Single Phase System

Pure Resistive Circuit in Series i

v Rv = Vm sin t

Instantaneous voltage is given by

tv

i sinR

V

Rm

Instantaneous current is given by

Im = Vm/R

The maximum value for current Im and maximum value for voltage Vm can be related as

I = V/R

The rms value for current Irms (simply I) and rms value for voltage Vrms (simply V) can be related as

Vm

Im

voltage

current

+

-

time

In circuit contains resistor , the V and I are in phase as in diagram below

I

V

waveform

phasor

p = i2R = (Im2/R)sin2 t

p = v2/R = (Vm2/R) sin2 t

p = vi = VmIm sin2 t

Power dissipated in the resistor

/2

0

m/2

0

2m)cos1(

4

I)(sin

2

Idtt

Vdtt

VP

mmav

Average value for power

VIVV

ttV mmm

2

I.

22

I2sin

2

1

4

I mm/2

0

m

voltage v

current i

power p

Wave in pure resistance circuit

v

i

Pure Inductive Circuit in Seriesi = Im sin t;If the current is

dt

diLevL

)sin( tIdt

dL m

tLIm cos

)2/sin()2/sin( tVtLI mm

The voltage is leading the current by /2 rad (90o) OR current is lagging behind the voltage by /2

Thus Vm= L Im

t

v, i

t

Vm

Im

/2

vi

Voltage and current waveform in a purely inductive circuit

Maximum voltage: Vm = LIm

Voltage r.m.s value .: V = LI

V/I = Vm/Im = L = XL

XL is measured in ohm and called as inductive

reactance=L

where = 2f

V

E

I90

90

I, XL I X

L

f

Changes of I and XL with frequency

Phasor for purely inductive circuit

P = vi = (Im sin t)(Vm sin (t + /2)

= VmIm sin t sin (t + /2)

= VmIm sin t [sin t cos /2 + cos t sin

/2 ]

= VmIm sin t cos t

= ½VmIm sin 2t

Power dissipated in purely inductive circuit

0cos2ωo8π

IV)2(sin

4

I /20

mm/2

0

m

dtt

VP

mav

Average power

Half cycle has cancelled the other half cycle that is why the average power is zero.

t

v, i, p

vi

p

++

--

CURRENT, VOLTAGE AND AVERAGE POWER WAVEFORM IN A PURELY

INDUCTIVE CIRCUIT.

Pure capacitive Circuit in Series

v

iv = Vm sin t If the voltage is given as

)cos)sin( tCVtVdt

dC

dt

dvCi mm

Then

)2/sin()2/sin( tItCV mm

Therefore Im= C Vm

In this case the current is leading a voltage by /2 ( 90o) OR voltage is lagging behind the current by /2.

v, i

t

v iVm

Im

/2

AC VOLTAGE AND CURRENT WAVEFORM IN PURELY

CAPACITIVE CIRCUITMaximum current value Im = CVm

r.m.s value .: I = CVratio V/I = Vm/Im = 1/C = XC

XC = is measured in ohm and called as capacitive reactance =1/C

I

V

/2

Phasor diagram

f

I, Xc

Xc

I

Changes of I and XC with frequency

p = vi = (Vm sin t)(Im sin (t + /2))

= VmIm sin t sin (t + /2)

= VmIm sin t [sin t cos /2 + cos t sin

/2 ]

= VmIm sin t cos t

= ½VmIm sin 2t

Power dissipated by capacitor

0cos2ωo8π

IV)2(sin

4

I 2ππ/0

mm/2

0

m

dttV

Pm

av

Average power

Half cycle has cancelled the other half cycle that is why the average power is zero.

arus ivoltan v

Kuasa p

CURRENT, VOLTAGE AND POWER IN PURELY CAPACITIVE CIRCUIT

vR

vL

i

v

vR = iR = ImR sin t in phase with ivL = iXL = ImXLsin (t + /2) = LIm sin (t + /2) leading the i by /2v = vR + vL = ImR sin t + LIm sin (t + /2) = ImR sin t + LIm cos tv = Vm sin (t + ) (1)Where (2)and = tan-1(L/R) (3)

