Code: AE10 Subject: ELECTRICAL ENGINEERING

PART - I

TYPICAL QUESTIONS & ANSWERS

OBJECTIVE TYPE QUESTIONS

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 self-starting 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 two-phase 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 two-winding single phase transformer has a voltage regulation of 4.5% at full-load and unity power-factor. At full-load and 0.80 power-factor 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 =0⁰

\ 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 whileis halved. (D) is constant but is halved.

Ans: B

N α V – I_aR or N α E_b

T α I_a F , F α I_a

\ T α I_a²

Q.27 A balanced three-phase, 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 ( 1-s) = 1500 (1-0.05) = 1425

\relative speed = 1500 – 1425 = 75 rpm

Q.28 An alternator is delivering rated current at rated voltage and 0.8 power-factor lagging case. If it is required to deliver rated current at rated voltage and 0.8 power-factor 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) split-phase 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 single-phase ac motor.

(C) a multi-phase motor. (D) a two phase motor.

Ans: D

Stepper motor works on 1-phase-ON or 2-phase –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 mixer-grinder 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 step-up 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

I₁= V₁ /R₁ = 120/600 = 0.2 (h = 100%, losses are zero \V₁ = V_R = I₁R₁)

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, 3-phase 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 (1-S) = N_S –N_S x S

1440 = N_s (1-S)

N_s = 1440 / (1-S)

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 3-phase 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 single-phase 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 3-phase winding on rotor and a single-phase winding on stator.

(B) a 3-phase winding on stator and a commutator winding on rotor.

(C) a 3-phase winding on stator and a single-phase winding on rotor.

(D) a single-phase 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

Ans: B

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

= 60/6 = 10 volts

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 Oil-filled 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

Specific heat of Nickel-Chrome is 440 J/kg°C to 450 J/kg°C

Q.52

The polarity test is not necessary for the single-phase 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.

Ans: D

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 short-circuit 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 speed-torque characteristics of a DC series motor are approximately similar to those of the _________motor.

(A) universal (B) synchronous

(C) DC shunt (D) two-phase

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 torque-speed characteristics of an a.c. operated universal motor has a ______characteristic and it______ be started under no-load 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 = i²R and i α f²

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) two-phase A.C. servomotor (B) two-phase 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) 2-phase servomotor (D) DC series motor

Ans: D

T_aα Φ.Ia and Φ α Ia ; therefore T_aα Ia²

Q.61 The synchronous speed for a 3 phase 6-pole 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 , P₁= 6,

P₂ = 4, s = 0.05,

Frequency f = N_sÍP/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 / P₂

Therefore, N=N_s- 120 f / P₂= 1800-120Í0.05Í60/4 = 1800-90 = 1710

Q.62 The eddy current loss in an a-c 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 f²

New loss α (2f)²

New loss α 4f²

\ 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 = 0⁰

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 no-load 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 3-phase, 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 (1-S) and N_S =120 f / p

=120 x 50 /4 = 1500 rpm

\N= 1500 (1-0.02) =1470 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 = (V²/R) or (V²/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_hfB_m^1.6 and W_e = k_ef²B_m².k

Therefore, hysteresis and eddy current losses will be decreased when frequency decreases.

Q.73 A synchronous motor is operating on no-load 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

N₂/ N₁ =E_b2/E_b1 x F₁ / F₂ ; F₂ = 0.95F₁ ; E_b2= 0.9E_b1

\ N₂ /1140 = 0.9 x 1/0.95

N₂ = 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
Counter e.m.f is produced, it neutralizes the reactive emf.

Q.76 The rotor power output of a 3-phase 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 = (1-s) input

\ Input = output/(1-s) = 15000 /1-0.04 = 15625

\ loss = 15625 -1500 = 625 watt

Q.77 The direction of rotation of hysteresis motor is reversed by

(A) Shift shaded pole with respect to main pole

(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, 4-phase 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 360⁰

\ 1-8 = -40 + N_r/40 N_r x 360⁰

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
In the hysterisis motor, the direction of rotation can be reversed by shifting the shaded pole region with respect to main pole. But not by changing supply lead because it has ac supply.

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) Per-unit 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_a²

Q.85 A 400kW, 3-phase, 440V, 50Hz induction motor has a speed of 950 r.p.m. on full-load. 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 (1-S)

950 = 120 x 50 (1-S)/6

S = 0.05

Q.86 Reduction in the capacitance of a capacitor-start 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°

Ans: B

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 I²R 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 R₂^// S) -(3I^/2 R₂^/)

= 3I^/2 R₂^/(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

R α 1/V

Regulation = (V₀– V_L ) / V₀ , if V_L is high the (V₀– V_L ) will be low.

Therefore R α 1/V

Code: AE10 Subject: ELECTRICAL ENGINEERING

PART - II

NUMERICALS

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

Now

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 pole-pair. 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 =

(5)

Substituting for in (3)

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

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

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

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=40KW-1KW=39KW

Rotor copper loss=sP_G=0.025*39=0.975KW

Gross mechanical output=(1-s)P_G=39-0.975=38.025KW

Net mechanical output=Gross mechanical output-friction and winding loss

= (38.025-2.000)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=2X60X0.25

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 full-load 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

I₂

a

R_e^/

q E₁=V₂a

X_e^/

I₂

a

V₁

E₂

E₁

X_eq

R_eq

200V

Fig. A3 Equivalent circuit referred to primary

Primary equt. resistance = r₁+ a²r_{2 =}a² [equt. resistance referred to secy]

Also primary equt. reactance =x₁ + a²x₂= a² [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

(E₁ + 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 = 200²

This gives

E₂ = 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:

I₀

V

b

I₁

a

I^’₂

X^’₂

R^’₂

a

R₁

X_m

X₁

R₁

Fig B3 shows the per-phase exact equivalent circuit of a 3-phase 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

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

Also

Also for a series motor

E_a = V-I_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

From (C) and (D)

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

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, 1-phase transformer has no-load 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 no-load 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, 1-F

No load: -

I_o = 0.64 A

W_o= 700W

Iron loss current = 700/2400 =0.2916 A

Now , I_o² = I_w²+ I_m²

Magnetizing component I_m = (I_o² – I_w²)^1/2

= (0.64² – 0.2916²)^1/2

= 0.5697 A

On load :-

I₂= 40 A

F₂ = 0.8 lag

cos F_o = W_o/(V_oI_o)

= 700/(2400x0.64)

= 0.455

F_o = cos ^–1 0.455

= 62.88^o

F₂ = cos^–1 0.8

= 36.86^o

Now, turn ratio K =V₁/V₂ = 240/ 2400 = 0.1

I₂¹ = KI₂ = 40x0.1 = 4 A

Angle between I₀ and I₂¹ = 62.88^o – 36.86^o

= 26.02^o

I₀ = 0.64 Ð–62.88^o

I₂¹ = 4Ð–36.86^o

I₁= I_o + I₂¹= 0.64 [cos( 62.88) – j sin (62.88) ] + 4[ cos(36.86) – jsin( 36.86) ]

= 4.583

I₁= 4.583 A

cos F₁ = 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

V = 240 V

R_a= 0.04 W, R_sh = 24W

E_g = V + I_aR_a

E_g = V + (I_L + I_sh) R_a

I_sh = V / R _sh = 240/24 = 10A.

Substituting the values in the above expression,

254 = 240 + (I_L + 10) 0.04

I_L = 340A

Q.12 The shaft output of a three-phase 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

Windage and Friction losses = 920W

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

= 87.81 %

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

Mech. loss = 2 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.

slip = (1000–950)/100 =0.05

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

P=120 * 50/1000=6;

(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 single-phase, 50 kVA, 2300/ 230 V, 60 Hz transformers are connected to form a 3-phase, 4000V / 230-V transformer bank. The equivalent impedance of each transformer referred to low-voltage is 0.012 + j 0.016 . The 3-phase transformer supplies a 3-phase, 120 kVA, 230 V, 0.85 power-factor (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 ; P₀ = 50kVA ; 2300/230V, 60Hz

no. of transformers are 3 (three) to form a 3 F transformer

4000/230V , P₀ = 120 kVA , 230V and cosF =0.85 lagging

Z₀₂ = (0.012 + j 0.016) Ω and

Z₀₁ = Z₀₂ /K²

(i) Schematic diagram to show transformer connection

(ii) Calculation of winding currents of transformer:

Given , P₀ = 120 kVA

P₀ =√ 3 V_LI_L cosF

I_L = I_ph ; V_L = √ 3 V_ph (star connection)

\ I_L = 120 x cos F x 10³ / √ 3 V_L cosF

= 120 x 10³ /√ 3 x 230

= 30.12 amp

\ secondary line current = 30.12 amp.

