Q.1 The two windings of a transformer is
(A) conductively linked. (B) inductively linked.
(C) not linked at all. (D) electrically linked.
Ans : B
Q.2 A salient pole synchronous motor is running at no load. Its field current is switched off. The motor will
(A) come to stop.
(B) continue to run at synchronous speed.
(C) continue to run at a speed slightly more than the synchronous speed.
(D) continue to run at a speed slightly less than the synchronous speed.
Ans: B
Q.3 The d.c. series motor should always be started with load because
(A) at no load, it will rotate at dangerously high speed.
(B) it will fail to start.
(C) it will not develop high starting torque.
(D) all are true.
Ans: A
Q.4 The frequency of the rotor current in a 3 phase 50 Hz, 4 pole induction motor at full load speed is about
(A) 50 Hz. (B) 20 Hz.
(C) 2 Hz. (D) Zero.
Ans: C
Q.5 In a stepper motor the angular displacement
(A) can be precisely controlled.
(B) it cannot be readily interfaced with micro computer based controller.
(C) the angular displacement cannot be precisely controlled.
(D) it cannot be used for positioning of work tables and tools in NC machines.
Ans: A
Q.6 The power factor of a squirrel cage induction motor is
(A) low at light load only.
(B) low at heavy load only.
(C) low at light and heavy load both.
(D) low at rated load only.
Ans: A
Q.7 The generation voltage is usually
(A) between 11 KV and 33 KV. (B) between 132 KV and 400 KV.
(C) between 400 KV and 700 KV. (D) None of the above.
Ans: A
Q.8 When a synchronous motor is running at synchronous speed, the damper winding produces
(A) damping torque.
(B) eddy current torque.
(C) torque aiding the developed torque.
(D) no torque.
Ans: D
Q.9 If a transformer primary is energised from a square wave voltage source, its output voltage will be
(A) A square wave. (B) A sine wave.
(C) A triangular wave. (D) A pulse wave.
Ans: A
Q.10 In a d.c. series motor the electromagnetic torque developed is proportional to
(A) _{}. (B) _{}.
(C) _{}. (D) _{}.
Ans: B
Q.11 In a 3 – phase induction motor running at slip ‘s’ the mechanical power developed in terms of air gap power _{} is
(A) _{}. (B) _{}.
(C) _{}. (D) _{}.
Ans: C
Q.12 In a 3 – phase induction motor the maximum torque
(A) is proportional to rotor resistance _{}.
(B) does not depend on _{}.
(C) is proportional to _{}.
(D) is proportional to _{}.
Ans: B
Q.13 In a d.c. machine, the armature mmf is
(A) stationary w.r.t. armature. (B) rotating w.r.t. field.
(C) stationary w.r.t. field. (D) rotating w.r.t. brushes.
Ans: C
Q.14 In a transformer the voltage regulation will be zero when it operates at
(A) unity p.f. (B) leading p.f.
(C) lagging p.f. (D) zero p.f. leading.
Ans: B
Q.15 The maximum power in cylindrical and salient pole machines is obtained respectively at load angles of
(A) _{}. (B) _{} .
(C) _{} . (D) _{}.
Ans: D
Q.16 The primary winding of a 220/6 V, 50 Hz transformer is energised from 110 V, 60 Hz supply. The secondary output voltage will be
(A) 3.6 V. (B) 2.5 V.
(C) 3.0 V. (D) 6.0 V.
Ans: C
Q.17 The emf induced in the primary of a transformer
(A) is in phase with the flux. (B) lags behind the flux by 90 degree.
(C) leads the flux by 90 degree. (D) is in phase opposition to that of flux.
Ans: C
Q.18 The relative speed between the magnetic fields of stator and rotor under steady state operation is zero for a
(A) dc machine. (B) 3 phase induction machine.
(C) synchronous machine. (D) single phase induction machine.
Ans: all options are correct
Q.19 The current from the stator of an alternator is taken out to the external load circuit through
(A) slip rings. (B) commutator segments.
(C) solid connections. (D) carbon brushes.
Ans: C
Q.20 A motor which can conveniently be operated at lagging as well as leading power factors is the
(A) squirrel cage induction motor. (B) wound rotor induction motor.
(C) synchronous motor. (D) DC shunt motor.
Ans: C
Q.21 A hysteresis motor
(A) is not a selfstarting motor. (B) is a constant speed motor.
(C) needs dc excitation. (D) can not be run in reverse speed.
Ans: B
Q.22 The most suitable servomotor for low power applications is
(A) a dc series motor.
(B) a dc shunt motor.
(C) an ac twophase induction motor.
(D) an ac series motor.
Ans: B
Q.23 The size of a conductor used in power cables depends on the
(A) operating voltage. (B) power factor.
(C) current to be carried. (D) type of insulation used.
Ans: C
Q.24 Out of the following methods of heating the one which is independent of supply frequency is
(A) electric arc heating (B) induction heating
(C) electric resistance heating (D) dielectric heating
Ans: C
Q.25 A twowinding single phase transformer has a voltage regulation of 4.5% at fullload and unity powerfactor. At fullload and 0.80 powerfactor lagging load the voltage regulation will be
(A) 4.5%. (B) less than 4.5%.
(C) more than 4.5%. (D) 4.5% or more than 4.5%.
Ans: C 

% R = V_{r} cos F + V_{x} sin F = V_{r} p.f = cos F =1 \ F =0^{0} \ kVA = kW & kVAR =0 No reactive power component Percentage regulation (%R) = V_{r} cos F ± V_{x} sin F When cos F = 0.8 lagging %R = V_{r} cos F + V_{x} sin F = V_{r} (0.8) + V_{x} (0.6) %R = (0.8)V_{r} +(0.6) V_{x} at p.f 0.8 lagging and %R = V_{r} at unity p.f 

Q.26 In a dc shunt motor the terminal voltage is halved while the torque is kept constant. The resulting approximate variation in speed _{} and armature current _{} will be
(A) Both _{} and _{} are doubled. (B) _{} is constant and _{}is doubled.
(C) _{} is doubled while_{}is halved. (D) _{} is constant but _{} is halved.
Ans: B 


N α V – I_{a}R or N α E_{b} T α I_{a} F , F α I_{a } \ T α I_{a}^{2} 
Q.27 A balanced threephase, 50 Hz voltage is applied to a 3 phase, 4 pole, induction motor. When the motor is delivering rated output, the slip is found to be 0.05. The speed of the rotor m.m.f. relative to the rotor structure is
(A) 1500 r.p.m. (B) 1425 r.p.m.
(C) 25 r.p.m. (D) 75 r.p.m.
Ans: D
N_{S} = 120f /P = 120 x 50 /4 =1500rpm N = N_{S} ( 1s) = 1500 (10.05) = 1425 \relative speed = 1500 – 1425 = 75 rpm 
Q.28 An alternator is delivering rated current at rated voltage and 0.8 powerfactor lagging case. If it is required to deliver rated current at rated voltage and 0.8 powerfactor leading, the required excitation will be
(A) less. (B) more.
(C) more or less. (D) the same.
Ans: B 


Over excitation gives leading power factor and under excitation gives lagging p.f .
Q.29 A ceiling fan uses
(A) splitphase motor.
(B) capacitor start and capacitor run motor.
(C) universal motor.
(D) capacitor start motor.
Ans: D 


To give starting torque and to maintain speed.
Q.30 A stepper motor is
(A) a dc motor. (B) a singlephase ac motor.
(C) a multiphase motor. (D) a two phase motor.
Ans: D 


Stepper motor works on 1phaseON or 2phase –ON modes of operation
Q.31 The ‘sheath’ is used in cable to
(A) provide strength to the cable.
(B) provide proper insulation.
(C) prevent the moisture from entering the cable.
(D) avoid chances of rust on strands.
Ans: A 


The sheath in underground cable is provided to give mechanical strength.
Q.32 The drive motor used in a mixergrinder is a
(A) dc motor. (B) induction motor.
(C) synchronous motor. (D) universal motor.
Ans: D 