Ifi = Im sin t

Then

22mm )( RIV L

Current and Voltage waveform in L-R serial circuit

VR in phase with I VL is leading in phase with I by

t

v, i

v

ivR

vL

VL

VRI

V

/2

Phasor diagram for I, VR, VL & V in R & L serial circuit

•I and VR overlap to represent the in phase•VL is vertical to represent the 90o leading out of phase•Resultant between VL and VR give the value of V

2L

2R VV(V RL

1 V/Vtan and

The representation of voltage can be written as V /

RI

XL Z

/2

Phasor diagram for I, R, XL & Z in R & L serial circuit

2222 )(( LRXRZ L

RLRX L /tan/tan 11

ZZImpedance is represented as

p = vi = Vm sin (t + ) Im sin t

= ½VmIm [cos - cos (2t - )]

= ½VmIm cos - ½VmIm cos (2t - )

component ½VmIm cos (2t - ) is zero

rmsrmsmm

mm IVIV

IVP 2

.2

cos2

1

Therefore the average value is only given by

CURRENT, VOLTAGE AND POWER WAVEFORMS FOR L-R SERIAL CIRCUIT

p

i

v

t

V,I

i

v

P

f (Hz)

R, XL, Z ()

R

Z

X

L

Phasor diagram

V

I

I cos

I sin

VR = V cos

We can also calculate the power from

P = I2R or P = VR2/R or VRI

All in r.m.s values I= Irms , VR=VRrms

From phasor diagram P=VRI=VIcos(active power)

Cos is a power factor

Reactive power (VAR) = VLI = VI sin

cos = VR/V = R/Z.

L=31.8mHV=100V50Hz

R=7.0

V L

V R

I

AZ

VI 2.8

2.12

100

A resistance of 7.0 is connected in series with a pure inductance of 31.8mH and the circuit is connected to a 100V, 50Hz, sinusoidal supply. Calculate (a) the circuit current. (b) the phase angle

ooL orlagR

X5555

0.7

0.10tantan 11

2.120.100.7 21

21

2222LXRZ

0.10108.315022 3fLX L

L=318mHV (50Hz)

R=75

V L

V R=150V

I

12510075 21

21

2222LXRZ

VVR 150

A pure inductance of 318mH is connected in series with a pure resistance of 75. The circuit is supplied from a 50Hz sinusoidal source and the voltage across the 75 resistor is found to be 150V. Calculate the supply voltage

VVVV LR 250200150 21

21

2222

VIXV LL 2001002

AR

VI 2

75

150

100103185022 3fLX L

check

VIZV 2501252

Z L=50 V

R=40

V L

V R

I=3A

VIRVR 120403

A coil, having both resistance and inductance, has a total effective impedance of 50 and the phase angle of the current through it with respect to the voltage across it is 45o lag. The coil is connected in series with a 40 resistor across a sinusoidal supply. The circuit current is 3A; by constructing a phasor diagram, estimate the supply voltage and the circuit phase angle

VIZV LrLr 150503

62500

lago25

707.01501202150120cos2 22222 LrLrRLrR VVVVV

904.0

250

707.0150120coscos

V

VV LrLrR

VV 250

V LrV

V R

VLr cosLr

I

R=12

L=0.1H

100V50Hz

IX LV

IRI

o69

A coil having a resistance of 12 W and a inductance of 0.1H is connected across a 100V, 50Hz supply. Calculate:

(a)The reactance and the impedance of the coil;(b)The current;(c)The phase difference between the current and the applied voltage:

617.212

4.31tan

R

X

AZ

VI 97.2

6.33

100

4.311.05022 LfX L

6.334.3112 2222LXRZ

A serial ac circuit ,ABCD, contains a resistor AB, an inductor BC with resistance R and inductance L and a resistor CD. When a current 6.5A flow in the circuit, voltage drops across those points are: VAB = 65 V; VBC = 124 V; VAC = 149 V. The supply voltage is 220 V at 50 Hz. Calculate

(a)Draw the circuit(b)Voltage drop VBC and its phase compared to the current(c)Resistance and inductance of the inductor