(iii) Primary current I₁ = k I₂

= 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 shaft-speed 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 full-load 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 full-load torque and (ii) at twice the full-load torque. (6)

Ans: In a DC shunt motor

V = 240V

R_a = 0.4Ω

N₁= 600rpm (full load speed)

I_a = 25A and, I_Sh is constant

R = 1 Ω added in series with armature

E_b1 = V - I_aR_a

= 240 -25 x 0.4

= 230 volts

E_b2
=V - I_a(R_a+ R)

= 240 -25 ( 0.4+ 1)

= 201 volts

Now N₁ / N₂ = E_b1 / E_b2 x F₂ / F₁ (F₁ = F₂ = constant)

N₂ = N₁ 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,

N₃ / N₁ = E_b3 / E_b1 x F₁ / F₃ (F₁ = F₂= F₃ = 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.

N₃ = N₁ 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, 4-pole, 50 Hz, 3-phase, 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) no-load speed (iii) full-load speed.

(iv) frequency of rotor current at full-load (v) full-load torque. (5 x 2 = 10)

Ans: Given :

V_L = 400 volts ; P = 4 nos, 50 Hz,

P₀ = 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

N₀ = N_S ( 1 – s) = 1500 ( 1- 0.01)

= 1485 rpm

iii. Full load speed at s_f = 0.04

N_fl = N_S (1-s_f) = 1500 ( 1-0.04)

= 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 P₀ / N_fl

= 9.55 x 7355 /1440

= 48.78 Nm

Q.19 A 2.2 kVA, 440 / 220 V, 50 Hz, step-down transformer has the following parameters referred to the primary side : and . The transformer is operating at full-load with a power-factor of 0.707 lagging. Determine the voltage regulation of the transformer. (10)

Ans:

Given : P₀ = 2.2 kVA , 440/220 V , 50 Hz

R_01
= 3Ω , X₀₁ = 4Ω , R_m= 2.5kΩ , X_m = X₀ = 2kΩ

cosF = 0.707 lagging

Therefore sinF = 0.707

Find Voltage regulation

cosF = 0.707 ; therefore F = 45^o

(Voltage drop) = I₂ (R₀₁ cosF + X₀₁ sinF)

and I₂ = P₀/ V₂ cosF

= 2.2 x 10³ x 0.707 / 220 x 0.707

= 1 x 10

I₂ = 10A

Voltage drop = 10 (3 x 0.707 + 4 x 0.707)

= 10 (4.950)

= 49.50 Volts

Therefore, Voltage regulation = ( voltage drop /V₂ ) x 100

= (49.50 / 220) x 100 = 22.50%

Q.20 A 9-kVA, 208 V, 3-phase, Y-connected, 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 full-load at 0.8 power-factor lagging at rated voltage. Calculate the voltage regulation for rated load at 0.8 power-factor (leading). (10)

Ans:

P₀ = √3V_L I_L cos F = 9kVA ; I_L = I_ph= P₀ / √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 = P₀ / √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 X₁)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 240-V, 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

N₁ = 850 rpm ; I_L = 72 Amp. At rated load

R_a = 0.242, R_sh = 95.2Ω

N₂ = 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 = N₁ F₁/ N₂ F₂

Therefore, F₁/ F₂ = (E_b1 / E_b2) x (N₂ / N₁)

= (223.19 / 227.80) x (1650 / 850) = 36826.35/19363

F₁/ F₂ = 1.90/1 = 19/10

Therefore, change in flux ∆F = (F₁ - F₂ )/ F₁ x 100

= 9/19 x 100 = 47.37%

Q.22 The power input to the rotor of a 3-phase, 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 = (1-S) P_i

= (1- 1/30) x 80 kW = 77.33kW

Q.23 The parameters of the equivalent circuit of a 150-kVA,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 short-circuit test (LV side) with full-load current flowing. What is the input power and its power factor? (8)

Ans: Given Rating = 150kVA \ P_o =150kVA

2400/240 V \ V₂ = 2400V; V₁ = 240 V

R₁ = 0.2Ω X₁= 0.45Ω

R₂ = 2 x 10^-3Ω X₂ = 4.5 x10^-3Ω

R_i = 10 kΩ X_m= 1.6kΩ

Find I₂o = ?, P₂o = ?, cos F₂o = ?,

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 = V₁ I₀ cos F₀

= 240 x I₀ cos Fo …(1)

I₀ = V₁/Z_m ; V_{1 =}240 volts; ;

Z_m = √(10² + 1.6²) = 10.13 kΩ (Z_m = √ R₁² + X_m²)

I₀ = 240 / 10.13 = 23.69 x 10^-3 Amp. Or 23.69 mA.

Power factor cos F₀ = R_i / Z_m

= 10/10.13

cos F₀ = 0.987

Now open circuit power (W_0i) = 240 x I₀ x cos F₀

= 240 x 23.69 x 10^-3 x 0.987

W_0i = 5.6 watt

(ii) R₀₁= R₁ + R₂ / K² (K = V₂ /V₁ )

= 0.2 + 2x10^-3/100

= 0.20002Ω

X₀₁= X₁ + X₂ / K²

= 0.45 + 4.5x10^-3/100

= 0.450045Ω

Z₀₁ = √ (R₀₁² + X₀₁²)

= 0.4924Ω

Full load primary current (I₁) = 150000/2400 = 62.5 Amp (max.)

Short circuit p.f. = R₀₁/ Z₀₁ = 0.20002 / 0.4924 = 0.406

V_SC = I₁Z₀₁ = 62.5 x 0.4924 = 30.78 Volt

\ power absorbed = I₁²R₀₁ = (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_LI_L cosF

\I_L = 800 x 10³/ √3 x3300 x 0.707

I_ph = I_L /√3 = 114.30 Amp.

Now, excitation e.m.f (E₀) will be :

E₀ =√ (V_ph cosF)²+ (V_ph sinF + I_phX_ph)²

=√ (3300 x 0.707)²+ (3300 x 0.707+ 114.3 x 18)²

=4.9712 x 10³ 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 no-load 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

N₂ = full load speed

Find I_af = ? ; & N₂ = ?

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 = N₀ ( fPZ/60A) (2)

E_b2 / N_{0 =} fPZ/60A = 247/1000 = 0.247

Now E_b2 at full load

E_b2 = N₀ ( fPZ/60A) = V- I_af R_a

E_b2 = N₂ x 0.247= 250- I_af x 0.6 (3)

Now T_a at no load

T_a0 = 9.55 E_b1I_a0 / N₀

= 9.55 x 247 x 5 / 1000 = 11.79 Nm

T₁ / T₂ = I₁ / I₂

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)

N₂ x 0.247 = 250 - 42.41 x 0.6

\ N₂ = 224.55/ 0.247 = 909.13 rpm

\ Full load speed = 909 rpm

Q.26 A 400Volts, 1450 rpm, 50 Hz, wound-rotor 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)

Ans:

V = 400V; N = 1450 rpm ; f =50 Hz

R₁= 0.3 W R₂^’ = 0.25 W ; X₁=X₂^’ = 0.6 W ; X₀ = 35W

Rotational losses =1500 W

Find starting torque T₁ = ? I_f =?

R₀₁= R₁ + R₂^/ = 0.3 + 0.25 =0.55 W

X₀₁= X₁ + X₂^/ = 0.6 + 0.6 =1.2 W

Z₀₁ = Ö (R₀₁² + X₀₁² )=Ö0.55² + 1.2² = 1.32W

S = R₂^/ / Ö{R₁² +(X₁ + X₂^/ )² } = 0.25 /Ö0.3² +( 0.12)²

= 0.25 / Ö1.53 = 0.2021

\ N_S = N/ (1-S) = 1450 / (1-0.2) = 1812 rpm

I₂^/ = I_f = V_ph / Ö{( R₁ + R₂^/ )² +( X₁ + X₂^/ )² }

= 400/Ö3 (V_ph = V_L / Ö3 let motor Y connected)

Ö (0.55)² + (1.2)²

Or I_f = V/Z₀₁ = 400/Ö3

1.32

= 175.16 amp.