The universal motor is suitable for AC & DC both supply systems.
Q.33 A 1:5 stepup transformer has 120V across the primary and 600 ohms resistance across the secondary. Assuming 100% efficiency, the primary current equals
(A) 0.2 Amp. (B) 5 Amps.
(C) 10 Amps. (D) 20 Amps.
Ans: A
I_{1}= V_{1} /R_{1} = 120/600 = 0.2 (h = 100%, losses are zero \V_{1} = V_{R} = I_{1}R_{1})
Q.34 A dc shunt generator has a speed of 800 rpm when delivering 20 A to the load at the terminal voltage of 220V. If the same machine is run as a motor it takes a line current of 20A from 220V supply. The speed of the machine as a motor will be
(A) 800 rpm. (B) more than 800 rpm.
(C) less than 800 rpm. (D) both higher or lower than 800 rpm.
Ans: C
N_{g}= E_{g} (60A / Fpz) E_{g} = V + I_{a} R_{a }; in generator
N_{m}= E_{b} (60A / Fpz) E_{b} = V  I_{a} R_{a }; in motor
E_{g} > E _{b} for same terminal voltage
Therefore, N_{g} > N _{m}
Q.35 A 50 Hz, 3phase induction motor has a full load speed of 1440 r.p.m. The number of poles of the motor are
(A) 4. (B) 6.
(C) 12. (D) 8.
Ans: A
N= N_{s} (1S) = N_{S} –N_{S} x S
1440 = N_{s} (1S)
N_{s} = 1440 / (1S)
N_{s} = (120 f/ p) = 120 x 50/p = 6000 p
N_{s} will be closer to N i.e 1440
When P=2 ; N_{s} = 3000 rpm , not close to N
When P=4 ; N_{s} = 1500 rpm , it is closer to N
Therefore P =4 for N=1440
Q. 36 In a 3phase synchronous motor
(A) the speed of stator MMF is always more than that of rotor MMF.
(B) the speed of stator MMF is always less than that of rotor MMF.
(C) the speed of stator MMF is synchronous speed while that of rotor MMF is zero. (D) rotor and stator MMF are stationary with respect to each other.
Ans: D
Because, Motor is magnetically locked into position with stator, the rotor poles are engaged with stator poles and both run synchronously in same direction Therefore, rotor & stator mmf are stationary w.r.t each other.
Q.37 In a capacitor start singlephase induction motor, the capacitor is connected
(A) in series with main winding.
(B) in series with auxiliary winding.
(C) in series with both the windings.
(D) in parallel with auxiliary winding.
Ans: B
To make single phase motor self start. We split the phases at 90 degree. Hence, motor behaves like a two phase motor.
Q.38 A synchro has
(A) a 3phase winding on rotor and a singlephase winding on stator.
(B) a 3phase winding on stator and a commutator winding on rotor.
(C) a 3phase winding on stator and a singlephase winding on rotor.
(D) a singlephase winding on stator and a commutator winding on rotor.
Ans: C
Synchros : The basic synchro unit called a synchro transmitter. It’s construction similar to that of a Three phase alternator.
Q.39 As the voltage of transmission increases, the volume of conductor
(A) increases. (B) does not change.
(C) decreases. (D) increases proportionately.
Ans: C
Decreases due to skin effect.
Q.40 The size of the feeder is determined primarily by
(A) the current it is required to carry.
(B) the percent variation of voltage in the feeder.
(C) the voltage across the feeder.
(D) the distance of transmission.
Ans: A
Size of conductor depends upon amount of current flow.
Q. 41 The boundary of the protective zone is determined by the
(A) Location of CT (B) sensitivity of relay used
(C) Location of PT (D) None of these
The boundary of the protective zone is determined by the sensitivity of relay used. If the relay is more sensitive, the protective zone will be increased.
Q.42 In a three phase transformer, if the primary side is connected in star and secondary side is connected in delta, what is the angle difference between phase voltage in the two cases.
(A) delta side lags by 30°. (B) star side lags by 30°.
(C) delta side leads by 30°. (D) star side leads by 30°.
Ans: C
This is vector group and has +30° displacement. Therefore, delta side leads by +30°.
Q.43 To achieve low PT error, the burden value should be ____________.
(A) low (B) high
(C) medium (D) none of the above
Ans: A
In a Potential transformer, burden should be in permissible range to maintain errorless measurement.
Q.44 Slip of the induction machine is 0.02 and the stator supply frequency is 50 Hz.
What will be the frequency of the rotor induced emf?
(A) 10 Hz. (B) 50 Hz.
(C) 1 Hz. (D) 2500 Hz.
Ans: C
Given : s = 0.02 ; f = 50 Hz
Therefore, frequency of rotor induced emf = s f
= 0.02 x 50 = 1.0 Hz
Q.45 A 4 pole lap wound dc shunt motor rotates at the speed of 1500 rpm, has a flux of 0.4 mWb and the total number of conductors are 1000. What is the value of emf?
(A) 100 Volts. (B) 0.1 Volts.
(C) 1 Volts. (D) 10 Volts.
Ans: D
Given N = 1500 rpm, F = 0.4 mWb, Z = 1000, P = 4, & A= 4
Therefore, E_{b} = NFPZ / 60 A
= 1500 x 0.4 x 4 x 1000 x 10^{3 }/ 60 x 4
Q.46 The synchronous reactance of the synchronous machine is ______________.
(A) Ratio between open circuit voltage and short circuit current at constant field current
(B) Ratio between short circuit voltage and open circuit current at constant field current
(C) Ratio between open circuit voltage and short circuit current at different field current
(D) Ratio between short circuit voltage and open circuit current at different field current
Ans. A
The Synchronous reactance of a synchronous machine is a total steady state reactance, presented to applied voltage, when rotor is running synchronously without excitation.
Therefore , X_{S} = E_{f} / I_{S}
= Emf of OC for same I_{f} / short circuit current
Q.47 A 3 stack stepper motor with 12 numbers of rotor teeth has a step angle of ____________.
(A) 12° (B) 8°
(C) 24° (D) 10°
Ans. D
Given m = 3, N_{r} = 12
Step angle = 360 / m x N_{r} = 360 /3 x 12 = 10°
Q.48 In case of a universal motor, torque pulsation is minimized by _________.
(A) load inertia (B) rotor inertia
(C) both rotor and load inertia (D) none of the above
Ans: C
In a universal motor, torque pulsation is minimized by rotor and load inertia.
Q.49 Oilfilled cable has a working stress of __________ kV/mm
(A) 10 (B) 12
(C) 13 (D) 15
Ans: D
This is defined by dielectric strength of mineral oil i.e. 15 kV/mm.
Q.50 Inverse definite minimum time lag relay is also called ___________
(A) pilot relay. (B) differential relay.
(C) over current relay. (D) directional overcurrent relay.
Ans: B
Inverse definite minimum time lag relay characteristic is inverse but minimum time is fixed. The operating time is inversely proportional to the magnitude of actuating quantity.
Q.51 Specific heat of nickel –chrome is _____________
(A) 0.112 (B) 0.106.
(C) 0.108. (D) 0.110.
Ans: None of these
Q.52
The polarity test is not necessary for the singlephase transformer
shown in Fig. 1 so as to correctly determine _____________of the transformer.
(A) shunt branch parameters.
(B) transformation ratio.
(C) series parameters.
(D) any of the above characteristics.
Polarity test is required for parallel operation of transformers to know the direction of current flow in secondary circuit w.r.t primary circuit.
Q.53 The
shortcircuit ratio of a typical synchronous machine is obtained from the OCC
and SCC curves of Fig.2 as
(A) _{}
(B) _{}
(C) _{}
(D) _{}
Ans: B
As shown in SCC curve the ratio of two field currents
Q.54 The speedtorque characteristics of a DC series motor are approximately similar to those of the _________motor.
(A) universal (B) synchronous
(C) DC shunt (D) twophase
Ans: A
Universal motor has same characteristics as DC series motor and also known as an a.c series motor.
Q. 55 The rotor frequency for a 3 phase 1000 RPM 6 pole induction motor with a slip of 0.04 is________Hz
(A) 8 (B) 4
(C) 6 (D) 2
Ans: D
Given: N=1000 rpm ; P= 6; s= 0.04;
and f = NÍP/ 120
= 1000Í6/120
= 50 Hz
Rotor frequency f_{r}=sÍf = 0.04Í50
= 2.0 Hz
Q.56 The torquespeed characteristics of an a.c. operated universal motor has a ______characteristic and it______ be started under noload condition.
(A) inverse, can (B) nearly inverse, can
(C) inverse, cannot (D) nearly inverse, cannot
Ans: C
If torque is zero then speed may exceed up to infinite, that is dangerous for machine and machine can be damaged.
N α 1/ T
Q.57 In the heating process of the ________type a simple method of temperature control is possible by means of a special alloy which loses its magnetic properties at a particular high temperature and regains them when cooled to a temperature below this value.
(A) Indirect induction over (B) core type induction furnace
(C) coreless induction furnace (D) high frequency eddy current
Ans: D
Magnetic property of alloy changes with change of the temperature and
Heat is produced due to eddy current = i^{2}R and i α f^{2}
Q.58 In order to reduce the harmful effects of harmonics on the A.C. side of a high voltage D.C. transmission system ______are provided.
(A) synchronous condensers (B) shunt capacitors
(C) shunt filters (D) static compensators
Ans: C
X_{c}= 1/ ωc
Q.59 An a.c. tachometer is just a ________with one phase excited from the carrier frequency.
(A) twophase A.C. servomotor (B) twophase induction motor
(C) A.C. operated universal motor (D) hybrid stepper motor.
Ans: D
This is a special purpose machine whose stator coil can be energized by electronically switched current.
Q.60 The torque, in a _____________is proportional to the square of the armature current
(A) DC shunt motor (B) stepper motor
(C) 2phase servomotor (D) DC series motor
T_{a }α Φ.Ia and Φ α Ia ; therefore T_{a }α Ia^{2}
Q.61 The synchronous speed for a 3 phase 6pole induction motor is 1200 rpm. If the number of poles is now reduced to 4 with the frequency remaining constant, the rotor speed with a slip of 5% will be _________.
(A) 1690 rpm (B) 1750 rpm
(C) 1500 rpm (D) 1710 rpm
Ans: D
Given : N_{s1} =1200 , P_{1}= 6,
P_{2} = 4, s = 0.05,
Frequency f = N_{s}ÍP/120
= 120Í6/120 = 60 Hz
rotor frequency f^{/} = s.f = 0.05 Í60 = 3.0 H_{z}
Now, N_{s2 }= 120 Í60 /4 = 1800 and Ns – N = 120 f / P_{2}
Therefore, N=N_{s} 120 f / P_{2 }= 1800120Í0.05Í60/4 = 180090 = 1710
Q.62 The eddy current loss in an ac electric motor is 100 watts at 50 Hz. Its loss at 100 Hz will be
(A) 25 watts (B) 59 watts
(C) 100 watts (D) 400 watts
Ans: D 


Eddy current losses α f^{2}
New loss α (2f)^{2}
New loss α 4f^{2}
\ 4 times
Q.63 The maximum power for a given excitation in a synchronous motor is developed when the power angle is equal to
(A) 0^{o} (B) 45^{o}
(C) 60^{o} (D) 90^{o}
Ans: A 


P = VI cosF
P_{max} = VI
\F = 0^{0}
Q. 64 A commutator in a d.c. machine
(A) Reduces power loss in armature.
(B) Reduces power loss in field circuit.
(C) Converts the induced a.c armature voltage into direct voltage.
(D) Is not necessary.
Ans: C 

As name suggests, it commutes ac into dc.
Q.65 The speed of a d.c. shunt motor at noload is
(A) 5 to 10% (B) 15 to 20%
(C) 25 to 30% (D) 35 to 40%
higher than its speed at rated load.
Ans: A 


T_{a} α F I_{a ,}, F = constant, \T α I_{a} N α E_{b} / F or N α E_{b} initially E_{b} less , so speed is less. 
Q.66 The efficiency of a transformer is mainly dependent on
(A) core losses. (B) copper losses.
(C) stray losses. (D) dielectric losses.
Ans: A 


Core loss has prominent value over other losses 
Q.67 When two transformers are operating in parallel, they will share the load as under:
(A) proportional to their impedances.
(B) inversely proportional to their impedances.
(C) 50%  50%
(D) 25%75%
Ans: A 


High rating transformer has higher impedance. kVA rating α Impedance of transformer 
Q.68 If the voltage is reduced to half, the torque developed by an induction motor will be reduced to
(A) _{} of original torque. (B) _{} of original torque.
(C) _{} of original torque. (D) _{} of original torque.
Ans: B T_{g} α V or T_{g} α P_{m} (rotor gross output) 
Q.69 A 3phase, 400 votts, 50 Hz, 100 KW, 4 pole squirrel cage induction motor with a rated slip of 2% will have a rotor speed of
(A) 1500 rpm (B) 1470 rpm
(C) 1530 rpm (D) 1570 rpm
Ans: B 


N = N_{S} (1S) and N_{S} =120 f / p =120 x 50 /4 = 1500 rpm 
Q.70 If the phase angle of the voltage coil of a directional relay is _{}, the maximum torque angle of the relay is
(A) _{} (B) _{}
(C) _{} (D) _{}
Ans: C 


Torque α Power Power α Voltage Therefore, It has same angle as ‘V’ has. 
Q.71 The voltage at
the two ends of a transmission line are 132 KV and its reactance is
40 ohm. The Capacity of the line is
(A) 435.6 MW (B) 217.8 MW
(C) 251.5 MW (D) 500 MW
Ans: A 


Line capacity is determined by power of line P = (V^{2}/R) or (V^{2}/Z) when cos F =1 
Q.72 A 220/440 V, 50 Hz, 5 KVA, single phase transformer operates on 220V, 40Hz supply with secondary winding open circuited. Then
(A) Both eddy current and hysteresis losses decreases.
(B) Both eddy current and hysteresis losses increases.
(C) Eddy current loss remains the same but hysteresis loss increases.
(D) Eddy current loss increases but hysteresis loss remains the same.
Ans: A 


W_{h} = k_{h}fB_{m}^{1.6} and W_{e} = k_{e}f^{2}B_{m}^{2}.k Therefore, hysteresis and eddy current losses will be decreased when frequency decreases. 
Q.73 A synchronous motor is operating on noload at unity power factor. If the field
current is increased, power factor will become
(A) Leading & current will decrease
(B) Lagging & current will increase.
(C) Lagging & current will decrease.
(D) Leading & current will increase.
Ans: A 


Initially synchronous motor is operating at no load and unity power factor. When field current increases, the excitation will increase. Therefore, p.f will be leading and current will be I CosF < I 
Q.74 A d.c. shunt motor runs at no load speed of 1140 r.p.m. At full load, armature reaction weakens the main flux by 5% whereas the armature circuit voltage drops by 10%. The motor full load speed in r.p.m. is
(A) 1080 (B) 1203
(C) 1000 (D) 1200
Ans: A 


N_{2 }/ N_{1} =E_{b2 }/E_{b1} x F_{1} / F_{2} ; F_{2} = 0.95F_{1} ; E_{b2 }= 0.9E_{b1} \ N_{2} /1140 = 0.9 x 1/0.95 N_{2} = 1080 
Q.75 The introduction of interpoles in between the main pole improves the performance of d.c. machines, because
(A) The interpole produces additional flux to augment the developed torque.
(B) The flux waveform is improved with reduction in harmonics.
(C) The inequality of air flux on the top and bottom halves of armature is removed.
(D) A counter e.m.f. is induced in the coil undergoing commutation.
Ans: D 


Q.76 The rotor power output of a 3phase induction motor is 15 KW and corresponding slip is 4%. The rotor copper loss will be
(A) 600 W. (B) 625 W
(C) 650 W (D) 700 W
Ans: B 


Rotor copper losses = rotor input rotor output and output = (1s) input \ Input = output/(1s) = 15000 /10.04 = 15625 
Q.77 The direction of rotation of hysteresis motor is reversed by
(B) Reversing supply lead
(C) Either A or B
(D) Neither A nor B
Ans: A 


This motor used single phase, 50Hz supply and stator has two windings. These are connected continuously from starting to running. 
Q.78 A 1.8°step, 4phase stepper motor has a total of 40 teeth on 8 pole of stator. The number of rotor teeth for their rotor will be
(A) 40 (B) 50
(C) 100 (D) 80
Ans: B 


Step angle ‘β’ = N_{S} – N_{r} / N_{S} N_{r} x 360^{0} \ 18 = 40 + N_{r}/40 N_{r} x 360^{0} N_{r} = 50 
Q.79 Low head plants generally use
(A) Pelton Turbines (B) Francis Turbine
(C) Pelton or Francis Turbine (D) Kaplan Turbines
Ans: A 


Q.80 The charging reactance of 50 Km length of line is 1500Ω. The charging reactance for 100Km length of line will be
(A) 1500 Ω (B) 3000 Ω
(C) 750 Ω (D) 600 Ω
Ans: B 


