Unit-3

Question Bank

Q.1 A delta connected load draw a current of 15A at lagging P.F. of .85 from 400V, 3-phase,

50Hz supply. Find R & L of each phase.

Ans. 42.98 , 53.7mH

Sol.

Given VP = VL= 400 0

IL = 15A

So IP = 15

3=8.66A

so Ip = 8.661.85Cos (lagging load)

= 8.66 31.79 A

ZP = 400 0

46.18 31.79 42.98 16.878.66 31.79

P

P

Vj

I

= RP+jXLP RP =

Lp = 16.87

2 2 50

LPX

f

Q. 2 A balanced 3-phase star connected load of 100 KW takes leading current of 80A, when

connected across 3-phase, 1100V, 50Hz supply. Find component of load per phase.

Ans. R = 5.2 , C = 532F

Sol.

Given P = 100KW IL = 80A

In case of star connected load IP = IL

IP = IL = 80A & VL = 1100V Vp= 1100

6353 3

LV V f = 50Hz

IL = IP = 3100 10

3 3 1100L

P

V Cos Cos

42.98 Ans.

53.7mH Ans.

Cos = 3100 10

.6563 1100 80

= 49, Sin = .754

load impedance/phase = VP/IP

= 1100 / 3

7.9480

R =Z Cos R = 7.94 .656

XC = Z Sin = 7.94.754 =5.986

Xc = 1

2 fC (As leading current is drawn)

C = 1

2 CfX =

1

2 50 5.986 =

R = 5.2 Ans.

532F Ans.

Q.3 Each phase of star connected load consists of non-inductive resistance of 50 in parallel

with a capacitance of 63.6F. Calculate the line current, total power absorbed, total kVA

and the power factor when this load is connected to a 381 V (line voltage), 3-phase, 50Hz

supply. Ans. 6.22A, 2.9kW, 4.1kVA, .707

Sol.

VL = 381V

Vp=381

03

=220 0= (220+j0)V.

Admittance of each phase is,

Yp=1

R+jC=

1

50+j31463.6106= (0.02+j0.02) S

Hence

Ip = Vp Yp = 220 (0.02+j0.02)=4.4+j4.4=6.22 45A

Pf=cos = cos45 =

IL=Ip =

Power factor = cos45 = 0.707 leading

Vp= (220+j0)V, Ip = (4.4+j4.4)A

Hence complex power/phase

S = VI*

S = (220+j0) (4.4j4.4)=968j968 = 1369 45 VA

Total apparent power in the 3 phase = 31369 VA =

Total power absorbed in the 3 phase = 3968W =

Q. 4 Calculate the active and reactive current components in each phase of a star connected,

10,000V, 3 phase alternator supplying 5000 kW at a power factor of 0.8. If the total current

remains the same when the load power factor is raised to 0.9, find the new output.

.707 Ans

6.22A Ans

4.107 kVA Ans.

2.904kW Ans.

Ans. 289A, 216A,

5625kW

Sol.

P = 3 L LV I Cos

IL = 3500 10

3613 cos 3 10,000 0.8L

PA

V

The active current = IL cos = 361 0.8 =

The reactive current = IL sin = 361 0.6 =

The new power at 0.9 P.F.

= 5000

0.90.8

=

Q. 5 A balanced 3-phase star connected load of 150kW takes a leading current of 100A with a

line voltage of 1100V, 50Hz. Find the circuit constants of the load per phase.

Ans. R = 5 XC= 3.9 , C =

816F

Sol.

Phase impedance

Zp= Phase Voltage

Phase current

= 1100 11

6.353 100 3

The Power per phase = RpIp2 =

3150 1000 50 103

W

Rp = 350 10

100 100

Xp = 2 2 3.9Z R

The reactive part is capacitive (leading current)

289 A Ans.

216 A Ans.

5625kW Ans.

5 Ans.

1

C

C = 610

314 3.9F

=

Q. 6 Three equal star connected inductors take 8kW at power factor 0.8 when connected to

460V, 3 phase, 3 wire supply. Find the line currents if one inductor is short circuited.

Ans. 21.7 83 A, 21.7 157 A, 37.6 53A

Sol.