Torque developed by rotor

T_g = (3 I₂^/ R₂^/ )/ S

2pN_S /60

Or T_g = (3 I₂^/ R₂^/ ){(1-S)/ S } Nm (N = (1-S)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 I₂^/ R₂^/ ){(1-S)/ S }- rotational losses Nm

2pN /60

(N = (1-S)N_S )

= 606.86 - 1500/151.67

= 596.97 Nm

Q.27 A universal motor (ac–operated) has a 2-pole 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 ; V₁ =100 volts

P_i =300W find X_a and f_m = ?

P₁ = V₁I₁cos f

\ cos f = P₁ / V₁I₁

= 300/ 100 x 4.6 = 0.652

E_bdc = V - I_aR_a or (NfPZ / 60A )

= 100 - 4.6 x 3.5

= 100 -16.1 = 83.9 volts

E_bac = Vcosf - I_aR_a

= 100 x 0.652 - 4.6 x 3.5 = 49.11 volts

And V² =((E_bac + I_aR_a )² + (I_a X_a)²

(4.6 X_a )² = 100² - (65-2)²

21.16 X_a² = 5749

X_a = 16.48 ohms

E_bdc = (Nf_mPZ / 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)

Ans:

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 10³ + 3.96 x 5000 x 10^-3 / 10 x 10³

V_s ={0.0195 x 20 x 5000 x 10³ + 3.96 x 5000 x 10^-3 / 10 x 10³ } + 10 x

10³ volts

= 1.2175 x 10⁴volts

= 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)

Ans:

P₀ = 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 10³ / 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 ) = ( V₁I₁ cosΦ - losses )/ V₁I₁ cosΦ

= ( V₁I₁ cosΦ - I₁²R₀₁- Wi )/ V₁I₁ cosΦ

= 1

Differentiating above equation for maximum efficiency

Therefore,

= 0 - - = 0

and

Therefore,

W_i= W_c

and load current, I_l=(√ W_i/R₀₂ )

Q.31 A 22 KV, 3 phase star-connected turbo- alternator with a synchronous impedance of 1.4 /phase is delivering 240 MW at unity p.f. to a 22 KV grid. If the excitation is increased by 25%, then the turbine power is increased till the machine delivers 280 MW. Calculate the new current and power factor. (10)

Ans:

Given : V_L = 33 kV, 3 phase, star connected Synchronous Generator

\ V_ph = 33/√3 = 19.1 kV

Z_ph = 1.4Ω

P₀₁ = 240 MW = 240 x 10⁶ Watt

cos F₁ = 1

Ψ₂ = 1.25Ψ₁ (Flux / excitation)

P₀₂ = 280 MW = 280 x 10⁶ Watt

Find I_L2 = ? ; cos F₂ =?

E_g1 /ph = V_ph1 + I_ph1Z_ph

= (33 x 10³ /√3) + {(240 x 10⁶ /√3 x 33 x 10³) x 1.4}

(P₀₁ =√3 V_L1 I_L1 cos F₁ & I_L1 = I_{ph 1})

= 24.98 x 10³ volts

≈ 25 kV

Now

Ψ₂ = 1.25Ψ₁, due to 25% increased excitation )

E_g2 _/ph = E_g1 _/ph *

\ E_g2 _/ph = 31.23kV

or E_g2 _/ph = V_ph + I₂Z_ph

I_{2 ph} = E_g2 -V/Z_ph = 12.125 x 10³ / 1.4

= 8.66 kA

Now P₀₂ =√3 V_L I_L2 cos F₂ ( I_L2 = I_{ph 2} : V_L = V_L1 = V_L2 ; P₀₂ = 280 MW )

cos F₂= P₀₂ /√3 V_L I_L2

= 280 x 10 ⁶ / √3 x 33 x 8.66 x 10⁶

= 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)

Ans:

Given R_a=0.55Ω,

R_T =0.75 additional resistance in series with armature

V = 250V

I_a = 30A at full load Then N₂ = 633 rpm

I_sh = Constt.

Find N₁ = ? at full load ; N₃ = ? at double load

N₁ α V - I_a1R_a

N₂ 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

N₁ = N₂ *1.106 = 700.5

N₁ = 701 r.p.m at full load without R_t

Speed N₃ at twice full load torque without R_t

\ ₌. 250 – 30 x 0.55 . = 233.5

250 – 60 * 0.55 217

\N₃ = N₁/ 1.076 = 701/1.076 = 651 rpm

Speed of motor at twice full load torque = 651 rpm

Q.33 A 4-pole, 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 equivalent-star 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

R₁= 1.6 Ω, X_1
=2.4Ω ; R₂^/ =0.48 Ω ; X₂^/₌1.2Ω

X_m = X₀ = 40 Ω

Rotational losses = 720 W; Wi = 0 Watt

N=1470 rpm

Find I₁ = ? ; 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 I₁ = V₁ / Z₀₁

V₁ =400 V; Let R_o is negligible

Z₀₁ = Z₁ + Z_{AB ;} where Z_ABis the impedance between point A and B

Z₁ = (R₁ + jX₁) = (1.6 + j 2.4)

Z_AB= j X_m [ R¹₂/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

\ Z₀₁ = (1.6 + j 2.4) + (16.9 + j11) = 18.5+ j13.4 = 22.84 ∟35.9^o

(i) \ I₁ = V₁/Z₀₁ = (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 = (1-S)P₂ ; Where P₂ is the power of air gap

= (1-0.02) 3I₂^׀²(R₂^// S)

= 0.98 x 3 x I²₁ R_AB

= 0.98 x 3 x (10.24)² + 16.9

= 5209.95

= 5210 Watt

Net Mech. Power = Total Mechanical Power – Rotational Losses

= 5210 – 720

P_mo= 4490 Watt

(iv) Net Torque = P_mo / ( 2 Π N/60)

= 4490 / (2 x 3.14 x 1470/60)

N = 1500 (1-0.02) = 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 W₁ = 0

(b) Stator ‘Cu’ losses = 3I₁² R₁ = 3 x (10.24)² x 1.6

= 3 x 167.77

= 503.31 Watt

= 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 2-pole 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_aR_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 10⁶

= 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_a² +X_a² = 28.84 Ω

\ R_a² +X_a² = 832 Ω

X_a² = 832 - R_a²

= 832 -5.5²

= 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 low-voltage 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 step-up one and is delivening 50 KVA at a power factor of 0.8 lagging. (12)

(b) Compute the efficiency under conditions of part (a). (4)

Ans:

Given : P₀ = 50 kVA, V₁ = 2300V , V₂ = 230V, f = 60 Hz

R₁ = 0.65Ω (H.V.Side), R₂ = 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) V₁ for rated V₂ when acts as step up transformer and delivering 50kVA at cos F = 0.8

(b) efficiency

At S.C test

Z₀₂ = V_SC / I₂ = 41.5 /21.7 = 1.912Ω & K =1/10

\Z₀₁ = Z₀₂ / K² = 1.912 x 10² = 191.2 Ω

R₀₁ = R₁ + R₂^/ = 0.65 + 0.0065/100 (K =1/10, H.V side is R₂)

R₀₁ = 0.13 Ω

R₀₂ = R₀₁ x K² = 0.13 / 100 = 0.0013 Ω

Now X₀₁= √ Z₀₁² – R²₀₁ = √ (191.2)² – (0.13)²

X₀₁= 191.2 Ω

& X₀₂= √ Z₀₂² – R²₀₂ = √ (1.912)² – (0.0013)²

X₀₁= 1.912 Ω

Total transformer voltage drop referred to secondary

VD₂ = I₂ ( R₀₂ cosF₂ + X₀₂ sinF₂)

= 21.7 ( 0.0013 x 0.8 + 1.92x 0.6) = 21.7 (1.153)

= 25.02 Volts

\V₁^/ = V₁ + VD₂ = 230 + 25.02 = 255.02 Volts.

\V₁ =255.02 volts for V₂ = 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 + I₁² R₁+ I₂² R₂

=

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 three-phase, 335-hp, 2000V, six pole, 60 Hz, Y-connected squirrel-cage 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)

Ans:

Given: 3 phase , 335 HP, 50 Hz, V = 2000V Induction motor

P₀= 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) R₁= 0.2Ω, R₂^/= 0.203 Ω

X₁ = X₂^/ = 0.707 Ω

x₀ = 77 Ω , R₀= 450 Ω

R₀₁= R₁+ R₂^/ = 0.2 + 0.203 = 0.403 Ω

X₀₁= X₁+ X₂^/ = 0.707 + 0.707 = 1.414 Ω

Z₀₁=√ R₀₁²+ X₀₁² = √ (0.403)²+ (1.414)² = 1.47 Ω

Load current I₂^/ = I_L

\ I₂^/ = V_Ph / (R₁ + R₂^/ / s) + j( X₁ + X₂^/)

= 1154.7 ∟0⁰ / (0.2 + 0.203/0.015) + j (0.707 + 0.707)

= 1154.7 ∟0⁰ /13.733 + j 1.414 = 1154.7 ∟0⁰ / 13.80 ∟5.88⁰

= 83.67 ∟-5.88⁰

\ I_L = 83.67 Amp.

