Electric Drives MCQs – Speed Control of Direct Current & Induction Motors MCQs ( Electric Drives ) MCQs
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Speed Control of Direct Current & Induction Motors MCQs ( Electric Drives ) MCQs – Electric Drives MCQs
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Speed Control of Shunt or Separately Excited Motors
1. The advantage of the double squirrel cage induction motor over a single cage rotor is that its _______
a) Efficiency is higher
b) Power factor is higher
c) Slip is larger
d) Starting current is lower
Answer: d
Explanation: The starting current flows through the outer cage which has higher resistance and hence starting current is lower. This is one the important advantage of the double squirrel cage induction motor over a single cage rotor.
2. A 16-pole, 3-phase, 60 Hz induction motor is operating at a speed of 150 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 20
b) 40
c) 30
d) 10
Answer: b
Explanation: Given a number of poles = 16. Supply frequency is 60 Hz. Rotor speed is 150 rpm. Ns=120×f÷P=120×60÷16 = 450 rpm. S=Ns-Nr÷Ns=450-150÷450=.666. F2=sf=.666×60=40 Hz.
3. Calculate the amplitude of the sinusoidal waveform z(t)=715sin(165πt+2π÷468).
a) 710
b) 715
c) 716
d) 718
Answer: b
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(Ωt+α) where Vm represents peak value, Ω represents angular frequency, α represents a phase difference.
4. R.M.S value of the sinusoidal waveform V=48sin(6.5πt+89π÷46.8).
a) 33.94 V
b) 33.56 V
c) 33.12 V
d) 33.78 V
Answer: a
Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ = 48÷2½ = 33.94 V and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm.
5. The short circuit test on a 3-φ induction motor is conducted at a rotor speed of _______
a) Zero
b) < Ns
c) > Ns
d) Ns
Answer: a
Explanation: Short-circuit test in an induction motor is also called a Blocked rotor test so it is conducted at zero speed. Net input power taken is equal to the variable losses.
6. If induction motor air gap power is 10 KW and mechanically developed power is 8 KW, then rotor ohmic loss will be _________ KW.
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=10-8=2 KW.
7. A 3-phase induction machine draws active power P and reactive power Q from the grid. If it is operated as a generator, then P and Q will be _________
a) Positive and Negative
b) Negative and Positive
c) Positive and Positive
d) Negative and Negative
Answer: b
Explanation: Induction generator will be delivering power to the bus to generate flux it will consume reactive power from the bus. Since active power is delivered the active power drawn will be negative but reactive power is absorbed and hence active power absorbed is positive.
8. The slope of the V-I curve is 39.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .81 Ω
b) .36 Ω
c) .75 Ω
d) .84 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 39.1° so R=tan(39.1°)=.81 Ω. The slope of the I-V curve is reciprocal of resistance.
9. If induction motor air gap power is 48 KW and gross developed power is 28 KW, then rotor ohmic loss will be _________ KW.
a) 10
b) 20
c) 30
d) 40
Answer: b
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=48-28=20 KW.
10. The power factor of a squirrel cage induction motor is ___________
a) Low at light load only
b) Low at heavy loads only
c) Low at the light and heavy loads both
d) Low at rate load only
Answer: a
Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low.
11. Calculate the active power in a .154 H inductor.
a) 22 W
b) 14 W
c) 45 W
d) 0 W
Answer: d
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90° = 0 W.
12. Calculate the time period of the waveform z(t)=24sin(πt+π÷4)+4sin(2πt+π÷6).
a) 2 sec
b) 3 sec
c) 4 sec
d) 1 sec
Answer: a
Explanation: The fundamental time period of the sine wave is 2π. The time period of z(t) is L.C.M {2,1}=2 sec. The time period is independent of phase shifting and time shifting.
13. Calculate the total heat dissipated in a resistor of 50 Ω when 1.4 A current flows through it.
a) 98 W
b) 92 W
c) 91 W
d) 93 W
Answer: a
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. P=I2R=1.4×1.4×50=98 W.
14. Calculate mark to space ratio if the system is on for 26.3 sec and off for 24.2 sec.
a) 1.086
b) 1.042
c) 1.214
d) 1.876
Answer: a
Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=26.3÷24.2=1.086.
Speed Control of Series Motor
1. A particular current is made up of two components: a 10 A and a sine wave of peak value 14.14 A. The average value of current is _________
a) Zero
b) 24.14 A
c) 10 A
d) 14.14 A
Answer: c
Explanation: Average value of DC current is 10 A. Average value of AC current is 0 A as it is alternating in nature. The average value of current is 10+0 = 10 A.
2. A 38-pole, 3-phase, 80 Hz induction motor is operating at a speed of 12 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 75.2
b) 76.1
c) 79.2
d) 79.6
Answer: b
Explanation: Given a number of poles = 8. Supply frequency is 50 Hz. Rotor speed is 720 rpm. Ns=120×f÷P=120×80÷38 = 252.63 rpm. S=Ns-Nr÷Ns = 252.63-12÷252.63 = .952. F2 = sf = .952×80 = 76.1 Hz.
3. Calculate the phase angle of the sinusoidal waveform i(t)=sin(.6πt+π÷.88).
a) 100π÷88
b) 100π÷8
c) 100π÷8
d) π÷88
Answer: a
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.