Characteristic reactance per km = 1500/50 = 30 ohms \ Characteristic reactance per 100km = 30 x 100 = 3000 ohms 
Q.81 Electric ovens using heating elements of _______________ can produce temperature upto 3000°C.
(A) Nickel (B) Graphite
(C) Chromium (D) Iron
Ans: C Chromium has high melting point. 
Q.82 In DC generators, armature reaction is produced actually by
(A) Its field current. (B) Armature conductors.
(C) Field pole winding. (D) Load current in armature.
Ans: D 


Because load current in armature gives rise to armature mmf which react with main field mmf. 
Q.83 Two transformers operating in parallel will share the load depending upon their
(A) Rating. (B) Leakage reactance.
(C) Efficiency. (D) Perunit impedance.
Ans: A 


Transformers having higher kVA rating will share more load. 
Q.84 As compared to shunt and compound DC motors, the series DC motor will have the highest torque because of its comparatively ____________ at the start.
(A) Lower armature resistance. (B) Stronger series field.
(C) Fewer series turns. (D) Larger armature current.
Ans: D 


T α F I_{a}^{ } (before saturation) F α I_{a}^{ } T α I_{a}^{ 2} 
Q.85 A 400kW, 3phase, 440V, 50Hz induction motor has a speed of 950 r.p.m. on fullload. The machine has 6 poles. The slip of the machine will be _______________.
(A) 0.06 (B) 0.10
(C) 0.04 (D) 0.05
Ans: D
N = N_{s} (1S) 950 = 120 x 50 (1S)/6 S = 0.05 
Q.86 Reduction in the capacitance of a capacitorstart motor, results in reduced
(A) Noise. (B) Speed.
(C) Starting torque. (D) Armature reaction.
Ans: C 


Reduction in the capacitance reduces starting voltage, which results in reduced starting torque. 
Q.87 Regenerative braking
(A) Can be used for stopping a motor.
(B) Cannot be easily applied to DC series motors.
(C) Can be easily applied to DC shunt motors
(D) Cannot be used when motor load has overhauling characteristics.
Ans: B 


Because reversal of I_{a} would also mean reversal of field and hence of E_{b} 
Q.88 At present level of technology, which of the following method of generating electric power from sea is most advantageous?
(A) Tidal power. (B) Ocean thermal energy conversion
(C) Ocean currents. (D) Wave power.
Ans: A 


At present level of technology, tidal power for generating electric power from sea is most advantageous because of constant availability of tidal power. 
Q.89 If the field circuits of an unloaded salient pole synchronous motor gets suddenly open circuited, then
(A) The motor stops.
(B) It continues to run at the same speed.
(C) Its runs at the slower speed.
(D) It runs at a very high speed.
Ans: B 


The motor continues to run at the same speed because synchronous motor speed does not depend upon load, Na f. 
Q.90 Electric resistance seam welding uses __________ electrodes.
(A) Pointed (B) Disc.
(C) Flat (D) Domed
Ans: B 

Disc type electrodes are used for electric resistance seam welding. 
Q.91 For LV applications (below 1 kV), ______________ cables are used.
(A) Paper insulated. (B) Plastic.
(C) Single core cables. (D) Oil filled.
Ans: C 

