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
1
so Ip = 8.66 Cos .85
(lagging load)
= 8.66 31.79 A
ZP =
400 0
VP

 46.18 31.79  42.98  j16.87 
I P 8.66 31.79
= RP+jXLP
42.98  Ans.
RP =
 Lp =
X LP
16.87


2 f 2  50
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.
53.7mH Ans.
Ans. R = 5.2  , C = 532µF
Sol.
Given P = 100KW IL = 80A
In case of star connected load IP = IL
IP = IL = 80A & VL = 1100V
IL = IP =
P
3VLCos

Vp=
VL
3

1100
100 103
3 1100  Cos
3
 635V
f = 50Hz
Cos  =
100  103
3  1100  80
 .656
 = 49°, Sin  = .754
 load impedance/phase = VP/IP
=
1100 / 3
 7.94
80
R =Z Cos 
 R = 7.94 × .656 
R = 5.2  Ans.
XC = Z Sin  = 7.94×.754 =5.986 
Xc =
1
(As leading current is drawn)
2 fC
1
1
=
=
 C=
2 fX C
2  50  5.986
532µF Ans.
Each phase of star connected load consists of non-inductive resistance of 50  in parallel
with a capacitance of 63.6µF. 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
Q.3
Sol.
VL = 381V
381
Vp=
3
0
=220 0 = (220+j0)V.
Admittance of each phase is,
Yp=
1
1
+jωC=
+j314×63.6×10–6= (0.02+j0.02) S
R
50
Hence
Ip = Vp Yp = 220 (0.02+j0.02)=4.4+j4.4=6.22 45 A
Pf=cos  = cos45° =
.707 Ans
6.22A Ans
IL=Ip =
Power factor = cos45° = 0.707 leading
Vp= (220+j0)V, Ip = (4.4+j4.4)A
Hence complex power/phase
S = V×I*
S = (220+j0) (4.4–j4.4)=968–j968 = 1369 45 VA
Total apparent power in the 3 phase = 3×1369 VA =
Total power absorbed in the 3 phase = 3×968W =
Q. 4
4.107 kVA Ans.
2.904kW Ans.
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.
Ans. 289A, 216A,
5625kW
Sol.
P=

IL =
3 VL I L Cos
P
3VL cos 

500  103
3 10, 000  0.8
 The active current = IL cos  = 361 × 0.8 =
 361A
289 A Ans.
216 A Ans.
The reactive current = IL sin  = 361 × 0.6 =
The new power at 0.9 P.F.
=
Q. 5
5000
 0.9 =
0.8
5625kW Ans.
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 =
816µF
Sol.
Phase impedance
Zp=
=
Phase Voltage
Phase current
1100
3  100
The Power per phase = RpIp2 =

Rp =

Xp =

11
3
 6.35
150  1000
 50 103W
3
50  103

100  100
5  Ans.
Z 2  R 2  3.9
The reactive part is capacitive (leading current)

1

C

106
C=
F
314  3.9
=
Q. 6
3.9  Ans.
816µF Ans.
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 53 A
Sol.
P=
I=
3 VL I L Cos
p
3VL cos 

8000
3  460  0.8
 12.55 A
The phase impedance
=
VP
460

 21.16
Ip
3 12.55
(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

I1 

V13

Z

460 120
21.16 37
=
–21.7 83 A Ans.
= –2.64–j21.5 A

I2 

V23

Z

460 120
21.16 37
=
21.7 157 A Ans.
= –19.97–j8.48 A

 
I 3    I1  I 2 



(As at N


I1 + I 2 + I 3 = 0)
= 19.97+2.64+j(8.48+21.5) A
= 22.61 + j29.98 =
37.6 53 A Ans.
Ans.AAns.
(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
60 V
Sol.
 L  314  0.552  173.3
R+j  L = 100+j173.3 = 200 60
(R+r)+j  L = (200+100) + j173.3 = 346.5 30 
Phase 123
V12 = 400 0 V, V23 = 400 120 V, V31= 400 120 V
VAB= (r+j  L )I12+RI23
= 200 60
400 120
V12
 200 
346.5 30
346.5 30
=231 30  231 150
=231 30  231 30 
o
0 Ans.
Phase 132
I23=
400 120
V23