P = 3 L LV I Cos

I = 8000

12.553 cos 3 460 0.8L

pA

V

The phase impedance

= 460

21.163 12.55

P

p

V

I

(Ip=IL for star connection)

cos = 0.8; = 37

Since Z3 is shorted 3 and N are at the same potential.

The three line voltages as phasors will be

V12 = 460 0

V23 = 460 120

V31= 460 120

Since 3 and N are at the same potential

131460 120

21.16 37

VI

Z

3.9 Ans.

816F Ans.

21.7 83 A Ans.

21.7 157 A Ans.

37.6 53A Ans.

Ans.AAns.

=

= 2.64j21.5 A

232460 120

21.16 37

VI

Z

=

= 19.97j8.48 A

3 1 2I I I

(As at N 1I

+ 2I

+ 3I

= 0)

= 19.97+2.64+j(8.48+21.5) A

= 22.61 + j29.98 =

(i.e.) the three line currents are 21.7A, 21.7A and 37.6 A

Q. 7 The circuit shown in which R = 200 , r = 100 and L = 0.552H is connected to a

symmetrical 3 phase, 400 V, 50 Hz system at points 1, 2 and 3. Find the p.d. between A, B

and C, when the phase sequence of the supply is (a) 1, 2, 3 (b) 1, 3, 2

Ans. (a) 0 , (b) 400

60V

Sol.

314 0.552 173.3L

R+j L = 100+j173.3 = 200 60

(R+r)+j L = (200+100) + j173.3 = 346.5 30

400 60V Ans.

Phase 123

V12 = 400 0 V, V23 = 400 120 V, V31= 400 120V

VAB= (r+j L )I12+RI23

= 200 12400 120

60 200346.5 30 346.5 30

V

=231 30 231 150

=231 30 231 30o

Phase 132

I23= 23

400 1201.154

346.5 30 346.5 30

V

90A

Hence,

VAB = 231 30 200 1.154 90

= 231 30 231 90

= 200+j115.5+j231

= 200+j346.5 =

Q. 8 A 3 wire 3 phase supply feeds a load consisting of three equal resistors. By how much is

the load reduced if one of the resistors be removed (a) when the load is star connected (b)

when the load is delta connected? Ans. (a)

50%, (b) 33.3%

Sol.

(a) Star connected

With all three resistors

IL = 3

LV

R=Ip

The total power = 3RIp2 = 3R

2 2

23

L LV V

R R .(i)

With one resistor removed

0 Ans.

Ip = 2

2

LV

R

The power =2RIp2 = 2R

2 2

24 2

L LV V

R R .(ii)

This is half of the above.

The reduction in power =

2 2 2

2 2

2 2100 100

L L L

L L

V V V

R R R

V V

R

(c) Delta connected

With all three resistors current in each resistor

Ip= PV

R

The Power = 3RIp2 = 3R

2 2

2

3L LV V

R R (i)

With one resistor removed

50% Ans.

Current in each resistor Ip=LV

R

The power =2RIp2 = 2R.

2 2

2

2L LV V

R R (ii)

Hence reduction in power

2 2 2

2 2

3 2

100 1003 3

L L L

L L

V V V

R R R

V V

R R

=

Q. 9 A 3-phase star-connected alternator feeds a 2000 hp delta-connected induction motor

having a P.F. of 0.85 and an efficiency of 0.93. Calculate the current and the active and

reactive components in (a) each alternator phase (b) each motor phase. The line voltage is

2200V. Ans. (a) 495A, 421A, 261A (b) 286A, 243A,

151A

Sol.

Motor input

= 6746 2000 746

1.6 100.93

HP

Watts

If IL is the line current we have

3 cosL LV I Input

IL = 61.6 10

3 2200 0.85

Power factor = 0.85 = cos

sin = 0.527

(a) Alternator (star connected)

Phase current = line current = 495A

Active component = I cos =

Reactive component = I sin =

(b) Induction motor (delta connected)

33.3% Ans.

495A Ans.

421A Ans.

261A Ans.

3.36 9.8 A Ans.