P.f = cos ( -5.88⁰) = -0.99 lagging

(ii) Torque generated (T_g) or developed = (9.55 x 3 I₂^/2 R₂^// 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 10³ 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 I₂^/2 + R₂ = 3 x (83.67)² x 0.203 = 4.26 kW

stator ‘Cu’ losses = 3 I₁^/2 R₁= 3 x (86.2)² x 0.2 = 4.458 kwatt

\ Total losses = 4.1 + 4.26 + 4.46 = 12.82 kW

I₁ = I₂^/2 + I₀= 83.67 + 2.53 = 86.20 Amp

I₀ = V_Ph/ √ R₀²+ X₀²

= 1154.7/ √ 450²+ 77²

= 2.529Amp

h = {246.40 / ( 246.40 + 12.82) }x100

= 95.05%

Q.37 A 2300-V, three phase, 60 Hz, star-connected 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 power-angle is 15 electrical degrees as measured by a stroboscope. Neglect ohmic resistance and determine:

(a) the induced excitation per phase.

(b) the line current

Ans:

Given : 2300V , 3 phase, 60Hz, Synch. Motor

X_S = 11Ω /ph , Star connected, V_Ph = 2300/ √3 V

P₀ = 200 hp = 200 x 735.5 = 147.1 kW

h = 90%

\ P_i = P₀ /0.9 = 163.44 kW

Power angle α = 15⁰ ( 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 10³ x 11 / 3 x 1327.9 x sin 15⁰

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 15⁰

= - 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 ∟38.32⁰

(ii) Line current = 52.28 Amp

(iii) & p.f cos F = cos 38.32⁰ = 0.78

Q.38 When a 250-V, 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)

Ans:

Given V_L = 250 V P₀ = 50 hp = 50 x 735.5 36.78 kw

N₁= 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 N₂ =?

Power developed (P_m) = ?

Efficency (h) = ?

(i) E_b1 = N₁(FPZ/60A)

And V = E_b1 + I_aR_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 = N₁ /N₂ (when F₁ = F₂)

N₂= N₁ 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

Total losses = Armature ‘Cu’ Losses + Field loss + Rotational losses

=(I_a² R_a^/ /1000) + 1.6 kW + 2.0kW

= (160² x 0.54/ 1000) + 1.6 +2.0

= 17.42 kW

(iii) Efficiency = (P_i– losses / P_in) x 100

= (26.18-17.42 / 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 : V₀ = 200 V ; I₀ = 4A ; W₁ = 120 W

S.C Test : V_SC= 60 V ; I_SC = 10A ; W_SC = 300 W

Primary equivalent secondary induced voltage

E₂^/ = E₂ / K

V₂^/ = V₂ /K

& I₂^/ = KI₂

From O.C Test:

V₀I₀ cosF₀ = W₀

\ cosF₀ = W₀ / V₀I₀ = 120 / 200 x 4 = 0.15

\ sinF₀ = 0.988

Now I_w = I₀ cosF₀ = 4 x 0.15 = 0.60 Amp.

I_u = I₀ sin F₀ = 4 x 0.988 = 3.95 amp.

R₀ = V₀ / I_w = 200 /0.6 = 333.33 Ω

X₀ = V₀ / I_u = 200 /3.95 = 50.63 Ω

From S.C test:

Z₀ = V_SC/ I_SC = 60 / 10 = 6Ω

and K = 200 /2000 = 1/10 = 0.1

Z₀₁ = Z₀₂/ k² = 60 / (1/10)² = 600Ω

Now I_SC² R₀₂ = W _SC

\ R₀₂ = W _SC/ I_SC² = 300/100 =3Ω

R₀₁ = R₀₂ / k² = 3/(1/10)² =300Ω

X₀₁ = √( Z₀₁² - R₀₁²) = √( 600² - 300²)

X₀₁= 519.62Ω

Q.40 The efficiency of a 3-phase 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 = 25.84⁰

E_R = IZ_S = 24 (0.2 +j2) volts

= (4.8 +j48) volts = 48.23∟84.29⁰

q =84.29⁰

Now at leading p.f

E_b/ph = V + I_aZ_S cos {180 – (q + F)} + j I_aZ_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_LI_Lcos F

= √3 VIacos F = √3 x 400 x 24 x 0.9

= 14,964.92 watt

total copper losses = 3 I² 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

N₁ = 600 rpm

I₁ = 30A = Ia₁

Find : N₂ & I₂ ; when R =20Ω added with R_a in series

E_b1 = V –I_a R_a (losses are negligible , \ I_aR_a =0)

E_b1 = V = 200v (I_a R_a= 0)

T₁ = 9.55 E_b1 I₁ / N₁ = 9.55 x 200 x30 / 600

= 573Nm

T α N²

\ T₁ / T₂ = N₁ ² / N₂ ²

\ N₂ ² / T₂ = N₁ ² / T₁ = 600² / 573

Or N₂ = 600 √ T₂ / 573 (1)

and N₁ / N₂ = E_b1 / E_b2 . F₂ /F₁ (F₂ = F₁ = constt)

\ N₁ / N₂ = E_b1 / E_b2

or N₂ = N₁ (E_b2 / E_b1)

N₂ = 600 (E_b2 / 200)

N₂ = 3E_b2 (2)

And

E_b2 =200 -20I₂ (3)

(E_b2 = V –I_a R_a)

E_b2 = N₂ / 3 from eqn. no. (2) put in eqn. no (3)

\ N₂ = 600-60I₂ (4)

T₁ / T₂ = N₁ ² / N₂ ² = I₁/ I₂

\ 600 ² / N₂ ² = 30 / I₂

30 / I₂ = 600 ² /( 600-60I₂)² (N₂ = 600-60I₂)

30 / I₂ = 600 x 600 / 60 x 60 ( 10-I₂)²

10I₂ = 3(100 + I₂² - 20 I₂)

10I₂ = 300 + 3I₂² - 60 I₂

\ 3I₂² -70 I₂ + 300 = 0

I₂ = 70 ± √ 4900 – 4 x 3 x300 = 5.66 or 17.66

I₂ = 5.66 amp

\ N₂ = 600-60I₂

= 600 – 60 x 5.66 = 260.55 rpm

N₂ = 260 rpm

I₂ = 5.66 amp, 17.66 amp is not possible for N₂

Q.42 A 3-phase 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 = R₂ / X₂

\ (T_st / T_max ) = 2a /(1+a²) = 0.5/1

2a /(1+a²) = 1/2

4a = 1+a²

a²- 4a + 1 = 0

a = 4 ± √ 16 – 4 x 1 x 1

a = 4 ± √ 12

a = 3.73 or 0 .2679

a = slip at max torque = 0.2679

Now, T_f /T_max = 1/2 = 2aS_f / a² + S_f² (S_f = full load slip)

2 x 0. 2679 x S_f /{(0.2679)²+ S_f² } = 1/2

0.5358 S_f / {(0.0717)+ S_f² } = 1/2

\ 1.071 x S_f =(0.0717)+ S_f²

S_f² + 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 2-pole 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 ; N₁ =5000 rpm ; I_a1= 4.6 amp

V_a = 100v = I_a1R_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_aR_a = 100 x 0.652 – 4.6 x 3.5

= 49.1 volts

and V² = ( E_bac + I_aR_a)² + (I_aX_a)²

100² = ( 49.1 + 4.6 x 3.5)² + (4.6X_a)²

(4.6X_a)² = 100² - ( 49 + 16.1)² = 10000 – (65.2)²

= 10,000 – 4251 = 5748.96

21.16 X_a² = 5748.96

\ X_a = √5748.96 / 21.16

= 16.48 Ω

and E_bdc = V- I_aR_a = 100 - 4.6 x 3.5= 83.9 V

and E_bdc = NF_mPZ / 60A

83.9 = 5000 F_m x 2 x 960/ 60 x 2

F_m = 83.9 / 8.0 x 10⁴ = 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

P₀ = 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 10⁶/ 132 x 1000 = 25000/132

Line loss = 3 I² 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 3-phase transformer bank consisting of a three one-phase transformer is used to step-down the voltage of a 3-phase, 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

V₁ = 6600V, I₁ = 10A,

V₂ = ? , I₂ = ?