 1.154 90 A
346.5 30 346.5 30
Hence,
VAB
= 231 30  200 1.154 90
= 231 30  231 90
= 200+j115.5+j231
= 200+j346.5 =
Q. 8
400 60 V Ans.
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 =
The total power = 3RIp2 = 3R×
With one resistor removed
VL
3R
=Ip
VL 2 VL 2
……….(i)

3R 2
R
Ip =
VL 2
2R
=2RIp2 = 2R×
The power
VL 2 VL 2
………….(ii)

4R2 2R
This is half of the above.
VL 2 VL 2
VL 2

 The reduction in power = R 2 2 R  100  2 R2  100 
VL
VL
R
(c) Delta connected
With all three resistors current in each resistor
Ip=
The Power
VP
R
= 3RIp2 = 3R
With one resistor removed
VL 2 3VL 2

………(i)
R2
R
50% Ans.
Current in each resistor Ip=
The power
=2RIp2
VL
R
VL 2 2VL 2
= 2R. 2 
………(ii)
R
R
3VL 2 2VL 2

R  100 
Hence reduction in power R
3VL 2
R
Q. 9
VL 2
R  100 =
3VL 2
R
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
=
HP  746


2000  746
 1.6  106 Watts
0.93
If IL is the line current we have
3VL I L cos   Input

33.3% Ans.
IL =
1.6  106
3  2200  0.85

495A Ans.
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)
421A Ans.
261A Ans.
Phase current Ip
Active component
Reactive component
Q. 10
=
495
3

286A Ans.
= Ip cos  =
= Ip sin  =
243 A Ans.
151 A Ans.
Three star-connected impedances Z1 = 20+j37.7  per phase are connected in parallel with
three delta-connected impedances Z2 = 30–j159.3  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
3
 230V
Z1= 20+j37.7 = 42.7 62.05 
I1=
230
62.05  5.39 62.05
42.7
= 2.52–j4.76 A
Z2 = 30–j159.3 
The equivalent star impedance

Z2=
Z2
 10  j 53.1  54 79.3 
3
I2 =
230
79.3  0.79  j 4.19 A
54
The total current
The P.F.
3.36 9.8 A Ans.
I = I1+I2 = 3.31–j0.57 =
= cos  = cos9.8° =
0.985 lagging Ans.
sin  = 0.17
 The power
= 3 × power per phase
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
394 VAR Ans.
= 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 = VL / 3
For Y connective
PY = 3×VPIP Cos  = 3VP×
2
P
= 3V
 VL2
For
Rp
Z p2
Rp
Z2
VP R p
.
Zp Zp
……………..(1)
connective
VP = VL
 P  3 VP  I PCos
= 3  VP .
2
= 3VL
Rp
VP R p
.
 3VP2 2
Zp Zp
Z
Rp
Z p2
2
V  Rp
 3 L 
2
 3  Zp
……………………(2)
By Comparison (1) & (2)
=
P
= 3PY 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;
cos  = p.f.=
kW 12

= 0.8. Hence sin  = 0.6.
kVA 15
3I LVL cos 
P=
Hence VL=
Vp=
S = 15 KVA
VL
Q=
3

P
3I L cos 
866
3
=

12 103
3 10  0.8
=
500 volt. Ans.
3VL I L sin  = 3 ×866×10×0.6 =
Ip=IL=10A; Zp=
Vp
Ip
866 volt. Ans.