4. Calculate the moment of inertia of the satellite having a mass of 79 kg and diameter of 83 cm.
a) 13.65 kgm2
b) 13.60 kgm2
c) 12.67 kgm2
d) 13.82 kgm2
Answer: b
Explanation: The moment of inertia of the satellite can be calculated using the formula I=mr2. The mass of the satellite and diameter is given. I=(79)×(.415)2=13.60 kgm2. It depends upon the orientation of the rotational axis.
5. A particular voltage is made up of two components: a 5 A and a cosine wave of peak value 7.8 A. The average value of voltage is _________
a) Zero
b) 35.14 A
c) 78 A
d) 5 A
Answer: d
Explanation: Average value of DC voltage is 5 A. Average value of AC current is 0 A as it is alternating in nature. The average value of current is 5+0 = 5 A.
6. Armature reaction is demagnetizing in nature due to a purely capacitive load in the synchronous generator.
a) True
b) False
Answer: b
Explanation: Due to a purely capacitive load, armature current is in phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in phase with the field flux. It will try to increase the net magnetic field.
7. Armature reaction is magnetizing in nature due to a purely resistive load in the synchronous generator.
a) True
b) False
Answer: b
Explanation: Due to a purely resistive load, armature current is in quadrature with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in quadrature with the field flux. It will try to increase the net magnetic field.
8. The slope of the V-I curve is 4.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .03 Ω
b) .08 Ω
c) .04 Ω
d) .07 Ω
Answer: b
Explanation: The slope of the V-I curve is resistance. The slope given is 4.9° so R=tan(4.9°)=.08 Ω. The slope of the I-V curve is reciprocal of resistance.
9. Calculate the velocity of the satellite if the angular speed is 87 rad/s and radius is 7.4 m.
a) 643.8 m/s
b) 642.4 m/s
c) 641.9 m/s
d) 643.2 m/s
Answer: a
Explanation: The velocity of the satellite can be calculated using the relation V=Ω×r. The velocity is the vector product of angular speed and radius. V=Ω×r = 87×7.4 = 643.8 m/s.
10. A 3-phase induction motor runs at almost 70 rpm at no load and 50 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?
a) 20 revolution per minute
b) 30 revolution per minute
c) 40 revolution per minute
d) 50 revolution per minute
Answer: a
Explanation: Supply frequency=50 Hz. No-load speed of motor = 70 rpm. The full load speed of motor=50 rpm. Since the no-load speed of the motor is almost 70 rpm, hence synchronous speed near to 70 rpm. Speed of rotor field=70 rpm. Speed of rotor field with respect to rotor=70-50= 20 rpm.
11. Calculate the active power in a 9.854 H inductor.
a) 4.98 W
b) 0 W
c) 8.59 W
d) 1 W
Answer: b
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P=VIcos90° = 0 W.
12. Calculate the reactive power in a 45 Ω resistor with 1.78 A current flowing through it.
a) 28.8 VAR
b) 23.4 VAR
c) 25.82 VAR
d) 0 VAR
Answer: d
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. Q=VIsin(0°)=0 VAR.
13. Calculate the value of the frequency if the inductive reactance is 72 Ω and the value of the inductor is 7 H.
a) 1.63 Hz
b) 1.54 Hz
c) 1.78 Hz
d) 1.32 Hz
Answer: a
Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation XL = 2×3.14×f×L. F = XL÷2×3.14×L = 72÷2×3.14×7 = 1.63 Hz.
14. Calculate the active power in a 2 Ω resistor with 8 A current flowing through it.
a) 125 W
b) 128 W
c) 123 W
d) 126 W
Answer: b
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. P=I2R=8×8×2=128 W.
Induction Motors – Basics Principles of Speed Control
1. In the chopper circuit, commutation times are _______
a) Current commutation is equal to voltage commutation
b) Current commutation is less as compared to that of voltage commutation
c) Current commutation is more as compared to that of voltage commutation
d) Both commutation techniques are not comparable
Answer: b
Explanation: In current commutation, commutation time=CVr÷Io. In voltage commutation, commutation time= CVs÷Io. Hence, the commutation time of the current commutation is less as compared to voltage commutation.
2. The generated e.m.f from 4-pole armature having 1 conductors driven at 1 rev/sec having flux per pole as 10 Wb, with wave winding is ___________
a) 30 V
b) 40 V
c) 70 V
d) 20 V
Answer: d
Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. In wave winding number of parallel paths are 2. Eb = 10×4×1×60÷60×2 = 20 V.
3. The unit of voltage is Pascal.
a) True
b) False
Answer: b
Explanation: The voltage is equal to one volt when 1 A of current flows through 1 Ω resistor. It is mathematically represented as I×R. It is expressed in terms of a volt(V).
4. Calculate the moment of inertia of the sphere having a mass of 8.4 kg and radius of 61 cm.
a) 3.124 kgm2
b) 3.125 kgm2
c) 4.545 kgm2
d) 5.552 kgm2″
Answer: b
Explanation: The moment of inertia of the egg can be calculated using the formula I=Σmiri2. The mass of egg and radius is given. I=(8.4)×(.61)2=3.125 kgm2. It depends upon the orientation of the rotational axis.
5. The most suitable control-motor application is __________
a) AC shunt motor
b) DC separately motor
c) AC one-phase induction motor
d) DC shunt motor
Answer: b
Explanation: DC separately motor has definite full-load speed, so they don’t ‘run away’ when the load is suddenly thrown off provided the field circuit remains closed. The speed for any load within the operating range of the motor can be readily obtained.