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

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V_PP /V_PS = N₁ / N₂

V_LP /V_LS = √3 V_PP/V_PS = √3 N₁/N₂ = 12 √3

V_LS = V_LP/ 12√3 = 6600 / 12 √3

V_LS = 315.33 Volts

I_PP / I_PS = N₂ / N₁

I_LP / I_LS = I_PP / √3 x 1/ I_PS =(1/√3 ) x ( N₂ / N₁)

I_LS = (1/√3) x ( N₁ / N₂) x I_LP = 12 √3 x 10 Amp.

= 120√ 3 Amp.

o/ρ kVA = V_LS x I_LS

= 6600 x 12√3 x 10

12 √3

o/ρ = 66kVA

(ii) ∆/Y

For Y : V_p = V_L/√3 , I_p = I_L

∆ : V_p = V_L , I_p = I_L/√3

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V_PP /V_PS = N₁ / N₂

V_LP /V_LS = V_PP/√3 V_PS = 1/√3 x N₁/N₂ = 12 /√3

V_LS =√3V_LP/ 12 = √3 x 6600 / 12

V_LS = 550√3 Volts

I_PP / I_PS = N₂ / N₁

I_LP / I_LS = √3I_PP / I_PS =√3 N₂ / N₁ = √3/12

I_LS = 12/√3 x 10 = 40√3 Amp.

o/ρ kVA = V_LS x I_LS

= √3 x 6600 x 12 x10

12 x √3

o/ρ = 66kVA

Q.46 A 3300V, delta-connected motor has a synchronous reactance per phase (delta) of 18Ω. It operates at a leading power factor of 0.707 when drawing 800kW from the mains. Calculate its excitation emf. (8)

√3 V_L I_a cos F = √3 x 3300 x I_a x 0.707 =80,000

\ Line current =198A, Phase current I_a =198 √3 =114.3A

Z_S = 18Ω : I_a Z_S = 114.3 x 18 = 2058 V

F = cos^-1 0.707 ; F = 45^o ; θ = 90^o

cos ( θ+F ) = cos 135^o = -cos 45^o = - 0.707

From Fig, we find

E_b² = 3000² + 2058² -2 x 3000 x 2058 x -0.707

E_b = 4973 V

Q.47 The magnetization characteristic of 4-pole DC series motor may be taken as proportional to current over a part of the working range, on this basis the flux per pole is 4.5 mWb/A. The load requires a gross torque proportional to the square of the speed equal to 30 Nm at 1000 rev/min. The armature is wave-wound and has 492 conductors. Determine the speed at the which the motor will run and current it will draw when connected to a 220V supply, the total resistance of the motor being 2.0 Ω.

(7)

Ans:

E_a = (f N Z / 60 ) x (P/A)

= (4.5 x 10^-3 x Ia) x N x 492 (4/2)

= 0.0738 N I_a (1)

The torque developed :

T = f I_aZ (P/A) = (1/2p) (4.5 x 10 ^-3 I_a) x 492 (4/2)

= 0.705 I_a² (2)

Further E_a = V - I_a (R_a + R_Se ) = 220 -2 I_a (3)

Substituting equation (1) in (3)

0.0738 N I_a = 220 - 2 I_a

I_a = 220 / 2 + 0.0738N (4)

Substituting the expression for I_a in equation (2)

T = 0.705 (220/2+0.0738 N)²

Given , T_L = K_L N²

From the given data K_L can be calculated as

K_L = 30/1000² = 3 x 10 ^-5 Nm/rpm

Under steady operation condition T_L =T

Or 3 x 10^-5 N² = 0.705(200 / 2+ 0.0738 N)²

N = 662.6 rpm

Substituting for N in equation (4)

I_a = 220 / (2 + 0.0738 x 663.2) =4.32 Amp.

Q.48 A 150kW, 3000V, 50Hz, 6-pole star-connected induction motor has a star-connected slip-ring rotor with a transformation ratio of 3.6 (stator/rotor). The rotor resistance is 0.1Ω/phase and its per phase leakage inductance is 3.61 mH. The stator impedance may be neglected. Find (i) the starting current and torque on rated voltage with short-circuited slip-rings, and (ii) the necessary external resistance to reduce the rated voltage starting current to 30A and the corresponding starting torque. (8)

Ans: X₂= 2Π x 50 x 3.61 x 10^-3 = 1.13Ω

K = 1/3.6

⇒ R₂^/ = R₂ / K² = 3.6² x 0.1 = 1.3Ω

X₂^/ = 3.6²x 1.13 = 14.7 Ω

(i) I_ST = V/ √ (R₂^/ )² + (X₂^/)² = (3000/√3) / √ (1.3 )² + (14.7)²

= 117.4 A

N_S = 120 x 50 /6 = 1000 rpm = 50/3 rps

T_ST = (3/2ΠN_S ) {V²/ (R₂^/ )² + (X₂^/)²}

= (3 / 2 Π x 50/3 ) x { (3000/√3)² / (1.3 )² + (14.7)²}

= 513 Nm

(ii) Let the new resistance be R for I_ST =30A

I_ST = V/ √ (R )² + (X₂^/)² = 3000√3 / √ (R )² + (14.7)²

R² =(3000 /√3 /30 ) - (14.7)²

= (100/√3)² -14.7²

R = 55.83Ω

Necessary external resistance = R-R₂^/ = 55.83 – 1.3 = 54.53Ω

Corresponding starting torque

T_ST1 = 3 / 2Π (50/3) x {(3000 /√3)² x R / (R )² + (14.7)²}

= 3 / 2Π (50/3) x {30 x 30 x 55.83 / 1}

T_ST = 1438.89 Nm

Q.49 An ac operated universal motor has a 2-pole armature with 960 conductors. At a certain load the motor speed is 5000 rpm and the armature current is 4.6A; the armature terminal voltage and input are respectively 100 V and 300 W. Calculate the following quantities assuming an armature resistance of 3.5 Ω.

(i) Effective armature reactance

(ii) Max. value of useful flux/pole. (8)

Ans:

The operating conditions in terms of voltage and current of the armature circuit are shown in the Fig :

100 x 4.6 cos f = 330 W

Or f = 49.3^o (lagging because of the reactive nature of the circuit)

(a) From the circuit the following can be written

100 x Ð49.3^o - E_a Ð0^o = 4.6Ð0^o

3.5 x j X_a

E_a is in phase with I_a

Or 65.2 + j75.8 - E_a = 16.1 + j 4.6 X_a

Equating real & imaginary parts

E_a = 65.2 - 16.1 = 49.1 V

X_a = 75.8 / 4.6 = 16.5 W

(b) E_a = 1/Ö2 {(f N Z / 60 ) x (P/A)}

f = Ö2 x 49.1 x 60

5000 x 960

= 0.868 mwb

Q.50 Define voltage regulation of a single phase transformer. The primary and secondary winding of a 40kVA, 6600/250V, single phase transformer have resistance of 10 ohm and 0.02 ohm respectively. The total leakage reactance is 35 ohm as referred to the primary winding. Find full load regulation at a pf of 0.8 lagging. (8)

Ans: Primary Voltage V₁ = 6600 V

Secondary Voltage V₂ = 250 V

Transformation ratio k = V₂ / V₁ = 250/6600 = 0.03788

Equivalent resistance of transformer referred to secondary,

R_o2 = k²R₁ + R₂ =( 0.03788)² x 10 + 0.02

= 0.03435Ω

equivalent leakage reactance of transformer referred to secondary,

X_O2 = K² X₀₁ = ( 0.03788)² x 35 = 0.05022Ω

Secondary rated current , I₂ = Rated kVA x 1000/ V₂

= 40 x 1000 /250 = 160 A

Power factor cos F = 0.8 and sin F = √1-0.8² = 0.6

Full load regulation = I₂ R₀₂ cos F + I₂ X₀₂ sin F x 100

E_b

= 160 x 0.3435 x 0.8 + 160 x 0.05022 x0.6 x 100

250

= 3.687 %

Q.51 A star connected synchronous motor at 187 kVA, 3-f, 2300V, 47A, 50Hz, 187.5 rpm has an effective resistance of 1.5 ohm and a synchronous reactance of 20 ohm per phase. Determine internal power developed by the motor when it is operating at rated current and 0.8 power factor leading. (6)

