j40 . Draw a phasor diagram showing Y in the complex plane.

S ω10

siemens 90ω

1

90ω

11

LjB

LLXB

L

LL

S ω0

siemens 90ω90ω1

11

CjB

CCX

B

C

CC


10

PARALLEL AC CIRCUITS (1)

• Above shows a parallel connected set of impedances to an ac source

• Total admittance of the circuit is the sum of the admittances of the parallel connected components, i.e.

• YT = Y1 + Y2 +… + Yn = 1/Z1 + 1/Z2 +…+ 1/Zn

• Example: Find the total admittance of a 20mH inductor, a 1k resistor and a 0.16µF capacitor connected in parallel to an ac voltage source of 15sin(25103t) V. Draw a phasor diagram showing the total admittance in the complex plane. What is the total admittance if the frequency of the voltage source is doubled?

+

-

YT

Z1 Z2


11


• The total impedance of a network is the reciprocal of its total admittance, ZT = 1/YT

• For a parallel network:

• Example: Three components (an inductor with impedance j10, a resistor with impedance 2, and another inductor with impedance j5) are connected in parallel. Find the total equivalent impedance of this network.

• Example: A capacitor with impedance –j100 and an inductor with impedance j500 are connected in parallel. Find this network’s total equivalent impedance.

n

nTT

ZZZ

YYYYZ

111

1

11

21

21


12


• The same voltage appears across every parallel-connected impedance

• So current can be found using Ohm’s law and KCL

e Zn

+

-

Z1 Z2

iT = i1 + i2 +…+ in

i1 =e/Z1

in =e/Zn

i2 =e/Z2

TTT

nT

nnn

eYZei

iiii

eYZei

eYZei

21

111


13


• Example: For the circuit below, find the current in each impedance. Show that iT = eYT. Draw a phasor diagram showing e, iT, i1, i2 and i3.

e = 60sin(ωt) VZ3 =

-j80

+

-

Z1 = 50

Z2 =j40

iT

i1 i3i2


14

AC CURRENT SOURCES (1)

• Like a dc current source, an ideal ac current source supplies the same constant current to whatever network is connected across its terminals

• For ac, the current is constant in the sense that its peak value does not change

• The same symbol is used for an ac current source as for a dc current source

• The direction of the arrow is just a phase reference, since current reverses direction every half cycle

• Reversing the arrow is the same as multiplying the current by -1, which is the same as adding or subtracting 180 from its phase angle

Ipθ+180 Ipsin(ωt + θ) A= Ipθ

-Ipθ ==


15

AC CURRENT SOURCES (2)

• Example: For the circuit below, find the voltage vT across the ac current source. Find the currents i1 and i2. Show that i = i1 + i2. Draw a phasor diagram showing i, vT, i1 and i2. Sketch the waveforms of i, i1 and i2 versus angle.

i2

i(t) = 0.24sin(5106t) A

vT

i1

100 2000pF


16

POWER IN CIRCUITS CONTAINING REACTANCE (1)

• Recall that the average power dissipated by a resistance carrying sinusoidal ac current can be found by

• Above can be used to find Pavg dissipated by a resistor in a circuit containing reactance, provided the values used in the computations are indeed those of the voltage across and/or current through the resistor itself

• In many practical circuits, it is sometimes necessary to compute Pavg when the resistance or resistor voltage/current are not known

effeff

2eff2

eff

22

avg

222

IVR

VRI

IV

R

VRIP pppp


17

POWER IN CIRCUITS CONTAINING REACTANCE (2)

• Above is an ac circuit with a general impedance Z• Assuming that the voltage supplied has angle

• Angle between voltage and current is - ( - ) = • The angle between the voltage applied to a network

and the total current supplied to it equals the angle of the impedance of the network

• Resistive component of impedance Z is R = |Z|cos ohms

• Since only resistance in network dissipates power

+e = Ep

|Z|

i = (Ep/|Z|) →

θφθ

φ

Z

E

Z

Ei pp

2

θcos

,

2

θcos

2

avg

2

22

avg

pp

ppppp

pp

EIP

EIZIEZI

ZIRIP


18

POWER FACTOR

• Equations on previous slide provide a means for computing the average power in terms of voltage e across whole network and total supply current i

• Cos is called the power factor

• For a purely resistive network, voltage and current are in phase, so power factor = cos0 = 1 → Pavg = EpIp / 2

• For a purely reactive network, voltage and current are separated by ±90, and cos ±90 = 0 → Pavg = 0, i.e no power is dissipated by purely reactive components (capacitors and inductors)

• Since Vp = 2Veff and Ip = 2Ieff we can derive

θcos

θcos2

22

effeff

effeffavg

IV

IVP


19

EXAMPLE OF POWER IN AC CIRCUITS

• In the circuit above:

• Find the power factor

• Use the power factor to find the average power dissipated in the network

• Verify that the power computed above is the same as the power computed by using the voltage across and current through the resistor

+e = 600

i→

j80

100

-j20

Series and Parallel AC Circuits 1 SERIES RL CIRCUITS (1) Circuit above is a series RL network...

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