9000 VAR Ans.
500V
=50 
10 A
40  Ans.
Rp = Zp cos  = 50×0.8 =
Xp= Zp sin  = 50×0.6=30 
Inductance/phase is L =
Q. 13
Xp
2 f
=
30
=
2  50
0.095H Ans.
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.
(i) Star Connection
Vp = 400/ 3 V
Ip=
Vp
Zp

400
3 100

4
3
A  IL
cos  = 1 as load is non-inductive resistance
P=
3VL I L cos 
= 3  400  4 1/ 3 
1600W
……(1)
(ii) Delta Connection
Vp = 400V ; Rp = 100 
Ip = 400 / 100 = 4A
IL = 4  3 A
4800W
P=
3 ×400×4× 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 = 2×4×100 =
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.
800 W
...(3)
Current in each phase = 400/100=4A
3200W
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. 65.79µF, 197.4µF
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
cos 2  0.95 2  18.2; tan18.2  0.3288
Motor kVAR2 = P tan 2 = 41.191×0.3288 = 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
=kVAR1–kVAR2= 25.52–13.54 = 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 =
V
V

 VC
X c 1 / C
6
C = I c / V  9.09 / 2  50  440  65.79 10 F =
(b) In star connection, voltage across each capacitor is = 440/ 3 V
65.79µF Ans.
4000
Current drawn by each capacitor, Ic=
Ic =
V
 VC
Xc
440 / 3
 15.74 A
or, 15.74 
440
3
 2  50  C
197.4µF Ans.
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 = 14.32µF
Sol.
Zp =
202  152  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)
(ii) P =
3VL I L cos 1  3  400  9.24  0.8 =
(iii) Power factor =
 IL =
9.24A Ans.
5,120 W Ans.
0.8 (lag) Ans.
 Motor VAR1 = 3VL I L sin 1  3  400  9.24  0.6  3,840
When capacitors are connected
Power factor cos2 =0.95, 2 =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 tan 2 =5120  0.3288=1684
Leading VAR supplied by the three capacitors is
= 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 =
Q.16
V
 VC  C  1.798 / 2  50  400  14.32 106 F =
1/ c
14.32µF Ans.
A symmetrical 3-phase, 3-wire supply with a line voltage of 173 V supplies two balanced 3phase 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° Ω
Vp = 173/ 3 = 100 V
Ip 
100 0
5 53.1
 20 53.1 A = IL
Current in each load = 10 53.1 A
(i) branch current taken by each load is
(ii) line current is
10 A Ans.
20 A Ans
(iii) combined power factor = cos 53.1° =
Q.17
0.6 (lag) Ans.
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 
Let
Van
= Vp = 208/
Van
= 120 0 V
3 = 120V
Ian = 120 0 /11 30.6  10.9 30.6 A
Z aa = 0.8+j0.6 = 1 36.87 
Voltage drop on line conductors is
Vaa  I an Z aa  10.9 30.61 36.87  10.9 6.3  10.83  j1.196 V
Van  Van  Vaa  (120  j 0)  (10.83  j1.196)
= 109.17  j1.196 
Q.18
109.00 .628 Ans.
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 =
(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 = Z aa
Hence, connections become as shown in
Z a0  (4  j8)  (100  j 70)  104  j 78  130 36.9 
Va0  400 / 3  231 V
I a0  231 0 /130 36.9 
1.78 36.9 A Ans.
Line drop Vaa  I ao  Z aa  1.78 36.9 8.94 63.4  15.9 26.5  14.2  j7.1 V
216.9 1.87 Ans.
o
Vao  Vao  Vaa  (231  j 0)  (14.2  j 7.1) = (216.8–j7.1) =
Phase voltage at load end, Vao = 216.9V
Phase current at load end, Ia0 = 1.78A
1158.8 W Ans.
Power supplied to load = 3× 1.78×216.9=
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 overexcited 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.
tan   3
(W1  W2 )
W1  W2
W1 = –2kW
W2 = 7kW
tan   3
(2  7)
9
 3
 3.1176
2  7
5
  tan 1 (3.1176)  72.2(lead )
cos   cos 72.2 
.305(lead ) Ans.
Input  W1  W2  2  7 
Q.20
5kW Ans.
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
5.54×103 =(400/ 3)×30×cos
cos =0.8,sin =0.6
In the second case
W2=VYBIR cos (90 –  )  400  30  sin   400  30  0.6 =
7.2kW Ans.