For low voltage applications single core cables are suitable. 
Q.92 No load current in a transformer:
(A) lags the applied voltage by 90°
(B) lags the applied voltage by somewhat less than 90°
(C) leads the applied voltage by 90°
(D) leads the applied voltage by somewhat less than 90°
The primary input current under no load conditions has to supply (i) iron losses in the core i.e hysteresis loss and eddy current loss (ii) a very small amount of Cu loss in the primary (there being no Cu loss in secondary as it is open)
Q.93 A transformer operates most efficiently at 3/4th full load. Its iron (P_{I}) and copper loss (P_{Cu}) are related as:
(A) _{} (B) _{}
(C) _{} (D) _{}
Ans: D
If P_{Cu} is the Cu loss at full load, its value at 75% of full load is
P_{Cu} x (0.75)^{2} = 9/16 P_{Cu}
At maximum efficiency, it equals the iron loss P_{I} which remains constant through out. Hence max. efficiency at
P_{I} = 9/16 P_{Cu}
Or P_{I }/ P_{Cu} = 9/16
Q.94 In a salient pole synchronous machine (usual symbols are used):
(A) _{} (B) _{}
(C) _{} (D) _{}
Ans: C
Since reluctance on the q axis is higher, owing to the larger air gap, hence x_{q} < x_{d}
Q.95 The armature of a dc machine is laminated to reduce:
(A) Eddy current loss (B) Hysteresis loss
(C) copper losses (D) friction and windage losses
Ans: A
Thinner the laminations, greater is the resistance offered to the induced e.m.f., smaller the current and hence lesser the I^{2}R loss in the core.
Q.96 The resistance representing mechanical output in the equivalent circuit of an induction motor as seen from the stator is:
(A) _{} (B) _{}
(C) _{} (D) _{}
Ans: A
Mechanical Power developed by the rotor (P_{m}) or gross power developed by rotor (P_{g})
= rotor input –rotor Cu losses
= (3I^{/2} R_{2}^{/ }/ S) (3I^{/2} R_{2}^{/ })
= 3I^{/2} R_{2}^{/ }(1/ S 1)
Q.97 A single phase Hysteresis motor
(A) can run at synchronous speed only
(B) can run at sub synchronous speed only
(C) can run at synchronous and super synchronous speed
(D) can run at synchronous and sub synchronous speed
Ans: A
The rotor revolves synchronously because the rotor poles magnetically lock up with the revolving stator poles of opposite polarity
Q. 98 The temperature of resistance furnaces can be controlled by changing the:
(A) applied voltage (B) number of heating elements
(C) circuit configuration (D) All of the above
Ans: D
Temperature of resistance furnaces can be controlled by changing either applied voltage or by number of heating elements or by circuit configuration.
Q.99 The line trap unit employed in carrier current relaying:
(A) offers high impedance to 50 Hz power frequency signal
(B) offers high impedance to carrier frequency signal
(C) offers low impedance to carrier frequency signal
(D) Both (A) & (C)
Ans: B
The line trap unit employed in carrier current relaying offers high impedance to carrier frequency signal.
Because carrier frequency range is 35 km – 500 kHz
X_{L} = 2Π f _{l}
Where f increases X_{L} will also increases
Q.100 For a line voltage V and regulation of a transmission line R
(A) _{}^{ }(B) _{}
(C) _{} (D) _{}
Ans: B
Regulation = (V_{0 }– V_{L} ) / V_{0} , if V_{L} is high the (V_{0 }– V_{L} ) will be low.
Therefore R α 1/V
Q.1 Calculate the voltage regulation of a transformer in which ohmic drop is 2% and the reactance drop in 5% of the voltage at 0.8 lagging power factor. (7)
Ans: The expression for % voltage regulation is
% voltage regulation=_{}_{} …(1)
where _{} = rated secondary voltage while supplying full load at a specified power factor.
And _{} = secondary voltage when load is thrown off
Equation (1)
can be written as
% Voltage regulation =_{}
_{} …(2)
The quantities within the brackets are given in the problem as 2%( percent ohmic drop) and 5% (percent reactance drop). Also _{} is the lagging power factor angle. The plus sign in Equation (2) is because of the lagging nature of current.
Here _{}_{} and hence _{}
Voltage regulation _{}
_{}
_{}
Hence voltage regulation _{}
Q.2 Derive the expression of torque produced in a d.c. motor. (7)
fluxdensity B force on conductors
Ans:
conductor current (I_{c})
Fig. C1 Torque production in dc machine
Fig. C1 shows the flux density wave in the air gap and the conductor current distribution in the developed armature for one polepair. The force on the conductors is unidirectional. Each conductor, as it moves around with the armature, experiences a force whose time variation is a replica of the flux density(B).
Therefore, the average conductor force
_{}_{} (1)
where B_{av} = average flux density over a pole.
_{} = active conductor length, and _{}=conductor current
Total force
_{}, where z=total number of conductors (2)
This force (and therefore torque) is constant because both the flux density wave and current distribution are fixed in space at all times. Now the torque developed is
_{} (3)
where _{} is the mean air gap radius
The flux/pole can be expressed as
_{} (4)
where _{}= polepitch = _{}
so
_{}
or
_{} (5)
Substituting for _{} in (3)
_{}
or
_{}
A lap winding is assumed here. It has A=p parallel paths such that the armature current I_{a} divides out into ‘A’ paths giving a conductor current of
_{}
Thus
_{} Nm
_{} where_{}=constant (6)
Q.3 A 230 V d.c. shunt motor with constant field drives a load whose torque is proportional to the speed. When running at 750 rpm it takes 30 A. Find the speed at which it will run if a 10 ohm resistance is connected in series with the armature. The armature resistance may be neglected. (7)
Ans: Fig. C2 shows a dc shunt motor
I_{a} R_{a} I_{f} 230 V
Using equation 6 of Question 2, the torque
T=_{}
Original variables are _{} and _{}
Final variables are _{}and _{}
Now _{}^{ }(1)
Here _{}as flux is constant
Since torque is proportional to speed
_{}from equation (1) (2)
And _{}^{ }(3)
Back emf _{}
Or _{}
Or _{}
Or _{}
This gives
_{}
Therefore
_{}
_{}
_{}
Q.4 The power input to a 500 V, 50 Hz, 6 pole 3 phase squirrel cage induction motor running at 975 rpm is 40 KW. The stator losses are 1 KW and the friction and windage losses are 2 KW. Calculate
(i) Slip (ii) Rotor copper loss
(iii) Mechanical power developed (iv) The efficiency. (7)
Ans: _{}
_{}
Power across air gap(P_{G}) = power input  stator copper loss
Thus P_{G}=40KW1KW=39KW
Rotor copper loss=sP_{G}=0.025*39=0.975KW
Gross mechanical output=(1s)P_{G}=390.975=38.025KW
Net mechanical output=Gross mechanical outputfriction and winding loss
= (38.0252.000)KW
=36.025KW_{}
_{}
_{}
Note: Assume that the core loss is included in friction and windage loss and the total loss under this head is 2.0 kW
Q.5 A 120V, 60Hz, 1/4hp universal motor runs at 2000 rpm and takes 0.6A when connected to a 120V d.c. source. Determine the speed, torque and power factor of the motor when it is connected to a 120V, 60 Hz, supply and is loaded to take 0.6A (rms) of current. The resistance and inductance measured at the terminals of the motor are 20 ohm and 0.25H respectively. (7)
Ans: Universal
Motor: (A) When connected to a d.c. source it runs at 2000RPM and
0.6A
takes 0.6A [Fig F1]
E_{b} 20W 120V DC Source
_{}
_{}
When connected an ac source [120 Volts, 60Hz supply]
it takes 0.6A of current
_{}
R_{motor}=20_{}
L_{motor}=0.25H
X_{motor}=2_{}X60X0.25
Or, X_{motor}=94.25_{}
From the phasor diagram
_{}
_{}
_{}
Assume, same flux for the same current (i.e. _{} and _{})
_{}
Therefore
_{}
Power factor, _{}
_{}
_{}
Q.6 For a 4 KVA, 200/400 V, 50 Hz, 1 – phase transformer, calculate the efficiency, voltage at the secondary terminals and primary input current when supplying a full – load secondary current at 0.8 lagging power factor.
The following are the test results:
Open circuit with 200 V applied to the L.V. side: 0.8 A, 70 W. Short circuit with 20 V applied to the H.V. side: 10 A, 60 W. (14)
Ans: The transformer is supplying fullload secondary current at 0.8 lagging power factor
Full load secondary current _{}
From the open circuit test, core losses = 70W
From the S.C. test, full load copper losses = 60W
(a) Efficiency
_{}
=_{}
=_{}
(b) voltage at the secondary terminals is determined as follows with the help of equivalent circuit of Fig A3
I_{2} q E_{1 }=_{
}V_{2}a I_{2} V_{1} E_{2} E_{1} X_{eq} R_{eq} 200V
Fig. A3 Equivalent circuit referred to primary
Primary equt. resistance = r_{1 }+ a^{2}r_{2 = }a^{2} [equt. resistance referred to secy]
Also primary equt. reactance =x_{1} + a^{2}x_{2 }= a^{2} [equt. reactance referred to secy]
Where _{}
From the short circuit test conducted on the secondary side.
_{}
_{}
_{}
Equt. Resistance referred to primary _{}_{}
Equt. reactance referred to primary _{}
_{}
(E_{1} + 20 x 0.15 x0.8 + 20 x 0.48 x 0.6)^{2} + (20 x 0.48 x 0.8 – 20 x 0.15 x 0.6)^{2} = 200^{2}
_{}
This gives _{}
E_{2} = 191.75 x 2 = 383.5
So voltage at secondary terminals =383.5V
Primary input current with full load secondary current =20A
Q.7 Draw the per phase approximate equivalent circuit of a 3 – phase induction motor at slip ‘s’ and derive the expression for electromagnetic torque developed by the motor. Derive also the condition for maximum torque and the expression for the maximum torque. (14)
Ans:
I_{0} V b I_{1} a I^{’}_{2} X^{’}_{2} R^{’}_{2} R_{1} X_{m} X_{1} R_{1}
Fig B3 shows the perphase exact equivalent circuit of a 3phase induction motor. The power crossing the terminals ’ab’ in Fig B3 is the electrical power input per phase minus the stator copper loss and iron loss; Thus it is the power that is transferred from the stator to the rotor via the air gap. It is also known as the power across the air gap.
The Power across the air gap _{}
Rotor speed is _{} mech.rad./s
Electromagnetic torque developed is obtained as
(1  s) _{}T = P_{m} = (1 – s) P_{G}
_{ }or _{}
The condition (that is, slip) for maximum torque is obtained by equating _{}to zero.
With the approximation of _{}= _{} in Fig. B3 or using the approximate equivalent circuit.
_{}
_{}_{}
Equating _{}to zero gives the slip at maximum torque as
_{}
Also substitution of s_{m} for s in (I) gives the maximum torque as
_{}_{}
Q.8 A 230 V d.c. series motor has an armature resistance of 0.2 _{} and series field resistance of 0.10 _{}. Determine:
(i) the current required to develop a torque of 70 Nm at 1200 rpm