6. In a DC series motor, the e.m.f developed is proportional to _______
a) N×Ia
b) N×Ia2
c) N×Ia3
d) N×Ia.5
Answer: a
Explanation: In a DC series motor, the e.m.f developed is equal to KmΦN. In a DC series, the motor field winding is connected in series with the armature so the flux in the field winding is proportional to current. Eb = KmΦN α Ia×N.
7. Calculate the value of the time period if the frequency of the signal is .07 sec.
a) 14.28 sec
b) 14.31 sec
c) 14.23 sec
d) 14.78 sec
Answer: a
Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷.07=14.28 sec.
8. The slope of the V-I curve is 13.89°. Calculate the value of resistance.
a) .247 Ω
b) .345 Ω
c) .231 Ω
d) .222 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 13.89° so R=tan(13.89°)=.247 Ω. It behaves like a normal resistor.
9. In a DC shunt motor, the e.m.f developed is proportional to ___________
a) Ia
b) Ia2
c) Ia3
d) Iao
Answer: d
Explanation: In a DC shunt motor, the e.m.f developed is equal to KmΦN. In a DC shunt, the motor field windings are connected separately and excited by a constant DC voltage. E = KmΦN α Ia°.
10. Calculate the power factor angle during the resonance condition.
a) 0°
b) 10°
c) 80°
d) 90°
Answer: d
Explanation: During the resonance condition, the reactive power generated by the capacitor is completely absorbed by the inductor. Only active power flows in the circuit. Net reactive power is equal to zero and Φ=0°.
11. Calculate the value of the duty cycle if the system is on for 5 sec and off for inf sec.
a) 0
b) .4
c) .2
d) .1
Answer: a
Explanation: Duty cycle is Ton÷Ttotal. It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷Ttotal=5÷inf=0.
12. Calculate the value of the frequency of the AC supply in India.
a) 0 Hz
b) 50 Hz
c) 49 Hz
d) 60 Hz
Answer: b
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. AC supply magnitude is variable. It changes with time so the frequency of AC supply is 50 Hz.
13. DC series motor cannot run under no load.
a) True
b) False
Answer: a
Explanation: DC series motor cannot be run under no load condition because at no-load speed of the motor is very which can damage the shaft of the motor. There should be some load that should be connected to it.
14. Calculate the value of the frequency if the time period of the signal is .2 sec.
a) 5 Hz
b) 4 Hz
c) 2 Hz
d) 3 Hz
Answer: a
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷T = 1÷.2 = 5 Hz.
Induction Motors – Controlling Speed Using Rotor Resistance
1. What is the formula for the maximum active power in the cylindrical rotor synchronous machine? (Eb represents armature emf, Vt represents terminal voltage, δ represents rotor angle, X represents reactance)
a) Eb×Vt×sinδ÷X
b) Eb×Vt2÷X
c) Eb2×Vt×sinδ÷X
d) Eb×Vt÷X
Answer: d
Explanation: The real power in the cylindrical rotor machine is Eb×Vt×sinδ÷X. It is inversely proportional to the reactance. The stability of the machine is decided by the maximum power transfer capability. It is maximum for delta angle=90°.
2. Cylindrical pole machines are more stable than salient rotor machines.
a) True
b) False
Answer: b
Explanation: Cylindrical pole machines are less stable than salient rotor machines because of the less short circuit ratio and less real power transfer capability. The air gap length in salient pole machines is more as compare to cylindrical rotor machines.
Braking Of Electric Motors MCQs
3. The unit of oscillating power is VAR.
a) True
b) False
Answer: a
Explanation: The oscillating power is the waste power in case of electric circuits. It is the energy trapped that keeps on oscillating. It is expressed in Volt Ampere reactive.
4. Calculate the power factor if the power angle is 36.86°.
a) .8
b) .4
c) .6
d) 1
Answer: a
Explanation: Power factor is the ratio of the real power to the apparent power. It measures the useful power contained in the total power.Φ=cos-1(36.86)=.8.
5. Calculate the reactive power in a 2.2 Ω resistor.
a) 1.2 VAR
b) 2.6 VAR
c) 0 VAR
d) 1.9 VAR
Answer: c
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. Q=VIsin0° = 0 VAR.
6. What is the unit of the distance?
a) m
b) m/s
c) atm/m
d) Volt
Answer: a
Explanation: Distance is the total length traced by the body. It is a scalar quantity. It is expressed in the meter. It is not a tensor quantity. It has no direction.
7. Calculate the reactive power in a 7.2545 Ω resistor.
a) 1.221 VAR
b) 2.611 VAR
c) 0 VAR
d) 1.997 VAR
Answer: c
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. Q=VIsin0° = 0 VAR.
8. The slope of the V-I curve is 180o. Calculate the value of resistance. The graph is parallel to the x-axis.
a) 1.47 Ω
b) 1.81 Ω
c) 0 Ω
d) 4.12 Ω
Answer: c
Explanation: The slope of the V-I curve is resistance. The slope given is 180° so R=tan(180°)=0 Ω. The slope of the V-I curve is resistance. It behaves as a short circuit.
9. Calculate the value of the duty cycle if the system is on for 1 sec and off for 0 sec.
a) 1
b) .4
c) .1
d) .6
Answer: a
Explanation: Duty cycle is Ton÷Ttotal. It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷Ttotal=5÷5=1.