Ans: Line voltage , V_L = 2300V

Line Current I_L = 47 A

Power Factor cos F = 0.8

Power supplied to the motor, P = √3 V_L I_L cosF

= √3 x 2300 x 47 x 0.8 = 1,49,780Watts

= 149.78 kW

Total Copper loss = 3 I²R

= 3 x (47)² x 1.5 = 9,940 watts or 9.94 kW

Internal power developed = P - 3 I²R

= 149.78 – 9.94 = 139.84 kW

Q.52 A 220V dc shunt motor takes 22A at rated voltage and run at speed of 1,000 rpm. Its field resistance is 100 ohm and armature resistance is 0.1 ohm. Compute the value of additional resistance required in armature circuit to reduce the speed to 800 rpm when (i) load torque is proportional to speed and (ii) when load torque varies as the square of the speed. (10)

Ans: In normal Condition

Line current , I_L1 = 22A

Shunt field current , I_sh = V/ R_sh = 220 /100 = 2.2A

Armature current, I _a1 = I_L1 –I_sh = 22 – 2.2 = 19.8A

Back emf, E_b1 = V – I_a1 R_a = 220 – 19.8 x 0.1 = 218.02 V

Speed N₁ = 1000 rpm

Let the additional resistance required in armature current be of R ohms to reduce the speed to 800 rpm when the load torque is proportional to speed.

T₂ = T₁ N₂/ N₁ = T₁ 800/ 1000 = 0.87

Or I_a2F₂ = 0.8I_a1F₁

Or I_a2 = 0.8 x 19.8 = 15.84 A

E_b = V – I_a2 (R + R_a ) = 220 – 15.84 (R + 0.1)

=218.416 – 15.84 R

Since E_b2/ E_b1 = N₂ /N₁

(218.416 – 15.84R) / 218.02 = 800 / 1000

R = 2.778 Ω

(ii) Let the additional resistance required in the armature current be of R^/ ohms to reduce the speed to 800 rpm when no load torque varies as the square of the speed.

T₃ = T₁ x (N₂ /N₁)² = T₁ x (800 /1000)² = 0.64 T₁

Or I_a3F₃ = 0.64I_a1F₁

I_a3 = 0.64 x 19.8 = 12.672 A

E_b3 = V – I_a3 (R^/ + R_a ) = 220 – 12.672 (R^/ + 0.1)

= 218.7328- 12.672 R^/

Since E_b3/ E_b1 = N₃ /N₁

(218.416 – 12.672 R^/) / 218.02 = 800 / 1000

R^/ = 3.497 Ω

Q.53 A 120V, 60 Hz, ¼ hp universal motor runs at 2000rpm and takes 0.6 Amp when connected to a 120V dc source. Determine speed, torque and power factor of the motor, when it is connected to a 120V, 60 Hz supply, and is loaded to take 0.6 Amp(rms). The resistance and inductance measured at terminals of the machine are 20 ohm and 0.25H respectively. (8)

Ans: When connected to DC supply

Supply Voltage, V = 120V

Current drawn, I = 0.6 A

Back e.m.f, E_b = V-IR

= 120 – 0.6 x 20 = 108V

Speed, N dc = 2000 rpm

When connected to AC supply

Supply Voltage , V = 120V

Current drawn, I = 0.6 A

Resistance drop = IR = 0.6 x 20 = 12V

Counter emf, E = √[V² – (IX)² ]- IR

= √[(120)² – (47.12)² ]-12

= 98.36V

Speed Nac =

N_dc x E

E_b

= 2000 x 98.36

108

= 1,821.5 rpm

Power factor , cos F = (E + IR)/V = (98.36 + 12)/ 120

= 0.92 lagging

Torque developed, T = E x I / (2Π N/60)

= 98.36 x 0.6 x 60/(2Π x 1821.5) = 0.31 Nm

Q.54 A single phase generator supplies an inductive load of 4800 KW at a power factor of 0.6 lagging by means of an overhead line which is 25 km long. The line resistance and inductance are respectively 0.02 and 0.58 m H per km. The voltage at the receiving end is to be kept constant at 10.5 KV. Find the sending end voltage and the voltage regulation of the line. (8)

Ans Given :

1F generator, inductive load, kVAR = 4800 kW

cos F = 0.6; distance = 25 km

R = 0.02Ω/km L = 0.58 mH/km V_R = 10.5 kV

Find V_S =? ; % regulation = ?

Drop in line ∆V = V_S - V_R

V_S - V_R = RP + XQ / V_R

Where R = total line resistance

P = Active Power transfer

Q = Reactive power transfer

V_S = sending end voltage

V_R = receiving end voltage

P = kW = Reactive power / tanF

= 4800 /1.333 = 3600kw

Q = 4800 kW

V_S = 0.02 x 25 x 3600 + 4800 x 2 x 3.14 x 50 x 0.58 x 10^-3 x 25 + V_R

10500

= 19.225 + V_R

= 19.225 + 10500

= 10.52 kV

% regulation = (V_S - V_R / V_S ) x 100

= 0.19%

Code: AE10 Subject: ELECTRICAL ENGINEERING

PART - III

DESCRIPTIVES

Q.1 Draw and explain the phasor diagram of a transformer on load at a lagging power factor. (7)

Ans: (D)

Fig. AA1

Fig. AA2

Fig AA2 shows the phase diagram of a transformer on a load at lagging power factor and corresponds to the equivalents circuit of transformer (Fig. AA1) in which all quantities are referred to the primary. Thus is the secondary terminal voltage referred to the primary [where over bar implies a phasor].

Where R₁and R₂ are primary and secondary resistances, N₁ and N₂ are primary and secondary number of turns and and are leakage reactances of primary and secondary windings.

Also

The voltage which is applied to the primary can be written as

Q.2 Explain with proper phasor diagrams the operation of a 3 phase synchronous machine with normal excitation at the following conditions:

(i) The machine is floating on the supply bus. (7)

(ii) The machine is working as a synchronous motor at no load. (7)

Ans:

Fig B1 shows the circuit diagrams and phasor diagrams of a synchronous machine at generating mode [Fig B1(a)& (c)] and motoring mode [Fig B1(b)&(d)] . The machine is assumed to be connected to an infinite busbar of voltage V_t andits resistance is taken to be zero. It is seen from [Fig B1(c)] that in the generating mode, the excitation emf E_f leads V_tby an angle ; on the other hand in the motoring mode E_f lags V_t.

It can be inferred from the above that, when the machine is floating, it is working neither as a generator nor as a motor and I_awill be equal to zero. and will be equal and have the same phase [Fig. B2]

When the machine is working as a synchronous motor at no load, the figures B1 (b) and (d) will hold good, but with a very small value for the current I_a; correspondingly the angles and will be small. The reason for this is, that at no load the machine will take that much power which is the sum of friction and windage losses. [Fig B3] and is small. will also be small.

Q.3 Draw the torque speed characteristics of a 3 phase induction motor and clearly indicate the effect of change in rotor resistance. (7)

Ans: The Thevenin equivalent of an induction motor circuit model is given in Figure D1

The torque developed can be expressed as

The torque speed characteristics [or, characteristics] can be obtained from Equation (A); for different values of rotor resistance are shown in Fig.D2

Fig. D2 Torque-slip characteristics of induction motor with
increasing values of rotor resistance

Effects of change in rotor resistance are:

(1) The starting torque is affected: it increases with increase in R₂, goes to a maximum of T_max and then decreases with further increase in R₂.

(2) T_max, remains the same but the slip at which it occurs is changed with R₂.

Q.4 For a small and sensitive servo mechanism give four reasons why a.c. servo motors are generally preferred to d.c. servo motor. (7)

Ans: A dc servomotor is often employed in a control system where an appreciable amount of shaft power is required. These servomotors have separately excited fields. They are either armature controlled with fixed field or field controlled with fixed armature current. For example the dc servomotor [Fig.E1] used in instrument employs a fixed permanent magnetic field, and the control signal as applied to the armature terminals.

Fig. E3

The two phase ac servomotor [Fig E2] on the other hand is ideally suited for low power control applications. The two phases are called control phase (phase a) and reference phase (phase m), the latter being excited at a fixed magnitude of synchronous a.c. voltage, both voltages being taken from the same source. The control phase voltage is shifted in phase by from the reference phase voltage by means of phase-shifting networks. The motor torque gets reversed by phase reversal of the control phase voltage.