Phase current Ip = 495

3

Active component = Ip cos =

Reactive component = Ip sin =

Q. 10 Three star-connected impedances Z1 = 20+j37.7 per phase are connected in parallel with

three delta-connected impedances Z2 = 30j159.3 per phase. The line voltage is 398V.

Find the line current, P.F, Power and reactive volt-amperes taken by the combination.

Ans. 3.36 98 A, .985 lagging, 2284W, 394

VAR

Sol.

The phase voltage

=398

2303

V

Z1= 20+j37.7 = 42.7 62.05

I1= 230

62.05 5.39 62.0542.7

= 2.52j4.76 A

Z2 = 30j159.3

The equivalent star impedance

Z2= 2 10 53.1 54 79.3

3

Zj

I2 = 230

79.3 0.79 4.1954

j A

The total current I = I1+I2 = 3.31j0.57 =

The P.F. = cos = cos9.8 =

sin = 0.17

The power = 3 power per phase

243 A Ans.

286A Ans.

151 A Ans.

0.985 lagging Ans.

2284W Ans.

= 3 230 3.36 0.985 =

The reactive VA = 3 VI sin

= 3 230 3.36 0.17 =

Q. 11 Prove that P = 3PY if load per phase same.

Ans. P =

3PY

Sol

Let

VL Line voltage

Rp resistive load per phase

Zp impedance per phase

Then VP = / 3LV

For Y connective

PY = 3VPIP Cos = 3VPP

p

V

Z.

p

p

R

Z

=

2

2

2 23 3

3

p pLP

p p

R RVV

Z Z

22

p

L

RV

Z ..(1)

For connective

VP = VL

3 P PP V I Cos

= 2

23 . . 3

p pPP P

p p

R RVV V

Z Z Z

=2

23

p

L

p

RV

Z(2)

By Comparison (1) & (2)

394 VAR Ans.

Q. 12 Three identical coils are connected in star to a three phase 50Hz supply. If the line current

is 10A, total power consumed is 12kW and volt ampere input is 15 KVA, find the line

voltage VL, phase voltage Vp, VAR input, resistance and inductance of each coil.

Ans. 866V, 500V, 9000VAR, 40 , 0.095H

Sol.

IL=10A; P = 12kW; S = 15 KVA

cos = p.f.= 12

15

kW

kVA = 0.8. Hence sin = 0.6.

P = 3 cosL LI V

Hence VL=312 10

3 cos 3 10 0.8L

P

I

=

Vp=866

3 3

LV =

Q = 3 sinL LV I = 3 866100.6 =

Ip=IL=10A; Zp=500

10

p

p

V V

I A =50

Rp = Zp cos = 500.8 =

Xp= Zp sin = 500.6=30

Inductance/phase is L = 2

pX

f=

30

2 50 =

Q. 13 Three 100 non-inductive resistance are connected in (a) star (b) delta across a 400-V,

50Hz, 3-phase mains. Calculate the power taken from the supply system in each case. In

the event of one of the three resistances getting open-circuited, what would be the value

of total power taken from the mains in each of the two cases?

Ans. 1600W, 4800W, 800W 3200W

Sol.

866 volt. Ans.

500 volt. Ans.

9000 VAR Ans.

0.095H Ans.

P = 3PY Ans.

40 Ans.

(i) Star Connection

Vp = 400/ 3 V

Ip=400 4

3 100 3

p

L

p

VA I

Z

cos = 1 as load is non-inductive resistance

P = 3 cosL LV I

= 3 400 4 1/ 3 (1)

(ii) Delta Connection

Vp = 400V ; Rp = 100

Ip = 400 / 100 = 4A

IL = 4 3 A

P = 3 4004 3 1 = (2)

When one of the resistors is disconnected

(i) Star Connection

The circuit no longer remains a 3-phase circuit but consists of two 100 resistors in series across a

400-V supply. Current in lines A and C is = 400/200=2A

Power absorbed in both = 2I2Rp = 24100 = ...(3)

Comparing (3) with (1) it is seen that by

disconnecting one resistor, the power

consumption is reduced by half.

(ii) Delta Connection

In this case, currents in AB and AC remain as

usual 120 out of phase with each other.

