Single phase ac circuit 02 - Webs · EEE3405 EEP II-Single Phase ac circuit_02 AC Series-Parallel...

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EEE3405 EEP II-Single Phase ac circuit_02

From the geometry of the power triangle,

22LQPS +=

In complex form,

S = P + jQL (for inductive circuit)

S = P – jQC (for capacitive circuit)

or S = VI* where I* is the conjugate of current I.



Example 17-5



Example 17-6

A generator supplies power to an electric heater, an inductiveelement, and a capacitor as shown in Fig 17-17(a).

(a) Find P and Q for each load.

(b) Find total active and reactive power supplied by the generator.

(c) Draw the power triangle for the combined loads and determinetotal apparent power.

(d) Find the current supplied by the generator.



VA8.3530811800j2400jQPS)c(

.)（indkVAR8.16002400Q

kW5.2P)b(

.)cap(VAR60024

120

X

VQ0P:Capacitor

.)ind(kVA4.26

120

X

VQ0P:Inductor

0QkW5.2P:Heater)a(

:Solution

0TTT

T

T

2

C

2

CC

2

L

2

LL

HH

∠=+=+=

=−==

====

====

==



QT=1.8kVAR

35.8O

PT=2.5kW

ST=3081VA

The power triangle is drawn as follow:

AV

SId T 7.25

120

3081)( ===



Power Factor

From the power triangle

P = VI cosθ = S cosθ (W)

Q = VI sinθ = S sinθ (VAR)

The quantities cosθ is defined as Power Factor FP

Fp = cosθ, cosθ = P/S.

Fp is expressed as a number(0 ≤ Fp ≤ 1) or a percent.

Power factor angle θ is the angle between applied voltage and

current or the angle between S and P

θ = cos-1(P/S)



For pure resistance, θ = 0o

For pure inductance, θ = 90o

For pure capacitance, θ = -90o

For RL circuit, θ lies between 0o to 90o

For RC circuit, θ lies between 0o to -90o

For RLC circuit, -90o ≤ θ ≤ 90o

Unity,Lagging and Leading Power Factor

for a pure resistive circuit, θ = 0, FP = cos θ = 1

for a load containing RL, the current lags

voltage.(Lagging)

for a load containing RC, the current leads voltage. (Leading)

Thus, an inductive circuit has a lagging power factor while a

capacitive circuit has a leading power factor.



Power Factor Correction

In figure (a), S=P, I=200A

However, in figure (b),

The size of electrical

apparatus required to supply

a load depends on its VA

requirements.

Ax

V

SI

kVA

QPS

3.333600

10200

200

160120

3

2222

===

=+=+=



Power factor correction : To cancel some or all of the reactive

component of power by adding reactance of the opposite type

to the circuit.

If the reactive power is completely cancelled, this

is referred to as unity power factor correction.

Procedure : For a circuit with a fixed active power P,

find out the reactive power of the circuit Q

add an opposite component so that the total

reactive power QT is reduced

find out the new FP.

2T

2FQP

PP

+=



Example 17-7

For the circuit of Fig 17-18(b), a capacitance with QC=160kVAR is added in parallel with the load as in Fig 17-19(a). Determine generator current I.



AC Series-Parallel Circuits (Ch. 18)

OBJECTIVES

• Compute impedance, currents and voltages in simple ac circuits

• Draw phasor diagrams for voltages and current

• Apply Ohm’s law to analyze simple series circuits

• Apply Kirchhoff’s laws in simple ac circuits

• Determine the series or parallel equivalent of any network consisting of a combination of resistors, inductors and capacitors.



RL series circuit

IVR=IR

VL=IXLVS

θ R

fL2tan

R

Xtan

IR

IXtan

,IandVbetweenanglePhase

impedance)fL2(RXRZwhere

IZ

XRI

)IX()IR(

VVV

1L1L1

S

222L

2L

L

2L

2

2L

2

2L

2RS

π===θ

=π+=+=

=+=

+=

+=

−−−

XL

RVS

VL

VR



RC series circuit

IVR=IR

VC=IXCVS

θ

fCR2

1tan

R

Xtan

IR

IXtan


impedance)fC2

1(RXRZwhere

IZ

XRI

)IX()IR(

VVV

1C1C1

S

222C

2C

C

2C

2

2C

2

2C

2RS

π===θ

=π

+=+=

=+=

+=

+=

−−−

RVS

VC

VR

XC



RLC series circuit

R

XXtan

IR

)XX(Itan


impedance)XX(RZwhere

IZ

)XX(RI

)IXIX()IR(

)VV(VV

CL1CL1

S

2CL

2

2CL

2

2CL

2

2CL

2RS

−=−=θ

=−+=

=−+=

−+=

−+=

−−

IVR=IR

VL=IXL

VS

θ

VC =IXC

VL-VC=I(X L-XC)

XL

RVS

VL

VR

XC VC



Ohm’s law for AC circuits

In Complex form,

Circuit Impedance

R-L in series R + jXL

R-C in series R – jXC

R-L-C in series R + j(XL – XC)

Z1, Z2, ….Zn in series Z1 + Z2 + …….. + Zn



Example 18-5

For the circuit shown in Fig 18-20, find

a) the total impedance

b) the circuit current

c) voltage across each element of the circuit

Draw the phasor diagram



Solution:

(a) ZT = 25 + j(200-225) = 25 – j25= 35.56∠-45o Ω

(b) I = V/ZT = 10 ∠0o/ 35.56∠-45o = 0.283 ∠45o A

(c) VR = IR = (0.283 ∠45o) (25) = 7.07 ∠45o V

VL = IXL = (0.283 ∠45o) (200 ∠90o) = 56.6 ∠135o V

VC = IXC = (0.283 ∠45o) (225 ∠-90o) = 63.6 ∠-45o VI

VR

VS

VL

VC

VC - VL



AC Parallel Circuits

The admittance Y of any impedance is the reciprocal of the

impedance Z.

-jXC

jXL

G

Impedance

BC = 1/-jXC = j/XCC

BL = 1/jXL = -j/XLL

G=1/RR

AdmittanceCircuit element



For n impedances connected in parallel

IZ

ZI

Y

YIor

VYZ

VI

Y.....YYY

x

T

T

xx

xx

x

n21T

==

==

+++=

V



Example 18-16

Refer to the circuit of Fig 18-45, find:

(a) the total impedance, ZT(b) the supply current, IT(c) the branch currents, I1, I2 and I3.



mA025.0)05)(010x05.0(EYI)b(

k020RY

1Z

mS005.0YYYY

mS9011000

jY

mS9011000

jY

mS005.0010x20

1Y)a(

:Solution

0003TT

0

TT

0321T

03

02

0031

∠=∠∠==

Ω∠===

∠=++=

∠==

−∠=−=

∠=∠

=

−

mA905)05)(90001.0(EYI

mA905)05)(90001.0(EYI

mA025.0IEYI)c(

00033

00022

0T11

∠=∠∠==

−∠=∠−∠==

∠===

Single phase ac circuit 02 - Webs · EEE3405 EEP II-Single Phase ac circuit_02 AC Series-Parallel...

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