Fig. E2 Control scheme for 2-phase servomotor

The ac servomotor offers several advantages over its d.c. counterpart:

(i) a drift-free ac amplifier is used in the control circuitry.

(ii) it has low rotor inertia and hence faster response.

(iii) it has rugged, maintenance-free rotor construction

(iv) it has no brushes that contact the commutator segments. The rotor can withstand a higher temperature as it does not involve any insulation.

Q.5 What are different types of resistance welding? (3)

Ans: In resistance welding a heavy current is passed through the joint to be welded and the heat caused by the resistance of the joint is sufficient to cause fusion of the metal. Three types of resistance welding exist and they are as follows:

(i) Butt Welding:This method is used for welding of rods, wires or small pipes; the two ends are pressed together mechanically to form a butt joint as shown in Fig G1

(ii) Spot Welding: For jointing two or even three sheets of metal by means of an overlapping joint, as shown in Fig G2, The sheets are held between two electrodes and current is passed between these electrodes, causing fusion at a single spot.

(iii) Seam Welding :Where a continuous joint is required between two overlapping sheets of metal, as in constructing a tank, the two electrodes used in spot welding are replaced by two wheels between which the work travels. However it is not advisable to make a continuous weld, as there is then a tendency for the heat gradually to build up as the welding progresses and cause burning and warping.

Q.6 What are different welding controls used in resistance welding? (4)

Ans: The three different controls commonly in use are as follows:

(i) Mechanical Control for Constant – time equipment: A cam- operated switch connected in the primary circuit of the welding transformer and driven from the welding machine provides a simple device.

(ii) Valve-operated Control for Constant-time: Devices: To overcome the difficulties encountered in the case of mechanically operated gears at high speeds, various valve-operated devices have been developed, for example, the mercury-vapour valve.

(iii) Current and energy- actuated control: With this control, current is allowed to flow until a predetermined amount of energy has been supplied to the weld.

Q.7 Explain the principle of high frequency induction heating. What factors control the depth of penetration of heat? Give the industrial application of this mode of heating. (7)

Ans: Induction heating processes make use of currents induced by electromagnetic action in the material to be heated. Sufficient currents that cause effective heating can be produced only in materials of low resistivity, however it is necessary to use a magnetic field of very high frequency.

For coreless induction furnaces, the depth of penetration(t) is given by the formula

where =specific resistance of the molten charge [200X10^-6ohm/cm for steel]

=frequency(Hz)

and =permeability (this factor has a value of 1.0 for molten steel)

For normal supply frequency, with the power factor in the range 0.8-0.85 and in sizes upto about 1 tonne, the vertical core-type furnace is widely used in foundries for melting and refining brass and other non –ferrous metals. On the other hand the coreless induction furnace is used for the production of very high grade alloy steels; in small sizes it is widely used for work on alloys and precious metals.

Q.8 What is the fundamental difference between thermal and nuclear power plants? (3)

Ans: In a thermal power plant heat is released in combustion of coal; this heat is used in a boiler to raise steam.

Here the coal is conveyed to a mill and crushed into a fine powder, this being termed as pulverization. The pulverized coal is blown into the boiler where it mixes with the supply of air for combustion. Heat is transferred to steam pipes located in the top region of the boiler, these being initially fed with hot water from the boiler feed pump; the hot water then gets converted to steam at high temperature and pressure. This steam is fed to the steam turbines which are the prime movers for electric generators.

In a nuclear power plant binding energy of a nucleus is released by fission, which means breaking the nucleus into smaller fragments. Here one gram of the nucleus of uranium is isotope ²³⁵U releases energy at the rate of 1 MW/per day whereas 2.6 tons of coal produces the same amount of power in a conventional thermal power plant. The process of fission is carried out in a nuclear reactor. Thus the nuclear reactor is a very efficient source of energy because a small amount of fissile material produces a large amount of energy.

Q.9 List the advantages of nuclear power plants over conventional thermal power plants. (4)

Ans:

(i) A nuclear power plant is completely free of air pollution.

(ii) It requires very little fuel in terms of volume and weight and therefore poses no transportation problems. The nuclear power plant may be sited, independently of nuclear fuel supplies, close to load centres. However safety considerations require that these plants be normally located away from populated areas.

(iii) When one pound of pure uranium is completely fissioned it will create as much heat as the burning of 1500 tons of coal.

(iv) Whereas the stock of coal is limited and the supply of coal will go down in the coming decades, it is the nuclear energy that is promising. Nuclear power plants will therefore be important sources of electrical power in future.

Q.10 Discuss briefly the solid state circuits used for the stator voltage control of induction motor derives. (7)

Ans : Fig H1 shows the block diagram for a scheme of stator voltage control. Here the ratio V/f is kept constants, with the frequency being varied by means of an inverter. This in turn will help in maintaining the field flux constant. The rectifier and inverter are rigged by means of the thyristors or GTOs or MOSFETs. Because of its high cost, this type of control is justified only for drives wherein rugged, maintenance-free characteristics of the induction motor are essential.

Q.11 Differentiate between feeder, distributor and service main. (3)

Ans: Fig K1 shows a typical distribution system consisting of various kinds of substations and feeders, finally ending up with consumer’s service lines. The sequence, as shown in the figure, is as follows:

Firstly sub-transmission lines at 66KV emanate from a transmission substation. A distribution transformer at the 66 KV substation steps down this voltage from 66KV to 11KV, that is, the primary feeder voltage. At the distribution substation transformer this (11KV) is further stepped down to 415 V which is the voltage of the consumer’s services lines.

Q.12 What are the advantages of high voltage transmission? Give its limitations also.

(4)

Ans: High voltage transmission is subdivided into HVAC and HVDC transmission systems.

(i) HVAC transmission: Advantages of HVAC transmission are as follows:

As the voltage is increased, the current carried by the conductors decreases. The i²R losses correspondingly get reduced. However the cost of transmission towers, transformers, switches and circuit breakers rapidly increases with increase in voltage, in the upper ranges of a.c. transmission voltages.

(ii) HVDC transmission:

Advantages They (HVDC lines) are economical for bulk power transmission. The voltage regulation problem is much less in DC since only IR drop is involved. There is easy reversibality and controllability of power flow through a DC link. Also there is considerable insulation economy.

Limitations: The systems are costly since installation of complicated converters and DC switchgear is expensive. The converters require considerable reactive power. Lack of HVDC circuit breakers hampers network operation. Moreover there is nothing like DC transformer; voltage transformation has to be provided on the a.c. sides of the system.

Q.13 Discuss the working principle of a direction over-current relay. (7)

Ans : Overcurrent relay: This relay operates when current through it (sample of power system current) satisfies the condition

where is said to be the pickup value of the relay. Such a relay is an over current relay and would operate in the shortest possible time (depending upon the type of hardware employed), and is called instantaneous overcurrent relay.

Thus if:

the relay trips the circuit broker

and

if the relay blocks or does not trip the circuit breaker.

The directional feature can be incorporated for the above o/c relay by means of an induction cup type of structure (Fig L1) one of the coils being current excited and the other, voltage excited.

The phasor diagram of a directional relay is shown in Fig. L2. , the voltage coil current lags the voltage applied by an angle depending upon the coil impedance. Since the fluxes are proportional to the coil currents, and have the same angle separation as coil currents, the relay torque can be expressed as

(A)

where = phase angle of voltage coil current

and = angle by which the current applied to the relay leads its voltage

The proportionality

(B)

where and can be treated as fluxes from the current and voltage coils and F_(AV),net (average) force or an induction type of disc.

Taking the leading angle as positive and defining (a relay parameter) as the value of, when the relay develops maximum torque, then

Substitution of this relation in Equn (A) gives

which indeed is the directional characteristic

Q.14 Explain the working of nickel cadmium cells with its merits and demerits over lead acid cell. (7)

Ans: Nickel- Cadmium cells: In a Nickel-Cadmium cell the positive plates are made of nickel hydroxide enclosed in finely perforated steel tubes or pockets, the electrical resistance being reduced by the addition of flakes of pure nickel or graphite. These tubes or pockets are assembled in nickelled-steel plates. The active material in this cell is Cadmium mixed with a little iron, the latter being used to prevent caking of the active material and hence an impairment to porosity.

The electrolyte is a solution of potassium hydroxide (KOH) having a relative density of about 1.15 to 1.20 depending upon the type of cell and the condition of service. The electrolyte does not undergo any chemical change; hence the quantity of electrolyte can be reduced to the minimum value necessitated by adequate clearance between the plates. The plates are separated by insulating rods and assembled in sheet-steel containers, the latter being mounted in non-metallic crates so as to insulate the cells from one another.