10. The phase difference between voltage and current in the choke coil.
a) 90°
b) 40°
c) 50°
d) 68°
Answer: a
Explanation: In the case of a choke coil, the voltage leads the current by 90° or the current lags the voltage by 90°. The phase difference between voltage and current is 90°.
11. Armature reaction is demagnetizing in nature due to a purely capacitive load in the synchronous generator.
a) True
b) False
Answer: a
Explanation: Due to a purely capacitive load, armature current is in opposite phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in opposite phase with the field flux. It will try to reduce the net magnetic field.
12. Armature reaction is magnetizing in nature due to a purely resistive load in the synchronous motor.
a) True
b) False
Answer: b
Explanation: Due to a purely resistive load, armature current is in quadrature with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in quadrature with the field flux. It will try to increase the net magnetic field.
Induction Motors Speed Control – Rotor Voltage Injection
1. A 32-pole, 3-phase, 70 Hz induction motor is operating at a speed of 112 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 40.2
b) 46.1
c) 40.1
d) 40.6
Answer: c
Explanation: Given a number of poles = 32. Supply frequency is 70 Hz. Rotor speed is 112 rpm. Ns = 120×f÷P=120×70÷32 = 262.5 rpm. S=Ns-Nr÷Ns = 262.5-112÷262.5=.573. F2=sf=.573×70=40.1 Hz.
2. A 20-pole, 3-phase, 90 Hz induction motor is operating at a speed of _______ rpm. The frequency of the rotor current of the motor in Hz is 20.
a) 418.56
b) 420.12
c) 421.23
d) 422.45
Answer: b
Explanation: Given a number of poles = 20. Supply frequency is 90 Hz. Ns=120×f÷P=120×90÷20 = 540 rpm. S=20÷90 = .222 Hz. Nr=(1-s)Ns = 420.12 rpm. Rotor speed is 420.12 rpm.
3. Calculate the amplitude of the sinusoidal waveform z(t)=.27sin(.369πt+.142π÷4.25).
a) .287
b) .270
c) .216
d) .287
Answer: b
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference. By comparing the waveform z(t) with the generalized sinusoidal expression we can see Vm=.27 and ω=.369π rad/s.
4. R.M.S value of the sinusoidal waveform q(t)=2.11cos(87.25t+78π÷478.23).
a) 1.49 V
b) 1.56 V
c) 1.12 V
d) 1.78 V
Answer: a
Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm. The r.m.s value is Vrms = 2.11÷2½ = 1.49 V.
5. The no-load circuit test on a 3-Φ induction motor is conducted at a rotor speed of _______
a) Zero
b) < Ns
c) > Ns
d) Ns
Answer: d
Explanation: No-load circuit test in an induction motor is also called an open circuit test so it is conducted at synchronous speed. Net input power taken is equal to the no-load rotational losses.
6. If induction motor air gap power is 67 KW and mechanically developed power is 26 KW, then rotor ohmic loss will be _________ KW.
a) 41
b) 42
c) 43
d) 44
Answer: a
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=67-26=41 KW.
7. Calculate the line voltage in star connection when phase voltage=45 V.
a) 77.9 V
b) 77.6 V
c) 77.2 V
d) 77.8 V
Answer: a
Explanation: The line voltage in case of star connection is 1.73 times of phase voltage. It leads the phase voltage by an angle of 30°. VL-L=1.73×45=77.9 V.
8. The slope of the V-I curve is 1.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .019 Ω
b) .036 Ω
c) .075 Ω
d) .084 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 1.° so R=tan(1.1°)=.019 Ω. The slope of the I-V curve is reciprocal of resistance.
9. If induction motor rotor power is 157.5 KW and gross developed power is 79.9 KW, then rotor ohmic loss will be _________ KW.
a) 77.5
b) 77.6
c) 76.9
d) 77.1
Answer: b
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=157.5-79.9=77.6 KW.
10. The power factor of a squirrel cage induction motor generally is ___________
a) .6-.8
b) .1-.2
c) .2-.4
d) .5-.7
Answer: a
Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low. The power factor of a squirrel cage induction motor generally is .6-.8.
11. Calculate the active power in a .89 H inductor.
a) 1.535 W
b) 0 W
c) 2.484 W
d) 1.598 W
Answer: b
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P=VIcos90° = 0 W.
12. Calculate the time period of the waveform i(t)=sin(πt+6π÷4)+sin(πt+5π÷6).
a) 2 sec
b) 4 sec
c) 5 sec
d) 3 sec
Answer: a
Explanation: The fundamental time period of the sine wave is 2π. The time period of i(t) is L.C.M {2,2}=2 sec. The time period is independent of phase shifting and time shifting.
13. Calculate the total heat dissipated in a rotor resistor of 14.23 Ω when .65 A current flows through it.
a) 6.45 W
b) 6.01 W
c) 6.78 W
d) 6.98 W
Answer: b
Explanation: The rotor resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. P=I2R=.65×.65×14.23=6.01 W.
14. Calculate mark to space ratio if the system is on for 4.3 sec and off for 78.2 sec.
a) .054
b) .047
c) .039
d) .018
Answer: a
Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff = 4.3÷78.2 = .054.