(ii) percentage reduction in flux when the machine runs at 2000 rpm at half the current. (14)
Ans:
DC Series Motor:
Back emf _{}
Define _{}
So _{}_{}
_{} Also _{}
_{}
So _{}
Also for a series motor
E_{a} = VI_{a} (R_{a}+R_{se})(B)
where V is the applied voltage and I_{a} is the current through armature and series field.
Here R_{a} =.2_{} and _{}
From (A) and (B)
_{}
_{}
also _{}
or
_{}_{}
From (C) and (D)
_{}
or _{}
_{}
This gives _{} or _{}
The second value (726.33A) is inadmissible.
(i) Hence the current required to develop a torque of 70Nm at 1200RPM is 40.33A
(ii) Machine runs at 2000RPM
_{}
_{}
_{}
From (C) _{}
Division of (F) by (E) gives
_{}
or
_{}
This gives _{}
or reduction in flux in the second case is 38.3% of the original flux.
Q.9 The effective resistance of a 3 – phase, Y – connected 50 Hz, 2200 V synchronous generator is 0.5 _{} per phase. On short circuit a field current of 40 A gives the full load current of 200 A. An emf (line to line) of 1100 V is produced on open circuit with the same field current. Determine the synchronous impedance. Also compute the power angle and voltage regulation at full – load 0.8 lagging p.f. (14)
Ans:
The occ and scc characteristics of the synchronous generator are given in
Fig. M4
Synchronous impedance _{}
Thus _{}
Percentage voltage regulation is defined as _{}
The phasor diagram is given in Fig. M5
Fig. M5 Phasor diagram IX_{s} cos f = 0.8 I_{a} = 200A q d f V_{t(rated)} E_{f}
Here _{}
_{}
_{}
where V_{t}(terminal voltage) and E_{f }(field voltage) are per phase values
_{} _{}
Percent regulation_{}_{}
_{}
_{}
_{}
_{} _{}
Thus power angle _{}
Q.10 A 100 KVA, 2400/240 V, 50 Hz, 1phase transformer has noload current of 0.64 A and a core loss of 700 W, when its high voltage side is energized at rated voltage and frequency. Calculate the two components of noload current. If this transformer supplies a load current of 40 amp at 0.8 lagging power factor at its low voltage side, determine the primary current and its power factor. Ignore leakage impedance drop. (12)
Ans:
100 KVA, 2400/240 V, 50Hz, 1F
No load: 
I_{o} = 0.64 A
W_{o}= 700W
Now , I_{o}^{2} = I_{w}^{2}^{ }+ I_{m}^{2}
Magnetizing component I_{m} = (I_{o}^{2} – I_{w}^{2})^{1/2}
= (0.64^{2} – 0.2916^{2})^{1/2}
I_{2}_{ }= 40 A
F_{2} = 0.8 lag
cos F_{o} = W_{o}/(V_{o}_{ }I_{o})
= 700/(2400x0.64)
= 0.455
F_{o} = cos ^{–1} 0.455
= 62.88^{o}
F_{2} = cos^{ }^{–1} 0.8
= 36.86^{o}
Now, turn ratio K =V_{1}/V_{2} = 240/ 2400 = 0.1
I_{2}^{1} = KI_{2} = 40x0.1 = 4 A
Angle between I_{0} and I_{2}^{1} = 62.88^{o} – 36.86^{o}
= 26.02^{o}
I_{0} = 0.64 Ð–62.88^{o}
I_{2}^{1} = 4Ð–36.86^{o}
I_{1}_{ }= I_{o} + I_{2}^{1}^{ }= 0.64 [cos( 62.88) – j sin (62.88) ] + 4[ cos(36.86) – jsin( 36.86) ]
= 4.583
I_{1}_{ }= 4.583 A
cos F_{1} = cos 40.37 =0.7618 lag
Q.11 A shunt generator has an induced emf of 254 V. When the generator is loaded, the terminal voltage is 240 V. Neglecting armature reaction, find the load current if the armature resistance is 0.04 ohm and the field circuit resistance is 24 ohms. (10)
Ans:
E_{g} = 254 V
R_{a}= 0.04 W, R_{sh} = 24W
E_{g} = V + I_{a}_{ }R_{a}
E_{g} = V + (I_{L} + I_{sh}) R_{a}
I_{sh} = V / R _{sh} = 240/24 = 10A.
254 = 240 + (I_{L} + 10) 0.04
I_{L} = 340A
Q.12 The shaft output of a threephase 60 Hz induction motor is 80 KW. The friction and windage losses are 920 W, the stator core loss is 4300 W and the stator copper loss is 2690 W. The rotor current and rotor resistance referred to stator are respectively 110 A and 0.15 _{}. If the slip is 3.8%, what is the percent efficiency? (12)
Ans:
P_{m}_{ =}output = 80 KW
Stator core loss = 4300 W
Stator copper loss = 2690W
Slip = 3.8%
Gross mech output = P_{m} + windage and friction losses
= 80 KW + 920 W
= 80.92KW
rotor input / rotor gross output = 1/(1–s)
rotor input = rotor gross output / (1–s) = 80.92 KW /(1–0.038) = 84.11 KW
we know that ;
stator input = rotor input + stator core loss + stator cu loss
= 84.11 KW +4300 W + 2690 W = 91.1 KW
% h = (rotor output / stator input) x 100
= (80/91.1) x 100 KW
Q.13 A 6 pole 3 phase induction motor develops 30 H P including mechanical losses totalling 2 H P, at a speed of 950 RPM on 550 volt, 50 Hz mains. If the power factor is 0.88 and core losses are negligible, calculate:
(i) The slip
(ii) The rotor copper loss
(iii) The total input power if the stator losses are 2 Kw
(iv) The line current. (6)
Ans:
P=6, 3j, output = 30 H.P
N = 950 rpm
V= 550 V
f = 50 Hz.
cosj = 0.88
slip, S = (N_{s}_{ }– N_{r})/N_{s}
N_{s} = 120 f / P = 120 x50/6 = 1000 r.p.m.
Rotor gross output = output + Mech. loss = 30 +2 =32 H.P.
Rotor cu loss/ Rotor gross output = S/(1–S)
Rotor cu loss = 0.05x32/0.95 = 1.684 H.P. = 1.684 x 0.746 = 1.323 kW
Rotor input = Rotor gross output/(1–S) = 32 /0.95 = 33.68 H.P.
Total input = Rotor input + cu loss + core loss
(33.68x745.7 W) + 2000 W + 0 = 27.115 KW
Line current = Total input/ (1.732 x 550 x 0.88)
= 32.34 ampere
Q.14 If the motor is fed from a 50 Hz 3 phase line, calculate:
(i) number of poles
(ii) slip at full load
(iii) frequency of rotor voltage
(iv) speed of rotor field wrt rotor
(v) speed of rotor field wrt to stator
(vi) speed of rotor field wrt stator field
(vii) speed of rotor at a slip of 10 percent. (6)
Ans:
(i) N_{s}= 120f/P
(ii) S= (1000 – 950)/1000 =0.05;
(iii) frequency of rotor voltage =Sf=0.05x50=2.5 Hz
(iv) Speed of rotor field w.r.t. rotor =(120Sf)/P =120x2.5/6= 50 rpm
(v) Speed of rotor field w.r.t stator = 950+50=1000 rpm;
(vi) Speed of rotor field w.r.t stator field =1000–1000=0 rpm;
(vii) Speed of rotor at a slip of 10%= N_{s} (1–S)=900 rpm;
Q.15 Three singlephase, 50 kVA, 2300/ 230 V, 60 Hz transformers are connected to form a 3phase, 4000V / 230V transformer bank. The equivalent impedance of each transformer referred to lowvoltage is 0.012 + j 0.016 _{}. The 3phase transformer supplies a 3phase, 120 kVA, 230 V, 0.85 powerfactor (lagging) load.
(i) Draw a schematic diagram showing the transformer connection.
(ii) Determine the winding currents of the transformer.
(iii) Determine the primary voltage (line to line) required. (3 x 3)
Ans: Given :
single phase ; P_{0} = 50kVA ; 2300/230V, 60Hz
no. of transformers are 3 (three) to form a 3 F transformer
4000/230V , P_{0} = 120 kVA , 230V and cosF =0.85 lagging
Z_{02} = (0.012 + j 0.016) Ω and
Z_{01} = Z_{02} /K^{2}
(i) Schematic diagram to show transformer connection
(ii) Calculation of winding currents of transformer:
Given , P_{0} = 120 kVA
P_{0} =√ 3 V_{L}I_{L} cosF
I_{L} = I_{ph} ; V_{L} = √ 3 V_{ph } (_{ }star connection)
\ I_{L } = 120 x cos F x 10^{3} / √ 3 V_{L} cosF
= 120 x 10^{3} /√ 3 x 230
= 30.12 amp
\ secondary line current = 30.12 amp.
(iii) Primary current I_{1 } = k I_{2}
= 230/4000 x 30.12 = 1.732 amp.
Primary line voltage = 4000 volts.
Q.16 A pair of synchronous machines, on the same shaft, may be used to generate power at 60 Hz from the given source of power at 50 Hz. Determine the minimum number of poles that the individual machines could have for this type of operation and find the shaftspeed in r.p.m. (4+4)
Ans: Motor & generator (synchronous machine) are coupled. Therefore,
N_{S(m)} = N_{S(g)}
N_{S(m)} = 120 f_{m} /P_{m} ; N_{S(g)} = 120 f_{g} / P_{g}
Where : N_{S(m)} = synchronous speed of motor
N_{S(g) } = synchronous speed of generator
f_{m} = frequency of motor power
f_{g} = frequency of generator power
P_{m} = motor poles
P_{g } = generator poles
\ 120 f_{m} /P_{m} = 120 f_{g} / P_{g}
120 x 50 / P_{m} = 120 x 60 /P_{g}
\ P_{g} /P_{m} = 6/5
P_{g} : P_{m} = 6 : 5
Therefore minimum requirement of poles for motor
P_{m} = 10 (5 x 2)
P_{g} = 12 (6 x 2)
Now synchronous speed or shaft speed = 1200 x 50 / 10 = 600 rpm
Q.17 A 240V dc shunt motor has an armature resistance of 0.4 ohm and is running at the fullload speed of 600 r.p.m. with a full load current of 25A. The field current is constant; also a resistance of 1 ohm is added in series with the armature. Find the speed (i) at the fullload torque and (ii) at twice the fullload torque. (6)
Ans: In a DC shunt motor
V = 240V
R_{a} = 0.4Ω
N_{1 }= 600rpm (full load speed)
I_{a} = 25A and, I_{Sh} is constant
R = 1 Ω added in series with armature
E_{b1 } = V  I_{a}R_{a}
= 240 25 x 0.4
= 230 volts
E_{b2 = }V  I_{a }(R_{a}+ R)
= 240 25 ( 0.4+ 1)
= 201 volts
Now N_{1} / N_{2} = E_{b1} / E_{b2} x F_{2 } / F_{1 } (F_{1 } = F_{2 } = constant)
N_{2} = N_{1} x E_{b2} / E_{b1} at full load torque
= 600 x 201/ 230
= 534.78
\ (i) speed of motor at full load = 535 rpm
Now,
N_{3} / N_{1} = E_{b3} / E_{b1} x F_{1 } / F_{3 } (F_{1 } = F_{2}= F_{3 } = constant)
And E_{b3} at twice the full load torque
\ I_{a2} = 2 I_{a} = 50 amp.
\ E_{b3} = 240 50 (1 +0.4) = 240 – 70 = 170 volts.
N_{3} = N_{1} x E_{b3} / E_{b1} = 600 x 170 /230 = 443.47 rpm
(ii) speed of motor at twice of load = 443 rpm
Q.18 A 400V, 4pole, 50 Hz, 3phase, 10 hp, star connected induction motor has a no load slip of 1% and full load slip of 4%. Find the following:
(i) Syn. speed (ii) noload speed (iii) fullload speed.
(iv) frequency of rotor current at fullload (v) fullload torque. (5 x 2 = 10)
Ans: Given :
V_{L} = 400 volts ; P = 4 nos, 50 Hz,
P_{0} = 10 HP = 735.5 x 10 = 7355 watt
i. Synchronous speed N_{S} = 120 f / p
= 120 x 50 / 4 = 1500 rpm
ii. No load speed at s = 0.01
N_{0} = N_{S} ( 1 – s) = 1500 ( 1 0.01)
= 1485 rpm
iii. Full load speed at s_{f} = 0.04
N_{fl} = N_{S} (1s_{f }) = 1500 ( 10.04)
= 1440 rpm
iv. Frequency of rotor current (f_{r}) = s_{f} .f = 0.04 x 50
= 2.0 Hz
v. Full load torque at shaft
T_{Sh} = 9.55 P_{0} / N_{fl}
= 9.55 x 7355 /1440
= 48.78 Nm
Q.19 A 2.2 kVA, 440 / 220 V, 50 Hz, stepdown transformer has the following parameters referred to the primary side : _{} _{} _{} and _{}. The transformer is operating at fullload with a powerfactor of 0.707 lagging. Determine the voltage regulation of the transformer. (10)
Ans:
Given : P_{0} = 2.2 kVA , 440/220 V , 50 Hz
R_{01 = } 3Ω , X_{01} = 4Ω , R_{m}= 2.5kΩ , X_{m} = X_{0} = 2kΩ
cosF = 0.707 lagging
Therefore sinF = 0.707
Find Voltage regulation
cosF = 0.707 ; therefore F = 45^{o}
(Voltage drop) = I_{2} (R_{01} cosF + X_{01} sinF)
and I_{2} = P_{0}/ V_{2} cosF
= 2.2 x 10^{3} x 0.707 / 220 x 0.707
= 1 x 10
I_{2} = 10A
Voltage drop = 10 (3 x 0.707 + 4 x 0.707)
= 10 (4.950)