Chemical reaction

The curve A in Fig M1 represents the terminal voltage of a nickel-cadmium cell during discharge at the 10-hour rate, while curve B shows the variation for a 3-hour rate.

The curve C represents the variation of the terminal voltage when the cell is charged at a rate 1.5 times the 10-hour discharge rate.

The advantages of the nickel-cadmium cell are:

a) Its mechanical construction enables it to withstand considerable vibration.

b) It is free from ‘sulphating’ or any similar trouble and can therefore be left in any state of charge without damage.

The disadvantages are:

a) Its cost is higher than that of the led-acid cell

b) Its average discharge p.d. is about 1.2V as compared to 2V for the led-acid cell, so that for a given voltage, the number of alkaline cells is about 67 percent larger than that of lead-acid cells.

Q.15 Discuss methods of laying underground cables. (7)

Ans: Methods of laying underground cables

Underground cables are generally required in urban areas where there is a high density of population. The highest load density and the most restrictive limitations on feeder layout occur simultaneously in the central cores of major cities. Load densities are usually in the range of 60 to 100 MW per square mile.

Some problems associated with underground cable systems are as follows:

urban area are densely populated with/pedestrian walk ways, as well as water, sewer, storm drain, phone and other utility systems in addition to electrical. So there is limited room. The electrical utility must stake its claim to the routes and space allocated to it with its own duct banks.

Secondly, duct banks are needed to protect underground cable from the constant dig-ins of other utilities, stress form settling, and heat and moisture in this environment. Thirdly, digging into the street in an urban area for new additions or repairs is very expensive. Traffic control and other requirements add to cost and also digging around all the other utilities. Thus electric utilities have no choice but to use duct banks and cable vaults to create their own “cable tunnel system“ under the streets in dense urban areas.

Finally branching of underground cable providing for “T” or “X” connections of paths, while possible, is not as simple or as inexpensive as in overhead lines.

Summarizing, the underground urban area systems have the following characteristics

· Layout is restricted to street grid

· Loads are large and invariably three phase

· Fixed cost is very high

· The cost of capacity shortfall is extremely high

Q.16 Write short notes on any TWO of the following :-

(i) Switched reluctance motors.

(ii) Parallel operation of transformers.

(iii) Selsyns. (7 x 2 = 14)

Ans:

(i) Switched reluctance motors: A synchronous motor with salient poles but no field winding is known as the reluctance motor. It is used for low power, constant speed applications where special arrangements for d.c. excitation would be cumbersome.

The principle of this motor is that the stator produces a rotating field in space and the rotor is noncylindrical such that the reluctance of the magnetic path offered by the rotor to the rotating field is a function of the space angle. Here the rotor has the tendency to align itself in the minimum reluctance position with respect to the synchronously rotating flux of the forward field. The motor is made self starting by the induction principle by providing short-circuited copper bars in the projecting parts of the rotor.

In the single phase reluctance motor the rotating field can be produced by one of the phase-splitting methods. The salient pole structure is given to the rotor by removing some of the teeth of an induction motor rotor as shown in Fig M2. The remaining teeth carry short-circuited copper bars to provide the starting induction torque. After starting, the rotor reaches near synchronous speed by induction action and is pulled into synchronism during the positive half-cycle of the sinusoidally varying synchronous torque.

This would only be possible if the rotor has low inertia and the load conditions are light. The torque speed characteristic of a typical reluctance motor with induction start is given in Fig M3. Here the starting torque is highly dependent upon the rotor position because of the projecting nature of the rotor. This phenomenon is known as “cogging”. For satisfactory synchronous motor performance the frame size to be used must be much larger than that for normal single- phase induction motor. This accounts for the high value of starting torque shown in Fig M3.

(ii) Parallel operation of transformers: when the load exceeds the capacity of an existing transformer, it may be economical to install another one in parallel rather than replacing it with a single larger unit. Also for reliability two smaller units in parallel are preferred. The cost of maintaining a spare is also less with two units in parallel.

For satisfactory parallel operation of transformers the following conditions have to be fulfilled:

(i) They must be connected with proper polarities; this is to ensure that the net voltage around the local loop is zero. A wrong polarity connection results in a dead short circuit.

(ii) 3 phase transformers must have zero relative phase displacement on the secondary sides and must be connected with proper phase sequence. Only the transformers of the same group can be paralleled. For example Y/Y and Y/ transformers cannot be paralleled as their secondary voltages will have a phase difference of 30⁰

(iii) The transformers must have the same voltage ratio to avoid no-load circulating current, when the transformers are in parallel on both primary and secondary sides.

(iv) There should exist only a limited disparity in the per unit impedances (on their own bases) of the transformers. The currents carried by the two transformers are proportional to their ratings if their p.u. impedances on their own ratings are equal.

Fig N1 shows two transformers paralleled on both sides with proper polarities but on no load. The primary voltages V₁ and V₂ are equal. If the voltage ratios of the two transformers are not identical, the secondary voltages E₁ and E₂ though in phase will not be equal in magnitude. The difference (E₁-E₂) will appear across the switch S. When the secondaries are paralleled by closing the switch, a circulating current appears even though the secondries are not supplying any load. The circulating current will depend upon the total leakage impedance of the two transformers and also the difference in their voltage ratios. Only a small difference in the voltage ratios will be tolerated.

Division of load between transformers in parallel. Equal voltage ratios When the transformers have equal ratios E₁=E₂ in Fig N1, the equivalent current of the two transformers would then be as shown in Fig N2 on the assumption that the exciting current can be neglected in comparison to the load current.

It follows from the sinusoidal steady-state circuit analysis that

and

also

Taking as the reference phasor and defining complex power as , the multiplication of on both sides of equations (1) and (2) give

where

The above equations , and are phasor relationships giving loadings in the magnitude and phase angle.

(iii) Selsyns: Selsyns or synchros are control system components which are used for transmission of small torques or motions electrically. They can be categorize into three kinds:

(a) for transmission of small torques electrically, as sinchro-transmitter-receiver pair (Fig P2)

(b) for indicating the difference in positions, as generator – transformer combination (Fig P3)

In the synchro-transmitter-receiver pair (Fig P2) which is a single phase selsyn, the stator has three windings like the polyphase induction motor. The rotor is similar to the rotor of a small alternator and of one winding. Fig P1 shows the cross –section of a single phase selsyn.

Although it shows three stator windings, it is still a single –phase device. Any a.c. current in the rotor will produce at “stand–still” three stator voltages which are in time phase. Two of these devices in the circuit of Fig P2 provides a system for transmission of motion.

It is assumed that the generator and motor are similar units and that for the initial conditions the voltages produced on the stator of the generator by the generator rotor are equal in magnitude and 180⁰ out of phase with those produced by the motor rotor in the motor stator. Under these conditions the stator currents will be zero and no torque will be present in either machine.

If the rotor of the generator is turned through an angle while the position of the rotor of the motor is left unchanged, a circulating current I_a will result in the stators. The current acting on the air gap flux will tend to restore the generator to its original position. The current will produce a torque in the motor which wil tend to cause the rotor of the motor to assume an angle corresponding to that occupied by the rotor of the generator. If the rotor of the motor is free to turn, it will follow the angular position of the generator rotor. The rotor of the motor therefore is an indicator of the position of the generator rotor or any device connected mechanically to it.

Selsyns as position indicators: If the circuit to the rotor of the motor is opened (Fig P3) a voltage will be produced in the rotor winding, the magnitude of which is a function of the angular position of the rotor of the motor w.r.t. that of the generator rotor. When the motor rotor is at a position of 90 electrical degrees from the position occupied and the rotor is electrically connected and free, the magnitude of the rotor voltage will be zero. Then any movement of the generator rotor will produce a voltage in the motor rotor which is a function of the angular position of the generator rotor (or other equipment coupled to it). This circuit has found use in many servomechanism systems. For this application the motor unit serves just as a control transformer.

A third useful application of the selsyn system is that of the differential selsyn shown in Fig P4; this is constructed like a wound rotor induction motor having a 1:1 ratio of turns. When the rotors of the motor and generator of Fig P3 are in a given position, the differential selsyn will adjust the position of the rotor to its stator such that minimum current flows in its windings. With the rotor of the motor locked in a particular position, a change in the position of the generator rotor will cause a corresponding change in the position of the rotor of the differential selsyn. In this way the circuit of Fig P4 performs the same function as that shown in Fig P3.