1600W

4800W

800 W

Current in each phase = 400/100=4A

Power consumption in both = 2 42 100 = ..(4)

Comparing (4) with (2) it is seen that when one resistor is disconnected, the power consumption is

reduced by one-third.

Q. 14 A 3-phase, 37.3kW, 440-V, 50Hz induction motor operates on full load with an efficiency of

89% and at a power factor of 0.85 lagging. Calculate the total kVA rating of capacitors

required to raise the full-load power factor at 0.95 lagging. What will be the capacitance

per phase if the capacitors are (a) delta-connected and (b) star-connected?

Ans. 65.79F, 197.4F

Sol.

Motor power input P = 37.3/0.89 = 41.191kW

Power Factor = 0.85 (lag)

cos 1=.85 1= 31.78 tan 1 = .6197

KVR1 = 41.191 tan 1 = 41.191 .6197= 25.52

2 2cos 0.95 18.2 ; tan18.2 0.3288

Motor kVAR2 = P tan 2 = 41.1910.3288 = 13.54

The difference in the values of kVAR is due to the capacitors which supply leading kVAR to partially

neutralize the lagging kVAR of the motor.

leading kVAR supplied by capacitors is

=kVAR1kVAR2= 25.5213.54 = 12

Since capacitors are loss-free, their kVAR is the same as kVA

kVA/capacitors = 12/3 = 4 VAR/capacitor=4

(a) In -connection, voltage across each capacitor is 440V

Current drawn by each capacitor Ic= 4000/440 = 9.09 A

Now, Ic = 1/c

V VVC

X C

C = 6/ 9.09 / 2 50 440 65.79 10cI V

F =

(b) In star connection, voltage across each capacitor is = 440/ 3 V

3200W

65.79F Ans.

Current drawn by each capacitor, Ic=4000

15.74440 / 3

A

Ic = c

VVC

X or,

44015.74 2 50

3C

Or, C =

Q.15 Three impedance coils, each having a resistance of 20 and a reactance of 15 , are

connected in star to a 400-V, 3- , 50Hz supply. Calculate (i) the line current (ii) power

supplied and (iii) the power factor.

If three capacitors, each of the same capacitance, are connected in delta to the same

supply so as to form parallel circuit with the above impedance coils, calculate the

capacitance of each capacitor to obtain a resultant power factor of 0.95 lagging.

Ans. (i) 9.24A (ii) 5,120W (iii) .8, C = 14.32F

Sol.

Zp = 2 2

20 15 25

cos 1= Rp/Zp = 20/25 =0.8 lag; sin 1 = 0.6 lag

When capacitors are not connected

(i) Ip = 400 / 25 3 = 9.24A =IL (star connected load) IL =

(ii) P = 13 cos 3 400 9.24 0.8L LV I =

(iii) Power factor =

Motor VAR1 = 13 sin 3 400 9.24 0.6L LV I 3,840

When capacitors are connected

Power factor 2 2cos =0.95, =18.2; tan 18.2=0.3288

Since capacitors themselves do not absorb any power, active power remains the same i.e. 5,120 W

even when capacitors are connected, what changes is the VAR.

Now VAR2 = P 2tan =5120 0.3288=1684

Leading VAR supplied by the three capacitors is

197.4F Ans.

9.24A Ans.

5,120 W Ans.

0.8 (lag) Ans.

= VAR1 VAR2 = 3840 1684 = 2156 VAR

VAR/Capacitor= 2156/3 = 719

For delta connection,

voltage across each capacitor is 400V Ic = 719/400 = 1.798A

Also Ic = 61.798 / 2 50 400 14.32 10

1/

VVC C

c

F =

Q.16 A symmetrical 3-phase, 3-wire supply with a line voltage of 173 V supplies two balanced 3-

phase laods; one Y-connected with each branch impedance equal to (6+j8) ohm and the

other -connected with each branch impedance equal to (18+j24) ohm. Calculate

(i) The magnitudes of branch current taken by each 3-phase load

(ii) the magnitude of the total line current and

(iii) the power factor of the entire load circuit

Draw the phasor diagram of the voltages and currents for the two loads.

Ans. (i) 10A (ii) 20A (iii) 0.6 lag

Sol.