= 49.50 Volts
Therefore, Voltage regulation = ( voltage drop /V_{2} ) x 100
= (49.50 / 220) x 100 = 22.50%
Q.20 A 9kVA, 208 V, 3phase, Yconnected, synchronous generator has a winding resistance of 0.1 ohm per phase and a synchronous reactance of 5.6 ohms per phase. Determine the voltage generated (exciting emf) by the machine when it is delivering fullload at 0.8 powerfactor lagging at rated voltage. Calculate the voltage regulation for rated load at 0.8 powerfactor (leading). (10)
Ans:
P_{0} = √3V_{L} I_{L} cos F = 9kVA ; I_{L} = I_{ph }= P_{0} / √3V_{L} cos F
V_{L} = 208 V ; 3F ; Y connected synch. Gen.
V_{L} = √3V_{P} ; I_{L} = I_{Ph} ; p.f = 0.8 V_{ph } = V_{L} / √3 = 208 / √3 = 120V
R_{a }= 0.1Ω /ph. , X_{a} = 5.6Ω /ph
Find E_{g} = ? , Regulation = ?
I = P_{0} / √3V_{L} cos F = (9 x 10 ^{3} x 0.8)/ (208 x 0.8 x 1.73) = 25 Amp.
E_{g} = √ (V_{P} cosF + I R_{a})^{2} + (V sinF + I X_{1})^{2}
= √ (120 x 0.8 + 25 x 0.1)^{2} + (120 x 0.6 + 25 x 5.6)^{2}
E_{g} = √ (96+2.5)^{2} +(72+140)^{2}
= 233.76 Volts
% regulation = (E_{g} V/ E_{g} ) x 100
={ (233.76 208)/ 233.76 } x 100
= 11.02%
Q.21 A
240V, 20 hP, 850 r.p.m., shunt motor draws 72A when operating under rated
conditions. The respective resistance of the armature and shunt field are
0.242 ohm and 95.2 ohms, respectively. Determine the percent reduction in the
field
flux required to obtain a speed of 1650 r.p.m., while drawing an armature
current of 50.4 A. (9)
Ans:
Given V = 240V
P_{i} = 20hp = 20 x 735.5 watt = 14.71 kW
Find Change in flux =?
I_{sh } = V/ R_{sh} = 240 / 95.2 = 2.5 Amp
N_{1} = 850 rpm ; I_{L } = 72 Amp. At rated load
R_{a} = 0.242, R_{sh} = 95.2Ω
N_{2} = 1650 rpm , I_{a2} = 50.4 Amp
\ E_{b1} = V – I_{a1} R_{a}
= 240 – 69.47 x 0.242 = 223.19 volt
and E_{b2} = V – I_{a2} R_{a}
= 240 – 50.4 x 0.242 = 227.80 volt
E_{b1} / E_{b2 } = N_{1} F_{1}/ N_{2} F_{2}
Therefore, F_{1}/ F_{2} = (E_{b1} / E_{b2 }) x (N_{2} / N_{1})
= (223.19 / 227.80) x (1650 / 850) = 36826.35/19363
F_{1}/ F_{2 } = 1.90/1 = 19/10
Therefore, change in flux ∆F = (F_{1}  F_{2} )/ F_{1} x 100
= 9/19 x 100 = 47.37%
Q.22 The power input to the rotor of a 3phase, 50 Hz, 6 Pole induction motor is 80 kW. The rotor emf makes 100 complete alternations per minute. Find
(i) the slip (ii) the motor speed and (iii) the mechanical power developed by the motor. (10)
Ans:
Given P_{i} = 80 kW ; 50Hz
P = 6
Rotor frequency f ^{/} =(100/60) = 5/3 = 1.67 Hz
S = f ^{/} / f = (5/3) / 50 = 0.033
Mechanical Power developed by motor = (1S) P_{i}
= (1 1/30) x 80 kW = 77.33kW
Q.23 The parameters of the equivalent circuit of a 150kVA,2400/240V transformer are:
R1=0.2ohm, R2=2 x 10^{– 3} ohm , X1=0.45 ohm, X2= 4.5 x 10^{– 3} ohm,
Ri =10 kohm, Xm = 1.6 kohm as seen from 2400 volts side.
Calculate:
(i) open circuit current, power and PF when LV side is excited at rated voltage. (8)
(ii) The voltage at which the HV side should be excited to conduct a shortcircuit test (LV side) with fullload current flowing. What is the input power and its power factor? (8)
Ans: Given Rating = 150kVA \ P_{o} =150kVA
2400/240 V \ V_{2} = 2400V; V_{1} = 240 V
R_{1} = 0.2Ω X_{1 }= 0.45Ω
R_{2} = 2 x 10^{3}Ω X_{2} = 4.5 x10^{3}Ω
R_{i} = 10 kΩ X_{m }= 1.6kΩ
Find I_{2}o = ?, P_{2}o = ?, cos F_{2}o = ?,
when L.V side is excited & H.V side is open circuit.
I_{fL} = ?, P_{i} = ?, cos F_{i} = ?,
when H.V side is excited & short circuit at L.V side
(i) Open Circuit power = V_{1} I_{0} cos F_{0 }
= 240 x I_{0} cos Fo …(1)
I_{0} = V_{1}/Z_{m} ; V_{1 = }240 volts; ;
Z_{m} = √(10^{2} + 1.6^{2}) = 10.13 kΩ (Z_{m} = √ R_{1}^{2} + X_{m}^{2})
I_{0} = 240 / 10.13 = 23.69 x 10^{3} Amp. Or 23.69 mA.
Power factor cos F_{0} = R_{i} / Z_{m}
= 10/10.13
cos F_{0} = 0.987
Now open circuit power (W_{0i}) = 240 x I_{0} x cos F_{0}
= 240 x 23.69 x 10^{3} x 0.987
W_{0i} = 5.6 watt
(ii) R_{01 }= R_{1} + R_{2} / K^{2}^{ } (K = V_{2} /V_{1} )
= 0.2 + 2x10^{3}/100
= 0.20002Ω
X_{01 }= X_{1} + X_{2} / K^{2}
= 0.45 + 4.5x10^{3}/100
= 0.450045Ω
Z_{01} = √ (R_{01}^{2} + X_{01}^{2})
= 0.4924Ω
Full load primary current (I_{1}) = 150000/2400 = 62.5 Amp (max.)
Short circuit p.f. = R_{01 }/ Z_{01} = 0.20002 / 0.4924 = 0.406
V_{SC} = I_{1}Z_{01} = 62.5 x 0.4924 = 30.78 Volt
\ power absorbed = I_{1}^{2}_{ }R_{01} = (62.5)^{2} x 0.20002 = 781.33 Watt
\ Input Power = Out put power + power absorbed
= 150000 + 781.33
= 150781.33 W
= 150.781 kW
Q.24 A 3300 Volts, delta connected motor has a synchronous reactance
per phase (delta) of 18 ohm. It operates at a leading power factor of 0.707
when drawing
800 kW from the mains. Calculate its excitation emf. (8)
Ans:
Given V_{L} =V_{ph} = 3300V ; cosF = 0.707 leading ;
P_{i} =800 kw = √3 V_{L}I_{L} cosF
\I_{L} = 800 x 10^{3}/ √3 x3300 x 0.707
I_{ph } = I_{L} /√3 = 114.30 Amp.
Now, excitation e.m.f (E_{0}) will be :
E_{0} =√ (V_{ph} cosF)^{2 }+ (V_{ph} sinF + I_{ph}X_{ph})^{2}
=√ (3300 x 0.707)^{2 }+ (3300 x 0.707+ 114.3 x 18)^{2}
=4.9712 x 10^{3 } or 4971.22 Volts
Q.25 A 250 Volts dc shunt motor has Rf=150 ohm and Ra = 0.6 ohm. The motor operates on noload with a full field flux at its base speed of 1000 rpm with Ia = 5 Amps. If the machine drives a load requiring a torque of 100 Nm, calculate armature current and speed of the motor. (8)
Ans: Given V = 250 V
R_{f} = 150 W
R_{a} = 0.6W
i. operate no load and full load flux
N_{o} = 1000 rpm & I_{ao} = 5 amp.
ii. On Load T_{af } = 100 Nm,
I_{af} = armature full load current
E_{b1} = Back emf at No load
E_{b2} = Back emf at full load
N_{2} = full load speed
Find I_{af } = ? ; & N_{2} = ?
At No load I_{ao} = 5 amp.
E_{b1 } = V I_{a} R_{a} (1)
= 250  5 x 0.6
= 250  3 = 247 volts
& E_{b2} = N_{0} ( fPZ/60A) (2)
E_{b2} / N_{0 =} fPZ/60A = 247/1000 = 0.247
Now E_{b2} at full load
E_{b2} = N_{0} ( fPZ/60A) = V I_{af} R_{a}
E_{b2} = N_{2} x 0.247= 250 I_{af} x 0.6_{ } (3)
Now T_{a} at no load
T_{a0} = 9.55 E_{b1 }I_{a0} / N_{0}
= 9.55 x 247 x 5 / 1000 = 11.79 Nm
T_{1} / T_{2} = I_{1} / I_{2}
T_{a0} / T_{af} = I_{a0} / I_{af}
11.75/100 = 5/I_{af}
\ I_{af} = 500/ 11.79 = 42.41 amp
Therefore, armature current at full load =42.41 amp
Put this value in equation (3)
N_{2} x 0.247 = 250  42.41 x 0.6
\ N_{2} = 224.55/ 0.247 = 909.13 rpm
\ Full load speed = 909 rpm
Q.26 A 400Volts, 1450 rpm, 50 Hz, woundrotor induction motor has the following circuit model parameters.
R1= 0.3 ohm R2=0.25 ohm
X1=X2=0.6 ohm Xm= 35 ohm
Rotational loss =1500 W. Calculate the starting torque and current when the motor is started direct on full voltage. (8)
V = 400V; N = 1450 rpm ; f =50 Hz
R_{1 }= 0.3 W R_{2}^{’} _{ }= 0.25 W ; X_{1 }=X_{2}^{’ } _{ }= 0.6 W ; X_{0 } = 35W
Rotational losses =1500 W
Find starting torque T_{1} = ? I_{f} =?
R_{01 }= R_{1} + R_{2}^{/} = 0.3 + 0.25 =0.55 W
X_{01 }= X_{1} + X_{2}^{/} = 0.6 + 0.6 =1.2 W
Z_{01} = Ö (R_{01}^{2} + X_{01}^{2} )=Ö0.55^{2} + 1.2^{2} = 1.32W
S = R_{2}^{/ } / Ö{R_{1}^{2} +(X_{1} + X_{2 }^{/} )^{2} } = 0.25 /Ö0.3^{2} +( 0.12)^{2}
= 0.25 / Ö1.53 = 0.2021
\ N_{S} = N/ (1S) = 1450 / (10.2) = 1812 rpm
I_{2}^{/} = I_{f} = V_{ph} / Ö{( R_{1} + R_{2}^{/} )^{2} +( X_{1} + X_{2}^{/} )^{2} }
= 400/Ö3 (V_{ph} = V_{L} / Ö3 let motor Y connected)
Ö (0.55)^{2} + (1.2)^{2}
Or I_{f} = V/Z_{01} = 400/Ö3
1.32
= 175.16 amp.
Torque developed by rotor
T_{g} = (3 I_{2}^{/} R_{2}^{/} )/ S
2pN_{S} /60
Or T_{g} = (3 I_{2}^{/} R_{2}^{/} ){(1S)/ S } Nm (N = (1S)N_{S} )
2pN /60
= 3 x (175.16) x 0.25 x (0.8/0.2)
(2 x 3.14 x 1450) /60
= 606.86 Nm
T_{Shaft} = (3 I_{2}^{/} R_{2}^{/} ){(1S)/ S } rotational losses Nm
2pN /60
(N = (1S)N_{S} )
= 606.86  1500/151.67
= 596.97 Nm
Q.27 A universal motor (ac–operated) has a 2pole armature with 960 conductors. At a certain load the motor speed is 5000 rpm and the armature current is 4.6 Amps, the armature terminal voltage and input power are respectively 100 Volts and 300 Watts.
Compute the following, assuming an armature resistance of 3.5 ohm.
(i) Effective armature reactance.
(ii) Maximum value of useful flux/pole. (8)
Ans: P = 2; Z = 960; N= 5000
I_{i} =I_{a}= 4.6 amp ; V_{1} =100 volts
P_{i} =300W find X_{a} and f_{m} = ?
P_{1} = V_{1}I_{1}cos f
\ cos f = P_{1} / V_{1}I_{1}
= 300/ 100 x 4.6 = 0.652
E_{bdc} = V  I_{a}R_{a } or (NfPZ / 60A )
= 100  4.6 x 3.5
= 100 16.1 = 83.9 volts
E_{bac} = Vcosf  I_{a}R_{a }
= 100 x 0.652  4.6 x 3.5 = 49.11 volts
And V^{2} =((E_{bac} + I_{a}R_{a } )^{2} + (I_{a} X_{a})^{2}
(4.6 X_{a} )^{2 } = 100^{2}  (652)^{2}
21.16 X_{a}^{2} = 5749
X_{a } = 16.48 ohms
E_{bdc } = (Nf_{m}PZ / 60A)
\ f_{m} = E _{ bdc } x 60 A/ NPZ
Flux
\ f_{m} = 83.9 x 60 x 2 / 5000 x 2 x 960 (A=P)
= 1.048 x 10^{3} wb
Q.28 A single phase 50 Hz generator supplies an inductive load of 5,000kW at a power factor of 0.707 lagging by means of an overhead transmission line 20 km long. The line resistance and inductance are 0.0195 ohm and 0.63 mH per km. The voltage at the receiving end is required to be kept constant at 10 kV. Find the sending end voltage and voltage regulation of the line. (8)
Given 1 f, 50 Hz ; cosf = 0.707 lagging
Transmission length = 20 km
Generator supply inductive load = 5000kW = kVAR
R = 0.0195 ohms/km , L = 0.63 mH
V_{R} = 10kV
Find sending end voltage V_{S} = ? & % regulation =?; distance =20km
DV = V_{S} V_{R} ( drop in line )
(V_{S} V_{R} )=RP + XQ/ V_{R} {Active Power (P) = Reactive Power (Q)}
= 0.0195 x 20 x 5000 x 10^{3} + 3.96 x 5000 x 10^{3} / 10 x 10^{3}
V_{s}^{ } ={0.0195 x 20 x 5000 x 10^{3} + 3.96 x 5000 x 10^{3} / 10 x 10^{3} } + 10 x
10^{3} volts
= 1.2175 x 10^{4 }volts
^{ }= 12.175 kVolts
% Regulation = (V_{S} V_{R} )/ V_{S} x 100
= 17.86%
Q.29 A 37.7 HP, 220 V d.c shunt motor with a full load speed of 535 rpm is to be braked by plugging. Estimate the value of resistance which should be placed in series with it to limit the initial current to 200 A. (8)
P_{0} = 37.7 HP = 37.7 x 735.5 W = 27.73 kW
N_{L} = 535 rpm braked by plugging ; I_{a} = 200A
V =220 v
Under plugging :
I_{a} = V + E_{b} / (R + R_{a} )
and E_{b} = 27.73 x 10^{3} / 200 (assume negligible losses)
= 138.65 Volts
200 = 220 + 138.65/(R + R_{a} )
R + R_{a} = 358.65 /200 = 1.79 ohm
R = value of added resistance in series with armature resistance
R_{a} = armature resistance
Q.30 The losses of a 30 kVA, 2000/200 V transformer are
Iron losses: 360 W
Full load copper losses : 480 W