Induction Motors Speed Control – Slip Energy Recovery
1. In the rotor voltage injection method, when an external voltage source is in phase with the main voltage then speed will ___________
a) Increase
b) Decrease
c) Remain unchanged
d) First increases then decrease
Answer: a
Explanation: In the rotor injection method, when an external voltage is in phase with the main voltage net voltage increases and the value of slip decreases and the value of rotor speed increases.
2. A 2-pole, 3-phase, ______ Hz induction motor is operating at a speed of 550 rpm. The frequency of the rotor current of the motor in Hz is 2.
a) 9.98
b) 9.71
c) 9.12
d) 9.37
Answer: d
Explanation: Given a number of poles = 2. Rotor speed is 550 rpm. Ns=120×f÷P=120×f÷2 = 60f rpm. S=Ns-Nr÷Ns. F2=sf. S=F2÷f. Supply frequency is 9.37 Hz.
3. Calculate the average value of the sinusoidal waveform x(t)=848sin(1.65πt+2π÷0.68).
a) 0
b) 78 V
c) 15 V
d) 85 V
Answer: a
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(Ωt+α) where Vm represents peak value, Ω represents angular frequency, α represents a phase difference. The average value of a sine wave is zero because of equal and opposite lobes areas.
4. R.M.S value of the periodic square waveform of amplitude 72 V.
a) 72 V
b) 56 V
c) 12 V
d) 33 V
Answer: a
Explanation: R.M.S value of the periodic square waveform is Vm and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the periodic square waveform is Vm.
5. In the rotor voltage injection method, when an external voltage source is in opposite phase with the main voltage then speed will ___________
a) Increase
b) Decrease
c) Remain unchanged
d) First increases then decrease
Answer: b
Explanation: In the rotor injection method, when an external voltage is in opposite phase with the main voltage net voltage decreases and the value of slip increases and the value of rotor speed decreases.
6. The rotor injection method is a part of the slip changing technique.
a) True
b) False
Answer: a
Explanation: Rotor injection method comes under slip changing technique. It uses an external voltage source to change the slip value. The load torque remains constant here.
7. The slip recovery scheme is a part of the synchronous speed changing technique.
a) True
b) False
Answer: a
Explanation: Slip recovery scheme comes under the synchronous speed changing technique. It uses an external induction machine to change the frequency value. The load torque remains constant here.
8. The slope of the V-I curve is 9.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .16 Ω
b) .26 Ω
c) .25 Ω
d) .44 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 9.1° so R=tan(9.1°)=.16 Ω. The slope of the I-V curve is reciprocal of resistance.
9. If induction motor air gap power is 1.8 KW and gross developed power is .1 KW, then rotor ohmic loss will be _________ KW.
a) 1.7
b) 2.7
c) 3.7
d) 4.7
Answer: a
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power = 1.8-.1=1.7 KW.
10. The power factor of a squirrel cage induction motor is ___________
a) High at light load only
b) High at heavy loads only
c) Low at the light and heavy loads both
d) Low at rate load only
Answer: b
Explanation: At heavy loads, the current drawn is high due to which active power component increases. Increase in active power component increases the power factor of the machine.
11. At low values of slip, the electromagnetic torque is directly proportional to ___________
a) s
b) s2
c) s3
d) s4
Answer: a
Explanation: At low values of slip, the electromagnetic torque is directly proportional to slip value. Due to heavy loading slip value decreases which increases the ratio of R2÷s.
12. Calculate the time period of the waveform v(t)=12sin(8πt+8π÷15)+144sin(2πt+π÷6)+ 445sin(πt+7π÷6).
a) 8 sec
b) 4 sec
c) 7 sec
d) 3 sec
Answer: b
Explanation: The fundamental time period of the sine wave is 2π. The time period of z(t) is L.C.M {4,1,2}=4 sec. The time period is independent of phase shifting and time shifting.
13. Calculate the total heat dissipated in a resistor of 44 Ω when 0 A current flows through it.
a) 0 W
b) 2 W
c) 1.5 W
d) .3 W
Answer: a
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. P=I2R=0×0×44=0 W.
14. The value of slip at which maximum torque occurs ________
a) R2÷X2
b) 4R2÷X2
c) 2R2÷X2
d) R2÷3X2
Answer: a
Explanation: The maximum torque occurs when the slip value is equal to R2÷X2. Maximum torque is also known as breakdown torque, stalling torque and pull-out torque.
Induction Motors – Controlling Speed Using Inductance
1. Calculate the voltage regulation in the synchronous machine if the no-load voltage is 12 V and the full load voltage is 15V.
a) -20%
b) -40%
c) -60%
d) -80%
Answer: a
Explanation: Voltage regulation is defined as the fluctuation in the load voltage when the load is varied from no-load to full load. V.R(%) = (No-load voltage-Full load voltage) ÷ Full load voltage=12-15÷15 = -20%.
2. Calculate the condition for maximum voltage regulation in the synchronous machine.
a) Φ=ϴs
b) Φ=2ϴs
c) Φ=4ϴs
d) Φ=8ϴs
Answer: a
Explanation: Voltage regulation is maximum in case of inductive load. The condition for maximum voltage regulation is when the power factor angle of the load becomes equal to the impedance angle. V.R=Rp.ucos(Φ)+Xp.usin(Φ). Differentiate V.R with respect to Φ and put it equal to zero. We will get tan(Φ) = Xp.u÷Rp.u=tan(ϴs) then Φ=ϴs.