The equivalent Y-load of the given -load is = (18+j24)/3 = (6+j8) . With this, the problem now

reduces to one of solving two equal Y-loads connected in parallel across the 3-phase supply as

shown in (a) Phasor diagram for the combined load for one phase only is given in (b).

Combined load impedance

= (6+j8)/2 = 3+j4

= 5 53.1

Vp = 173/ 3 = 100 V

14.32F Ans.

100 0

20 53.15 53.1

pI

A = IL

Current in each load = 10 53.1 A

109.00 .628 Ans.

(i) branch current taken by each load is

(ii) line current is

(iii) combined power factor = cos 53.1 =

Q.17 Three identical impedance of 30 30 ohms are connected in delta to a 3-phase, 3-wire,

208 V abc system by conductors which have impedances of (0.8+j0.63) ohm. Find The

magnitude of the line voltage at the load end. Ans. 109.00

V

Sol.

The equivalent ZY of the given Z is 30 30 /3 = 10 30 = (8.66+j5) . Hence, the load

connections become as shown in

Zan = (0.8+j0.6)+(8.66+j5)

= 9.46+j5.6 = 11 30.6

Van = Vp = 208/ 3 = 120V

Let Van = 120 0 V

Ian = 120 0 /11 30.6 10.9 30.6 A

aaZ = 0.8+j0.6 = 1 36.87

Voltage drop on line conductors is

10.9 30.6 1 36.87 10.9 6.3 10.83 1.196aa an aaV I Z j V

(120 0) (10.83 1.196)a n an aaV V V j j

= 109.17 1.196j

Q.18 A balanced delta-connected load having an impedance ZL= (300+j210) ohm in each phase is

supplied from 400V, 3-phase supply through a 3-phase line having an impedance of Zs =

10 A Ans.

20 A Ans

0.6 (lag) Ans.

1.78 36.9 A Ans.

216.9 1.87o

Ans.

(4+j8) ohm in each phase. Find the total power supplied to the load as well as the current

and voltage in each phase of the load. Ans. 1158.2W, 1.78A, 216.9V

Sol.

The equivalent Y-load of the given -load is

= (300+j210)/3 = (100+j70) Zs = 4+j8 = 8.94 63.4= a aZ

Hence, connections become as shown in

0 (4 8) (100 70) 104 78 130 36.9aZ j j j

0 400 / 3 231aV V

0 231 0 /130 36.9aI

Line drop 1.78 36.9 8.94 63.4 15.9 26.5 14.2 7.1a a a o a aV I Z j V

(231 0) (14.2 7.1)ao a o a aV V V j j = (216.8j7.1) =

Phase voltage at load end, Vao = 216.9V

Phase current at load end, Ia0 = 1.78A

Power supplied to load = 3 1.78216.9=

Incidentally, line voltage at load end Vac = 216.9 3 = 375.7V

Q. 19 Two wattmeters are used for measuring the power input and the power factor of an over-

excited synchronous motor. If the reading of the meters are (2.0kW) and (+7.0 kW)

respectively, calculate the input and power factor of the motor. Ans. 5kW, .305

Sol.

1 2

1 2

( )tan 3

W W

W W

W1 = 2kW

W2 = 7kW

( 2 7) 9

tan 3 3 3.11762 7 5

1tan ( 3.1176) 72.2 ( )lead

cos cos72.2

1158.8 W Ans.

.305( )lead Ans.

1 2 2 7Input W W

Q.20 A wattmeter reads 5.54 kW when its current coil is connected in R phase and its voltage

coil is connected between the neutral and the R phase of a symmetrical 3-phase system

supplying a balanced load of 30 A at 400 V. What will be the reading on the instrument if

the connection to the current coil remain unchanged and the voltage coil be connected

between B and Y phases? Take phase sequence RYB. Draw the corresponding phasor

diagram.

Ans. 7.2kW

Sol.

As seen from

W1=VRIR cos

Or 35.5410 =(400/ 3)30cos

cos =0.8,sin =0.6

In the second case

W2=VYBIR cos (90 ) 400 30 sin 400 30 0.6 =

5kW Ans.

7.2kW Ans.