Calculate the efficiency at unity power factor for (i) full load and (ii) half load. Also determine the load for maximum efficiency; also compute the iron and copper losses for this maximum efficiency condition. (12)
Ans:
Given: 30 kVA, 2000/ 200 Volts, Wi =360 W
Wc = 480 W, cosΦ=1
(i) at full load unity p.f.
Total losses = 360+480 = 840 W
At Full load, output at unity p.f. = 30Í1= 30 kW
Therefore, Efficiency = (30/(30+0.84)) Í 100
= 97.28 %
(ii) At half load and unity p.f.
Wc = 480Í( ½)^{2} = 120 W
Wi = 360 W
_{ }Total losses = 360+120 = 480 W
At half load, output at unity p.f = 30/2Í1= 15 kW
Therefore, Efficiency = (15/(15+0.48)) Í 100
= 96.90 %
(iii) The load for maximum efficiency and condition for max. efficiency.
Efficiency ( ή ) = ( V_{1}I_{1} cosΦ  losses )/ V_{1}I_{1} cosΦ
= ( V_{1}I_{1} cosΦ  I_{1}^{2}R_{01} Wi )/ V_{1}I_{1} cosΦ
= 1_{}
Differentiating above equation for maximum efficiency
Therefore,
_{}= 0  _{} _{}= 0
and
_{} = _{}
Therefore,
W_{i }= W_{c}
and load current, I_{l }=_{}(√ W_{i }/R_{02} )
Ans:
Given : V_{L } = 33 kV, 3 phase, star connected Synchronous Generator
\ V_{ph} = 33/√3 = 19.1 kV
Z_{ph} = 1.4Ω
P_{01} = 240 MW = 240 x 10^{6} Watt
cos F_{1} = 1
Ψ_{2} = 1.25Ψ_{1} (Flux / excitation)
P_{02} = 280 MW = 280 x 10^{6} Watt
Find I_{L2} = ? ; cos F_{2} =?
E_{g1} /ph = V_{ph1} + I_{ph1 }Z_{ph}
= (33 x 10^{3} /√3) + {(240 x 10^{6} /√3 x 33 x 10^{3}) x 1.4}
(P_{01} =√3 V_{L1} I_{L1 } cos F_{1 } & I_{L1} = I_{ph 1})
= 24.98 x 10^{3} volts
≈ 25 kV
Now
_{}
Ψ_{2} = 1.25Ψ_{1 }, due to 25% increased excitation )
E_{g2} _{/ph} = E_{g1} _{/ph} * _{}
\ E_{g2} _{/ph} = 31.23kV
or E_{g2} _{/ph} = V_{ph} + I_{2 }Z_{ph}
I_{2 ph} = E_{g2} V/Z_{ph} = 12.125 x 10^{3} / 1.4
= 8.66 kA
Now P_{02} =√3 V_{L} I_{L2 } cos F_{2 } ( I_{L2} = I_{ph 2 } : V_{L } = V_{L1} = V_{L2} ; P_{02} = 280 MW )
cos F_{2 }= P_{02} /√3 V_{L} I_{L2}
= 280 x 10 ^{6} / √3 x 33 x 8.66 x 10^{6}
= 280 / 495
= 0.565 leading
Q.32 A 250 V DC shunt motor has an armature resistance of 0.55 _{} and runs with a full load armature current of 30A. The field current remaining constant, if an additional resistance of 0.75 _{} is added in series with the armature, the motor attains a speed of 633 rpm. If now the armature resistance is restored back to 0.55_{}, find the speed with (i) full load and (ii) twice full load torque. (12)
Given R_{a }=0.55_{ }Ω,
R_{T} =0.75 additional resistance in series with armature
V = 250V
I_{a} = 30A at full load Then N_{2} = 633 rpm
I_{sh} = Constt.
Find N_{1} = ? at full load ; N_{3} = ? at double load
N_{1} α V  I_{a1}R_{a}
N_{2} α V  I_{a2 }(R_{t}+R_{a}) I_{a1} = I_{a2} = 30 Amp; I_{f} =constt.
\ _{}
\ _{= } 250 – 30 x 0.55 . = 233.5
250 – 30 (0.55 + 0.75) 211
_{}
N_{1} = N_{2} *1.106 = 700.5
N_{1} = 701 r.p.m at full load without R_{t}
Speed N_{3} at twice full load torque without R_{t}
_{}
\ _{= }. 250 – 30 x 0.55 . = 233.5
250 – 60 * 0.55 217
_{}
\N_{3} = N_{1 }/ 1.076 = 701/1.076 = 651 rpm
Speed of motor at twice full load torque = 651 rpm
Q.33 A 4pole, 3 phase, 400 V, 50 Hz, induction motor has the following parameters for its circuit model (rotor quantities referred to the stator side) on an equivalentstar basis:
_{} and _{} = 40 _{}. Rotational losses are 720 W. Neglect stator copper losses. For a speed of 1470 rpm, calculate the input current, input power factor, net mechanical power output, torque and efficiency.
(12)
Ans:
Given:
P = 4
V_{L }= 400 V
f = 50 H_{Z}
R_{1 }= 1.6 Ω, X_{1 = }2.4_{ }Ω ; R_{2}^{/} =0.48 Ω ; X_{2}^{/}_{ = }1.2_{ }Ω
X_{m} = X_{0} = 40 Ω
Rotational losses = 720 W; Wi = 0 Watt
N=1470 rpm
Find I_{1} = ? ; cos Φ = ? ; P_{mo} = ?; T_{S} = ? ; h=?
N_{S} = 120 f / P = 120 x 50 / 4 = 1500 rpm
Therefore, s = (N_{s} – N)/ N_{S} = (1500 – 1470)/1500 = 0.02
Input current I_{1} = V_{1} / Z_{01}
V_{1} =400 V; Let R_{o} is negligible
Z_{01} = Z_{1} + Z_{AB ;} where Z_{AB }is the impedance between point A and B
Z_{1} = (R_{1} + jX_{1}) = (1.6 + j 2.4)
Z_{AB }= j X_{m} [ R^{1}_{2}/S + jX^{/}_{2} ] / { (R^{/}_{2}/S) + jX^{/}_{2} + j X_{m}}
= j40 (0.48 / 0.02 + j1.2) / {0.48 / 0.02 + j (1.2 + 40)}
= {j40 (24 + j 1.2)}/ [24 + j (41.2)]
= {(j960 – 48) x (24 – j41.2)}/{(24 + j41.2)(24 – j41.2) }
= {j960 x 24 + 960 x 41.2 – 48 x 24 + j 48 x 41.2}/{ (24)^{2}+ (41.2)^{2}}
= j 23040 +38400 + j 1977.6
= (38400 + j 25017.6 )/ (576 + 1697.44)
= 16.89 + j 11
Z_{AB} = 16.9 + j11 = 20.16 ∟33^{o}
\ Z_{01} = (1.6 + j 2.4) + (16.9 + j11) = 18.5+ j13.4 = 22.84 ∟35.9^{ o}
(i) \ I_{1} = V_{1}/Z_{01} = (400/√3 ) ∟0^{ o }/ 22.84 ∟35.9^{ o} = 10.24 ∟35.9^{ o}
(ii) p.f = cos Φ
= cos 35.9^{o}
= 0.81
(iii) Mech. total power = (1S)P_{2 } ; Where P_{2} is the power of air gap
= (10.02) 3I_{2}^{׀}^{ 2 }(R_{2}^{/ }/ S)
= 0.98 x 3 x I^{2}_{1} R_{AB}
= 0.98 x 3 x (10.24)^{2} + 16.9
= 5209.95
= 5210 Watt
Net Mech. Power = Total Mechanical Power – Rotational Losses
P_{mo }= 4490 Watt
(iv) Net Torque = P_{mo} / ( 2 Π N/60)
= 4490 / (2 x 3.14 x 1470/60)
N = 1500 (10.02) = 1470 rpm
= 4490 x 60 / 6.28 x 1470 = 29.18 Nm
T_{S} = 29.18 Nm
(v) output power = 4490
(a) Stator ‘core’ losses W_{1} = 0
(b) Stator ‘Cu’ losses = 3I_{1}^{2} R_{1} = 3 x (10.24)^{2} x 1.6
= 3 x 167.77
= 503.31 Watt
(c) Rotor ‘Cu’ losses = 3I_{2 }^{/2} R_{2} = SP_{2}
= 0.02 x 3 x (10.24)^{2} x 16.9
= 106.32 Watt
(d) Rotational losses = 720 Watt
Therefore: Efficiency (h) = (output / output + losses) x 100
= 4490 x 100/ 4490 + (0+503.31+106.32+720)
= 77.39 %
Q.34 A universal motor has a 2pole armature with 1020 conductors. When it is operated on load with a.c. supply with an armature voltage of 150, the motor speed is 5400 RPM. The other data is:
Input power : 360 W
Armature current : 5.2 A
Armature resistance: 5.5 _{}
Compute (i) the effective armature reactance and (ii) maximum value of armature flux per pole. (10)
Ans: Given:
P=2; V_{a} = 150 V; N = 5400 rpm
Z=1020; P_{i} = 360 MW; I_{a} = 5.2A;
R_{a} = 5.5Ω
Find X_{a} =?; F_{max} per pole=?
f = NP/120 = 5400 x 2 /120 = 90Hz
E _{abc} = V_{a }  I_{a}R_{a}
= 150 – 5.2 x 5.5
= 121.4 V
Now E_{b} = NFPZ/60A
\ F_{m} = E_{b} x 60 A / NPZ
= 121.4 x 60 x 2 / 5400 x 2 1020
= 1.4568 x 10 ^{4 }/ 1.102 x 10^{6}
= 1.322 x 10^{2} wb per pole
Now I_{a} = V_{a} /Z_{a}
\ Z_{a} = V_{a} / I_{a} = 150 /5.2 = 28.84 Ω
Z_{a} =√R_{a}^{2} +X_{a}^{2} = 28.84 Ω
\ R_{a}^{2} +X_{a}^{2} = 832 Ω
X_{a}^{2} = 832  R_{a}^{2}
= 832 5.5^{2}
= 801.75
\ X_{a} =2 8.32 Ω
Q.35 A 50 KVA, 2300/230 V, 60 Hz transformer has a high voltage winding resistance of 0.65 _{} and a lowvoltage winding resistance of 0.0065 _{}. Laboratory tests showed the following results:
Open circuit test: V = 230 V, I =5.7A, P = 190 W
Short circuit test: V=41.5 V, I=21.7 A, P=No wattmeter was used.
(a) Compute the value of primary voltage needed to give rated secondary voltage when the transformer is connected as a stepup one and is delivening 50 KVA at a power factor of 0.8 lagging. (12)
Given : P_{0} = 50 kVA, V_{1} = 2300V , V_{2} = 230V, f = 60 Hz
R_{1} = 0.65Ω (H.V.Side), R_{2} = 0.0065Ω (L.V Side)
Lab test: O.C (H.V.Side) : V= 230 V, I =5.7A, P = 190 watt
S.C (L.V.Side), : V= 41.5 V, I =21.7A, P = ? watt
Find (a) V_{1 } for rated V_{2} when acts as step up transformer and delivering 50kVA at cos F = 0.8
(b) efficiency
At S.C test
Z_{02} = V_{SC} / I_{2} = 41.5 /21.7 = 1.912Ω & K =1/10
\Z_{01} = Z_{02} / K^{2} = 1.912 x 10^{2} = 191.2 Ω
R_{01} = R_{1} + R_{2}^{/} = 0.65 + 0.0065/100 (K =1/10, H.V side is R_{2})
R_{01} = 0.13 Ω
R_{02} = R_{01} x K^{2} = 0.13 / 100 = 0.0013 Ω
Now X_{01 }^{ }= √ Z_{01}^{2} – R^{2}_{01} = √ (191.2)^{2} – (0.13)^{2}
X_{01 }^{ }= 191.2 Ω
& X_{02 }^{ }= √ Z_{02}^{2} – R^{2}_{02} = √ (1.912)^{2} – (0.0013)^{2}
X_{01 }^{ }= 1.912 Ω
Total transformer voltage drop referred to secondary
VD_{2} = I_{2} ( R_{02} cosF_{2} + X_{02} sinF_{2})
= 21.7 ( 0.0013 x 0.8 + 1.92x 0.6) = 21.7 (1.153)
= 25.02 Volts
\V_{1}^{/} = V_{1} + VD_{2} = 230 + 25.02 = 255.02 Volts.
\V_{1} =255.02 volts for V_{2} = 2300 volts rated value as step up.
Efficiency of Transformer = output/ Input
=
50 x 1000 x 0.8
.
output + losses (core Losses + Cu losess)
=
40000 .
40000 + 190 + I_{1}^{2} R_{1}+ I_{2}^{2} R_{2}
=
40000 .
40000 + 190 + (5.7)^{2} x 0.65+ (21.7)^{2} x 0.0065
(note: low voltage winding is short circuited)
=
40000 .
40000 + 32.49 x 0.65+ 470.89 x 0.0065
=
40000 .
40000 + 21.12 + 3.06
= 99.93%
Q.36 A threephase, 335hp, 2000V, six pole, 60 Hz, Yconnected squirrelcage induction motor has the following parameters per phase that are applicable at normal slips:
_{}
_{}
_{}
The rotational losses are 4100 watts. Using the approximate equivalent circuit, compute for a slip of 1.5%.
a. the line power factor and current.
b. developed torque.
c. efficiency. (8+4+4)
Given: 3 phase , 335 HP, 50 Hz, V = 2000V Induction motor
P_{0 }= 335 x 735.5 watt Rotational losses = 4100 watt; Slip (s) = 0.015
P = 6 nos., Y connected
\ I_{L} = I_{ph} & V_{L} =√3 V_{ph}
V_{ph} = V_{L} / √3 = 2000 / √3 = 1154.70 volts
Find : p.f = ? , Line current I_{L} = ? , T_{g} = ? , and h = ?
Sol: (i) R_{1 }= 0.2Ω, R_{2}^{/}_{ }= 0.203 Ω
X_{1} = X_{2}^{/} = 0.707 Ω
x_{0} = 77 Ω , R_{0 }= 450 Ω
R_{01 }= R_{1 }+ R_{2}^{/} = 0.2 + 0.203 = 0.403 Ω
X_{01 }= X_{1 }+ X_{2}^{/} = 0.707 + 0.707 = 1.414 Ω
Z_{01 }=√ R_{01}^{2}_{ }+ X_{01}^{2} = √ (0.403)^{2}_{ }+ (1.414)^{2} = 1.47 Ω
Load current I_{2}^{/} = I_{L}
\ I_{2}^{/} = V_{Ph} / (R_{1} + R_{2}^{/} / s) + j( X_{1} + X_{2}^{/})
= 1154.7 ∟0^{0} / (0.2 + 0.203/0.015) + j (0.707 + 0.707)
= 1154.7 ∟0^{0} /13.733 + j 1.414 = 1154.7 ∟0^{0} / 13.80 ∟5.88^{0}
= 83.67 ∟5.88^{0}
\ I_{L} = 83.67 Amp.
P.f = cos ( 5.88^{0}) = 0.99 lagging
(ii) Torque generated (T_{g}) or developed = (9.55 x 3 I_{2}^{/2} R_{2}^{/ }/ s)/ N_{S}
{N_{S} = 120 f /p = 120x50/6 = 1000 rpm}
= (9.55 x 3 x (83.67)^{2} x 0.203 / 0.015)/ 1000
T_{g} = 2.714 x 10^{3} Nm
(iii) Efficiency of machine = output / output + losses
Total losses = Rotational losses + rotor ‘Cu’ losses + stator ‘Cu’ losses
Rotational losses = 4100 watt = 4.1 kW
rotor ‘Cu’ losses = 3 I_{2}^{/2} + R_{2} = 3 x (83.67)^{2} x 0.203 = 4.26 kW
stator ‘Cu’ losses = 3 I_{1}^{/2} R_{1 }= 3 x (86.2)^{2} x 0.2 = 4.458 kwatt
\ Total losses = 4.1 + 4.26 + 4.46 = 12.82 kW
I_{1} = I_{2}^{/2} + I_{0 }= 83.67 + 2.53 = 86.20 Amp
I_{0} = V_{Ph }/ √ R_{0}^{2}_{ }+ X_{0}^{2}