3. Zero voltage regulation can be only achieved in leading power factor load.
a) True
b) False
Answer: a
Explanation: Zero voltage regulation only occurs during a leading power factor load. Condition for zero voltage regulation occurs when Φ+ϴs > 90o. For example – Capacitive load.
4. R.M.S value of the sinusoidal waveform v(t)=211sin(7.25t+7π÷78.3).
a) 149.19 V
b) 156.23 V
c) 116.57 V
d) 178.64 V
Answer: a
Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ = 211÷2½ = 149.19 V and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm.
5. Calculate the peak value of sinusoidal waveform if the r.m.s value is 21 V.
a) 29.69 V
b) 48.74 V
c) 69.23 V
d) 25.74 V
Answer: a
Explanation: R.M.S value of the sinusoidal waveform is Vpeak÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vr.m.s×2½. Vpeak = Vr.m.s×2½=21×1.414=29.69 V.
6. If induction motor air gap power is 2 KW and mechanically developed power is 1 KW, then rotor ohmic loss will be _________ KW.
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=2-1=1 KW.
7. Calculate the line voltage in the delta connection when phase voltage=45 V.
a) 46 V
b) 47 V
c) 45 V
d) 78 V
Answer: c
Explanation: The line voltage in case of delta connection is phase voltage. Line current leads the phase current by an angle of 30°. VL-L=Vph = 45 V.
8. The slope of the V-I curve is 35.48°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) 0.452 Ω
b) 0.462 Ω
c) 0.752 Ω
d) 0.712 Ω
Answer: d
Explanation: The slope of the V-I curve is resistance. The slope given is 35.48° so R=tan(35.48°)=.712 Ω. The slope of the I-V curve is reciprocal of resistance.
9. If induction motor rotor power is 157.5 KW and gross developed power is 79.9 KW, then rotor ohmic loss will be _________ KW.
a) 77.5
b) 77.6
c) 76.9
d) 77.1
Answer: b
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=157.5-79.9=77.6 KW.
10. Calculate the line current in the delta connection when phase current=17 A.
a) 29.44 A
b) 24.64 A
c) 23.48 A
d) 26.56 A
Answer: a
Explanation: The line voltage in case of delta connection is phase voltage. Line current leads the phase current by an angle of 30°. IL-L = 1.73×Iph = 29.44 A.
11. Calculate the active power in a 168.12 H inductor.
a) 65 W
b) 0 W
c) 68 W
d) 64 W
Answer: b
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90° = 0 W.
12. Calculate the maximum value of slip when rotor resistance is 2 Ω and rotor reactance is 3 Ω.
a) 0.66
b) 0.33
c) 0.44
d) 0.99
Answer: a
Explanation: The maximum torque occurs when the slip value is equal to R2÷X2. Maximum torque is also known as breakdown torque, stalling torque and pull-out torque. The maximum value of slip is R2÷X2=2/3=.66.
13. Calculate the total heat dissipated in a rotor resistor of 21 Ω when .81 A current flows through it.
a) 13.77 W
b) 12.56
c) 16.78 W
d) 13.98 W
Answer: a
Explanation: The rotor resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. P=I2R=.81×.81×21=13.77 W.
14. Calculate the active power in an 8.12 F capacitor.
a) 89 W
b) 41 W
c) 0 W
d) 48 W
Answer: c
Explanation: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P=VIcos90° = 0 W.
Induction Motors – Controlling Speed by Adjusting the Stator Voltage
1. The stator voltage control method is a part of the slip changing technique.
a) True
b) False
Answer: a
Explanation: Stator voltage control method comes under slip changing technique. The load torque remains constant here. SV2=constant.
2. Calculate the value of new slip using the given data: V1=12 V, S1=.1, V2=5.
a) 0.576
b) 0.247
c) 0.487
d) 0.987
Answer: a
Explanation: This question is based on the concept of stator voltage control method. The load torque remains constant. S1V12=S2V22=constant. S2=.576.
3. Calculate the average value of the sinusoidal waveform y(t)=4.56cos(77.64πt+4800π÷68).
a) 41 V
b) 0 V
c) 48 V
d) 78 V
Answer: b
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference. The average value of a sine wave is zero because of equal and opposite lobes areas. Since sine wave is an odd function then the net area of the waveform over a period is Net area = A+(-A) = 0. The average value is Net area÷Time=0.
Dynamics Of Electric Drives MCQs
4. R.M.S value of the periodic square waveform of amplitude 20 V is _______
a) 20 V
b) 18 V
c) 17 V
d) 13 V
Answer: a
Explanation: R.M.S value of the periodic square waveform is Vm and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the periodic square waveform is Vm. Vm=20 V.
5. Calculate the time period of the waveform y(t)=sin(.1πt)+cos(.2πt).
a) 20 sec
b) 30 sec
c) 40 sec
d) 10 sec
Answer: a
Explanation: The fundamental time period of the sine and cosine wave is 2π. The time period of y(t) is L.C.M {20,10}=20 sec. The time period is independent of phase shifting and time shifting.
6. The V/F control is a part of the synchronous speed changing technique.
a) True
b) False
Answer: b
Explanation: V/F control comes under the synchronous speed changing technique. Speed above or below synchronous speed can be achieved using V/F control.
7. The slope of the V-I curve is 56.489°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) 1.5 Ω
b) 1.6 Ω
c) 2.5 Ω
d) 1.4 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 56.489° so R=tan(56.489°)=1.5 Ω. The slope of the I-V curve is reciprocal of resistance.