A B

CD

Current= I = 6.5 ASupply voltage = V = 220 V Z = V/I = 220/6.5 = 33.8 (1)VAB = 65 V; R1 = 65/6.5 = 10 (2)VBC = 124 V; ZL = 124/6.5 = 19.1 (3)VAC = 149 V; ZAC = 149/6.5 = 22.9 (4)

From Equation (3):-ZL

2 = RL2 + XL

2

19.12= 364.8 =RL2 + XL

2 (5)

From Equations (2) and (4):-ZAC

2 = (R1 + RL)2 + XL2

22.92=524.4= (10 + RL)2 + XL2

(6)From Equations (5) and (6):-

RL = 3.0 (7)

Substitute RL in (5)XL = 18.9

(8)Dari (1)

Z2 = (R1 + RL + R2)2 + XL2

33.82=(10+3.0 + R2)2 + 18.92

R2 = 15

VR1 = 6.5 x 10 = 65 VVR2 = 6.5 x 15 = 97.5 VVRL = 6.5 x 3.1 = 20.2 VVXL = 6.5 x 18.8 = 122.2 V

VVVVV XLRRLBD 7.1692.122)5.972.20()( 22222

(b)

o46)5.972.20(

2.122tan 1

(b) Resistance of inductor = RL = 3.0 XL = 18.9 = 2fLL = 18.9/2f = 18.9/2 x 50 = 60.1 mH

vR

vC

i

If i = Imsint

vR = iR = ImR sin t --- in phase with i

t)dt(C

Idt

Cm

c sini1

v

)2/sin()2/sin(cos

tXItC

It

C

ICm

mm

C

1XC

f2

v = vR + vC

= (Im/C) sin (t - /2) + ImR sin t

v = Vm sin (t - )

22m )C/1(RV )CR/1(tan 1

where

; and;

Vc is lagging by 90o refer to I

v

vC

vR

it

/2

/ 3/2 2/

CURRENT AND VOLTAGE WAVEFORMS IN R-C SERIAL CIRCUIT

VR

VC

V

I

-

/2

2C

2R VVV

= tan-1 (VC/VR)

VCan be written as

R

XC

Z

I

-

/2

2C

2 XRZ

22 )C/1(R

= 2f

Z-

Phasor diagram Impedance triangle

where

= tan-1 (XC/R)

p = vi = Vm sin (t - ) Im sin t

= ½VmIm [cos - cos (2t - )]

= ½VmIm cos - ½VmIm cos (2t - )]

POWER

The average of component [½VmIm cos (2t - )] is zero, therefore

coscos2

.2

cos21

rmsrmsmm

mm IVIV

IVP

p

i

v

t

CURRENT, VOLTAGE AND POWER WAVEFORMS FOR

R-L SERIAL CIRCUIT

Phasor diagram

We can also calculate the power from

P = I2R or P = VR2/R or VRI

All in r.m.s values I= Irms , VR=VRrms

From phasor diagram P=VRI=VIcos(active power)

Cos is a power factor

Reactive power (VAR) = VCI = VI sin

Cos = VR/V = R/Z.Z

XC

R

f

I

I cos

I sin

V

230V

0.5A

C=8 F

R

A capacitor of 8.0F takes a current of 1.0 a when the alternating voltage applied across it is 230 V. Calculate:

(a)The frequency of the applied voltage;(b)The resistance to be connected in series

with the capacitor to reduce the current in the circuit to 0.5A at the same frequency;

(c)The phase angle of the resultants circuit.

HzCX

fC

5.862301082

1

2

16

CfI

VX C 2

1230

0.1

230

(a)

21

224605.0

230CXR

I

VZ

398230460 2222CXZR

o

Z

R30

460

398coscos 11

(b)

(c)

Leading by 30o

A

V

WlampofcurrentRated 5.7

100

750

VVC 270

222 100230 CV

222CR VVV

230V60Hz

I

C

R

V C

100V

V RI

VV C

A metal-filament lamp, rated at 750W, 100V, is to be connected in series with a capacitor across a 230V, 60Hz supply. Calculate:

(a)The capacitance required(b)The phase angle between the current and

the supply voltage

435.0230

100cos

V

VR

I

V

CfX C

C 2

1

'1264o

FVf

IC

C

7.73270602

5.7

2

(b)