= 1154.7_{ }/ √ 450^{2}_{ }+ 77^{2}
= 2.529Amp
h = {246.40 / ( 246.40 + 12.82) }x100
= 95.05%
Q.37 A 2300V, three phase, 60 Hz, starconnected cylindrical synchronous motor has a synchronous reactance of 11 _{} per phase. When it delivers 200 hp, the efficiency is found to be 90% exclusive of field loss, and the powerangle is 15 electrical degrees as measured by a stroboscope. Neglect ohmic resistance and determine:
(a) the induced excitation per phase.
(b) the line current _{}
(c) the power factor (8+4+4)
Given : 2300V , 3 phase, 60Hz, Synch. Motor
X_{S} = 11Ω /ph , Star connected, V_{Ph } = 2300/ √3 V
P_{0} = 200 hp = 200 x 735.5 = 147.1 kW
h = 90%
\ P_{i} = P_{0} /0.9 = 163.44 kW
Power angle α = 15^{0} ( electrical)
Find: induced excitation / ph (E_{g })_{ }= _{ }?
Line Current I_{a} = ?
Power Factor Cos F =?
(i) P_{i} = 3 x E_{g} /ph x V_{Ph} ) x Sin α
X_{S}
\ E_{g} /ph = P_{i} X_{S} / 3x V_{Ph} x Sin α
= 163.44 x 10^{3} x 11 / 3 x 1327.9 x sin 15^{0}
E_{g} /ph = 1743.68 volts (Due to over excitation)
Given Z_{S} = 0 + j 11 and R=0
E_{R} = ( V_{Ph} – E_{g} cos α) + j 1743.68 sin 15^{0}
=  356.36 + j 451.296
\ I_{a} = E_{R} /Z_{S} = (356.36 + j 451.30) / (0+j11)
= 41.02 + j 32.42
I_{a} = 52.28 ∟38.32^{0}
(ii) Line current = 52.28 Amp
(iii) & p.f cos F = cos 38.32^{0} = 0.78
Q.38 When a 250V, 50 hp, 1000 rpm d.c shunt motor is used to supply rated output power to a constant torque load, it draws an armature current of 160A. The armature circuit has a resistance of 0.04 _{} and the rotational losses are equal to 2 KW. An external resistance of 0.5 _{} is inserted in series with the armature winding. For this condition compute
(i) the speed
(ii) the developed power
(iii) the efficiency assuming that the field loss is 1.6 K.W (8+4+4)
Given V_{L} = 250 V P_{0} = 50 hp = 50 x 735.5 36.78 kw
N_{1 }= 1000 rpm I_{a} = 160 amp. , R_{a}= 0.04Ω, R = 0.5 Ω ,
Rotational losses = 2 kw ; Field losses = 1.6 kW
Find : speed after series resistance R in armature circuit N_{2} =?
Power developed (P_{m}) = ?
Efficency (h) = ?
(i) E_{b1} = N_{1}(FPZ/60A)
And V = E_{b1 } + I_{a}R_{a}
E_{b1} = 250 – 160 x0.04 = 243.60 volts
Now E_{b2} when R = 0.5 connected in series with armature
\ E_{b2} = 250 160 x (0.04 +0.5)
= 163.6 volts
Now E_{b1} / E_{b2} = N_{1} /N_{2} (when F_{1} = F_{2})
N_{2 }= N_{1 } x E_{b2} / E_{b1} = 1000 x 163.6 / 243.6 = 672 rpm
(ii) Now Input power developed in armature = E_{b2} I_{a}
= 163.6 x 160 = 26.18 kW
=(I_{a}^{2} R_{a}^{/} /1000) + 1.6 kW + 2.0kW
= (160^{2} x 0.54/ 1000) + 1.6 +2.0
= 17.42 kW
(iii) Efficiency = (P_{i }– losses / P_{in}) x 100
= (26.1817.42 / 26.18) x100
= 33.46%
Q.39 The following data were obtained on a 20KVA, 50Hz, 2000/200V distribution transformer
Open Circuit Test (on L.V. side): 200V, 4A, 120W
Short Circuit Test (on H.V. side): 60V, 10A, 300W
Draw the approximate equivalent circuit of the transformer referred to H.V. Side. (8)
Ans:
Given : 20 kVA, 50Hz , 2000/ 200V
O.C Test : V_{0} = 200 V ; I_{0} = 4A ; W_{1} = 120 W
S.C Test : V_{SC}= 60 V ; I_{SC} = 10A ; W_{SC} = 300 W
Primary equivalent secondary induced voltage
E_{2}^{/} = E_{2} / K
V_{2}^{/} = V_{2} /K
& I_{2}^{/} = KI_{2}
From O.C Test:
V_{0}I_{0} cosF_{0} = W_{0}
\ cosF_{0} = W_{0} / V_{0}I_{0} = 120 / 200 x 4 = 0.15
\ sinF_{0} = 0.988
Now I_{w} = I_{0} cosF_{0} = 4 x 0.15 = 0.60 Amp.
I_{u } = I_{0} sin F_{0} = 4 x 0.988 = 3.95 amp.
R_{0} = V_{0} / I_{w} = 200 /0.6 = 333.33 Ω
X_{0} = V_{0} / I_{u} = 200 /3.95 = 50.63 Ω
From S.C test:
Z_{0 } = V_{SC }/ I_{SC} = 60 / 10 = 6Ω
and K = 200 /2000 = 1/10 = 0.1
Z_{01 } = Z_{02 }/ k^{2} = 60 / (1/10)^{2} = 600Ω
Now I_{SC}^{2 } R_{02} = W _{SC}
\^{ } R_{02} = W _{SC}/ I_{SC}^{2} = 300/100 =3Ω
R_{01} = R_{02} / k^{2} = 3/(1/10)^{2} =300Ω
X_{01} = √( Z_{01}^{2}  R_{01}^{2}) = √( 600^{2}  300^{2})
X_{01}= 519.62Ω
Q.40 The efficiency of a 3phase 400V, star connected synchronous motor is 95% and it takes 24A at full load and unity power factor. What will be the induced e.m.f. and total mechanical power developed at full load and 0.9 power factor leading? The synchronous impedance per phase is (0.2+j2)Ω. (9)
Ans:
Given : 3F, 400V star connected synchronous motor
Output = 95% of input
V_{Ph} = 400/√3
at p.f = 1 ; I_{a} = 24 amp. ; V = V_{Ph} = 230.94 volt
Z_{S} = (0.2 + j2) Ω = 2∟84.29 Ω
Find : E_{b} at 0.9 p.f leading
Mechanical power developed ?
cos F = 0.9
\ F = 25.84^{0}
E_{R} = IZ_{S} = 24 (0.2 +j2) volts
= (4.8 +j48) volts = 48.23∟84.29^{0}
q =84.29^{0}
Now at leading p.f
E_{b/ph} = V + I_{a}Z_{S} cos {180 – (q + F)} + j I_{a}Z_{S} sin {180 – (q  F)}
\ E_{b/ph} = 231+24x2cos {180 – (84.29 + 25.84)} + j24x2
sin
{180 – (84.29  25.84)}
=231 + 48 cos (69.87) + j 48 x 0.938
= 231 + 16512 + j45
E_{b/ph} = 247.5 + j 45
E_{b/ph} = 251.55∟10.3
Synchronous motor input power = √3 V_{L}I_{L}cos F
= √3 VIacos F = √3 x 400 x 24 x 0.9
= 14,964.92 watt
total copper losses = 3 I^{2} R_{a}
= 3 x (24)^{2} x 0.2 = 3 x 576 x 0.2
= 345.6 watt
\ Mechanical output developed = Input – losses
= 14964.92 345.6
= 14619.32 watt
Q.41 A 200V shunt motor with a constant main field drives a load, the torque of which varies at square of the speed, when running at 600 r.p.m., it takes 30A. Find the speed at which it will run and the current it will draw, if a 20Ω resistor is connected in series with armature. Neglect motor losses. (9)
Ans:
Given: V =200v, shunt motor
N_{1} = 600 rpm
I_{1} = 30A = Ia_{1}
Find : N_{2} & I_{2} ; when R =20Ω added with R_{a} in series
E_{b1} = V –I_{a} R_{a} (losses are negligible , \ I_{a}R_{a} =0)
E_{b1} = V = 200v (I_{a} R_{a }= 0)
T_{1} = 9.55 E_{b1} I_{1 } / N_{1} = 9.55 x 200 x30 / 600
= 573Nm
T α N^{2}
\ T_{1} / T_{2} = N_{1} ^{2} / N_{2} ^{2}
\ N_{2} ^{2} / T_{2} = N_{1} ^{2} / T_{1 } = 600^{2} / 573
Or N_{2} = 600 √ T_{2} / 573 (1)
and N_{1} / N_{2} = E_{b1} / E_{b2} . F_{2} /F_{1} (F_{2} = F_{1} = constt)
\ N_{1} / N_{2} = E_{b1} / E_{b2}
or N_{2} = N_{1} (E_{b2} / E_{b1})
N_{2} = 600 (E_{b2} / 200)
N_{2} = 3E_{b2} (2)
And
E_{b2} =200 20I_{2 } (3)_{ }
(E_{b2} = V –I_{a} R_{a})_{ }
E_{b2 } = N_{2} / 3 from eqn. no. (2) put in eqn. no (3)
\ N_{2} = 60060I_{2 } (4)
T_{1} / T_{2} = N_{1} ^{2} / N_{2} ^{2} = I_{1}/ I_{2}
\ 600 ^{2} / N_{2} ^{2} = 30 / I_{2}
30 / I_{2 } = 600 ^{2} /( 60060I_{2 })^{2} (N_{2} = 60060I_{2})
30 / I_{2 } = 600 x 600 / 60 x 60 ( 10I_{2 })^{2}
10I_{2 } = 3(100 + I_{2}^{2}  20 I_{2})
10I_{2 } = 300 + 3I_{2}^{2}  60 I_{2}
\ 3I_{2}^{2} 70 I_{2} + 300 = 0
I_{2} = 70 ± √ 4900 – 4 x 3 x300 = 5.66 or 17.66
6
I_{2}_{ } = 5.66 amp
\ N_{2} = 60060I_{2}
= 600 – 60 x 5.66 = 260.55 rpm
N_{2} = 260 rpm
I_{2} = 5.66 amp, 17.66 amp is not possible for N_{2}
Q.42 A 3phase induction motor has a starting torque of 100% and a maximum torque of 200% of full load torque. Find
(i) Slip at maximum torque.
(ii) Full load slip.
(iii) Neglect the stator impedance (8)
Ans: Given : (T_{st} / T_{f} ) = 1 & (T_{max} / T_{f} ) = 2
\ (T_{st} / T_{max} ) = 1/2 = 0.5
let a = R_{2} / X_{2}
\ (T_{st} / T_{max} ) = 2a /(1+a^{2}) = 0.5/1
2a /(1+a^{2}) = 1/2
4a = 1+a^{2}
a^{2 } 4a + 1 = 0
a = 4 ± √ 16 – 4 x 1 x 1
2
a = 4 ± √ 12
2
a = 3.73 or 0 .2679
a = slip at max torque = 0.2679
Now, T_{f} /T_{max} = 1/2 = 2aS_{f} / a^{2} + S_{f}^{2 } (S_{f}^{ } = full load slip)
2 x 0. 2679 x S_{f} /{(0.2679)^{2 }+ S_{f}^{2} } = 1/2
0.5358 S_{f} / {(0.0717)^{ }+ S_{f}^{2} } = 1/2
\ 1.071 x S_{f} =(0.0717)^{ }+ S_{f}^{2}
S_{f}^{2} + 1.071S_{f } + 0.0717 =0
S_{f} = 0.9995 or 0.07145
\ full load slip = 0.07145
Q.43 A universal motor (a.c. operated) has a 2pole armature with 960 conductors. At a certain load the motor speed is 5000 r.p.m. and the armature current is 4.6A. The armature terminal voltage and input are respectively 100 V and 300 W. Compute the following, assuming an armature resistance of 3.5Ω.
(i) Effective armature reactance
(ii) Max. value of useful flux per pole. (8)
Ans: Given: P=2 ; Z =960 ; N_{1} =5000 rpm ; I_{a1}= 4.6 amp
V_{a} = 100v = I_{a1}R_{a}
P_{i} = 300 watt
R_{a} = 3.5 ohms
Find X_{a} = ?
F_{m} / pole = ?
P_{i} = VI cos F
\ cos F = P_{i} / VI = 300 / 100 x 4.6 = 300 / 460
= 0.652
Now E_{bac} = V cos F  I_{a}R_{a } = 100 x 0.652 – 4.6 x 3.5
= 49.1 volts
and V^{2} = ( E_{bac} + I_{a}R_{a})^{2} + (I_{a}X_{a})^{2}
100^{2} = ( 49.1 + 4.6 x 3.5)^{2} + (4.6X_{a})^{2}
(4.6X_{a})^{2} = 100^{2}  ( 49 + 16.1)^{2} = 10000 – (65.2)^{2}
= 10,000 – 4251 = 5748.96
21.16 X_{a}^{2} = 5748.96
\ X_{a} = √5748.96 / 21.16
= 16.48 Ω
and E_{bdc} = V I_{a}R_{a} = 100  4.6 x 3.5= 83.9 V
and E_{bdc} = NF_{m}PZ / 60A
83.9 = 5000 F_{m} x 2 x 960/ 60 x 2
F_{m} = 83.9 / 8.0 x 10^{4} = 1.048 x 10^{3} wb
(i) \ effective armature reactance = 16.48 Ω
(ii) Max. flux per pole = 1.048 x 10^{3}
Q.44 Using normal Π method, find the sending end voltage and voltage regulation of a 250Km, 3 phase, 50Hz transmission line delivering 25MVA at 0.8 lagging p.f. to a balanced load to 132KV. The line conductors are spaced equilaterally 3m apart. The conductor resistance is 0.11Ω/Km and its effective diameter is 1.6 cm. Neglect leakage. (8)
Ans: Given:
3F, 50Hz, d = 250km
P_{0} = 25mVA, cos F = 0.8 (lagging), V_{R }= 132 KV
V_{R/ph }= 132000/ √ 3 V
Spacing between conductors = 3m
R = 0.11 Ω/km
Dia (D) = 1.6 x 10^{2} m
Find : V_{s}= ? and % regulation =?
I_{ph = }P_{o}_{ }/ V_{R/ph}
I = 25 x 10^{6}/ 132 x 1000 = 25000/132
Line loss = 3 I^{2} R = 3 x (25000 / 132)^{2} x 0.11 x 250
Resistance / phase = 0.11 x 250 = 27.5 Ω
V_{S/ph} = V_{R/ph} + IZ_{ };_{ } ( Z = R/Phase)
= (132000/ √ 3 ) + (25000/132) x 27.5
= 76210.236 + 5208.333
= 81418.569
= 81.42 KV
% Regulation = (V_{S} – V_{R}/ V_{S}) x 100
=((81.42 76.21)/ 81.42 )x 100
= 6.39 %
Q.45 A 3phase transformer bank consisting of a three onephase transformer is used to stepdown the voltage of a 3phase, 6600V transmission line. If the primary line current is 10A, calculate the secondary line voltage, secondary line current and output kVA for the following connections:
(i) Y/Δ and (ii) Δ /Y. The turns ratio is 12. Neglect losses. (8)
Ans: (i) Y/∆
Given turn ratio = 12
V_{1} = 6600V, I_{1} = 10A,
V_{2} = ? , I_{2} = ?
o/ρ = kVA
I_{PP} = Phase current in primary winding
I_{LP } = Line current in primary winding
For Y : V_{p} = V_{L}/√3 , I_{p} = I_{L}
∆ : V_{p} = V_{L} , I_{p} = I_{L}/√3