8. If induction motor air gap power is 28.63 KW and gross developed power is 18.8 KW, then rotor ohmic loss will be _________ KW.
a) 9.83
b) 10.55
c) 15.54
d) 4.74
Answer: a
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=28.63-18.8=9.83 KW.
9. Calculate the value of inductive reactance if F=50 Hz and L=12 H.
a) 3768 Ω
b) 2578 Ω
c) 2477 Ω
d) 2456 Ω
Answer: a
Explanation: Inductive reactance can be calculated using the relation XL=2×3.14×f×L. The value of inductive reactance is XL=2×3.14×50×12=3768 Ω.
10. Calculate the value of capacitive reactance if F=60 Hz and C=14 H.
a) 189.5 mΩ
b) 252.4 mΩ
c) 244.5 mΩ
d) 244.8 mΩ
Answer: a
Explanation: Capacitive reactance can be calculated using the relation Xc=1÷2×3.14×f×L. The value of capacitive reactance is Xc=1÷2×3.14×50×12=189.5 mΩ.
11. Calculate the time period of the waveform v(t)=.6sin(.1πt+8π)+17cos(2πt+π)+ 4tan(.1πt).
a) 88 sec
b) 20 sec
c) 70 sec
d) 43 sec
Answer: b
Explanation: The fundamental time period of the sine wave is 2π. The time period of v(t) is L.C.M {20,1,20}=20 sec. The time period is independent of phase shifting and time shifting.
12. Calculate the value of resistance if total heat dissipated is 78 W when 5 A current flows through it.
a) 3.12 Ω
b) 2.24 Ω
c) 1.45 Ω
d) 5.13 Ω
Answer: a
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. R=P÷I2=3.12 Ω.
13. The value of slip at the starting of an induction motor is ________
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: The maximum torque occurs when the slip value is equal to R2÷X2. At the starting, the rotor speed is equal to zero so slip value is 1.
14. The value of the slip of an induction motor during full load condition is ________
a) 0.99
b) .1
c) .8
d) 0
Answer: d
Explanation: During full load condition the speed of the rotor is nearly synchronous speed. The value of slip is nearly zero during full load condition.
Induction Motors – Controlling Speed by Adjusting the Supply Frequency
1. 40 V, 7 A, 70 rpm DC separately excited motor having a resistance of 0.3 ohms excited by an external dc voltage source of 4 V. Calculate the torque developed by the motor on half load.
a) 18.10 N-m
b) 4.24 N-m
c) 40.45 N-m
d) 52.64 N-m
Answer: a
Explanation: Back emf developed in the motor during the full load can be calculated using equation Eb = Vt-I×Ra = 37.9 V and machine constant Km = Eb÷Wm which is equal to 5.17. Torque can be calculated by using the relation T = Km× I = 5.17×3.5 = 18.10 N-m.
2. Calculate the active power developed by a motor using the given data: Eb = 5.5 V and I = .5 A.
a) 2.75 W
b) 2.20 W
c) 5.30 W
d) 5.50 W
Answer: a
Explanation: Power developed by the motor can be calculated using the formula P = Eb×I = 5.5×.5 = 2.75 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
3. Calculate the value of the angular acceleration of the motor using the given data: J = .1 kg-m2, load torque = 45 N-m, motor torque = 55 N-m.
a) 100 rad/s2
b) 222 rad/s2
c) 300 rad/s2
d) 400 rad/s2
Answer: a
Explanation: Using the dynamic equation of motor J×(angular acceleration) = Motor torque – Load torque: .1×(angular acceleration) = 55-45=10, angular acceleration = 100 rad/s2.
4. Calculate the moment of inertia of the tennis ball having a mass of 7 kg and diameter of 152 cm.
a) 3.55 kgm2
b) 4.47 kgm2
c) 2.66 kgm2
d) 1.41 kgm2
Answer: c
Explanation: The moment of inertia of the tennis ball can be calculated using the formula I=mr2×.5. The mass of the ball and diameter is given. I=(7)×.5×(.76)2=2.66 kgm2. It depends upon the orientation of the rotational axis.
5. Calculate the moment of inertia of the thin spherical shell having a mass of 7.8 kg and diameter of 145.6 cm.
a) 2.72 kgm2
b) 5.96 kgm2
c) 5.45 kgm2
d) 2.78 kgm2
Answer: a
Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr2×.66. The mass of the thin spherical shell and diameter is given. I=(7.8)×.66×(.728)2=2.72 kgm2. It depends upon the orientation of the rotational axis.
6. Calculate the time period of the waveform y(t)=87cos(πt+289π÷4).
a) 2 sec
b) 37 sec
c) 3 sec
d) 1 sec
Answer: a
Explanation: The fundamental time period of the cosine wave is 2π. The time period of y(t) is a 2π÷π=2 sec. The time period is independent of phase shifting and time shifting.
7. Calculate the useful power developed by a motor using the given data: Pin = 10 W, Ia = .6 A, Ra=.2 Ω. Assume frictional losses are 2 W and windage losses are 3 W.
a) 4.928 W
b) 1.955 W
c) 1.485 W
d) 1.488 W
Answer: a
Explanation: Useful power is basically the shaft power developed by the motor that can be calculated using the formula Psh = Pdev-(rotational losses). Pdev = Pin-Ia2Ra = 10-.62(.2)=9.92 W. The useful power developed by the motor is Psh = Pdev-(rotational losses) = 9.92 –(5) = 4.928 W.
8. The slope of the V-I curve is 86.5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) 16.34 Ω
b) 15.88 Ω
c) 48.43 Ω
d) 54.57 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 16.34° so R=tan(16.34°)=16.34 Ω. The slope of the I-V curve is reciprocal of resistance.
9. Calculate the active power in a 7481 H inductor.
a) 1562 W
b) 4651 W
c) 0 W
d) 4654 W
Answer: c
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W. Voltage leads the current in case of the inductor.
10. Calculate the active power in a 56 F capacitor.
a) 6.45 W
b) 0 W
c) 15.45 W
d) 14.23 W
Answer: b
Explanation: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P=VIcos90 = 0 W. Current leads the voltage in case of the capacitor.
11. Calculate the active power in .an 18.064 H inductor.
a) 4.48 W
b) 17.89 W
c) 0 W
d) 25.45 W
Answer: c
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W.
12. Calculate the active power in a 1.7 Ω resistor with 1.8 A current flowing through it.
a) 5.5 W
b) 5.1 W
c) 5.4 W
d) 5.7 W
Answer: a
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. P=I2R=1.8×1.8×1.7=5.5 W.
13. Calculate the value of the time period if the frequency of the signal is .48 sec.
a) 2 Hz
b) 3 Hz
c) 7 Hz
d) 9 Hz
Answer: a
Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷.48=2 Hz.
14. Choose the correct in the case of V/F control.
a) Ns-Nr=constant
b) Ns*Nr=constant
c) Ns%Nr=constant
d) Ns+Nr=constant
Answer: a
Explanation: In variable frequency control Ns-Nr remains constant. V/f control is part of the synchronous speed changing technique. It is the most used technique in controlling the induction motor.
Induction Motors – Voltage/Frequency Control
1. Calculate the minimum value of the active power in the cylindrical rotor synchronous machine? (Eb represents armature emf, Vt represents terminal voltage, δ represents rotor angle, X represents reactance)
a) Eb×Vt×sinδ÷X
b) 0
c) Eb2×Vt×sinδ÷X
d) Eb×Vt÷X
Answer: b
Explanation: The real power in the cylindrical rotor machine is Eb×Vt×sinδ÷X. It is inversely proportional to the reactance. The stability of the machine is decided by the maximum power transfer capability. Its minimum value occurs for delta=0°.
2. Cylindrical pole machines are less stable than salient rotor machines.
a) True
b) False
Answer: a
Explanation: Cylindrical pole machines are less stable than salient rotor machines because of the less short circuit ratio and less real power transfer capability. The air gap length in cylindrical pole machines is less as compare to salient rotor machines.
3. The unit of area is m2.
a) True
b) False
Answer: a
Explanation: Area is defined as the product of length and breadth. The length and breadth are expressed in the meter. The unit of area is m2.
4. Calculate the power factor if the power angle is 45°.
a) .707
b) .407
c) .608
d) 1
Answer: a
Explanation: Power factor is the ratio of the real power to the apparent power. It measures the useful power contained in the total power cosΦ=cos(45°)=.707.
5. Calculate the reactive power in a 451.26 Ω resistor.
a) 1 VAR
b) .6 VAR
c) 0 VAR
d) .9 VAR
Answer: c
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. Q=VIsin0° = 0 VAR.
6. What is the unit of displacement?
a) m/s
b) m
c) atm/m
d) Volt
Answer: a
Explanation: Displacement is the difference between the final and initial point. It is a vector quantity. It is expressed in the meter. It is not a tensor quantity. It has direction.
7. Calculate the reactive power in a 725.45 Ω resistor.
a) 122.1 VAR
b) 261.1 VAR
c) 0 VAR
d) 199.7 VAR
Answer: c
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. Q = VIsin0° = 0 VAR.
8. The slope of the I-V curve is 180°. Calculate the value of resistance.
a) inf Ω
b) .81 Ω
c) 45 Ω
d) 41.2 Ω
Answer: a
Explanation: The slope of the I-V curve is the reciprocal of the resistance. The slope given is 180° so R=1÷tan(180°)=inf Ω. The slope of the V-I curve is resistance. It behaves as an open circuit.
9. Calculate the value of the duty cycle if the system is on for 1 sec and off for inf sec.
a) 0
b) .89
c) .148
d) .46
Answer: a
Explanation: Duty cycle is Ton÷Ttotal. It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷Ttotal=1÷inf = 0.
10. The phase difference between voltage and current in the purely inductive coil.
a) 20°
b) 90°
c) 40°
d) 88°
Answer: b
Explanation: In the case of a purely inductive coil, the voltage leads the current by 90° or the current lags the voltage by 90°. The phase difference between voltage and current is 90°.
11. Armature reaction is purely demagnetizing in nature due to a capacitive load in the synchronous motor.
a) True
b) False
Answer: a
Explanation: Due to a capacitive load, armature current is in opposite and 90° phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in 180° phase and 90° phase with the field flux. The nature of the armature reaction is partially demagnetizing and cross magnetizing.
12. Armature reaction is cross-magnetizing in nature due to a resistive load in the synchronous generator.
a) True
b) False
Answer: a
Explanation: Due to a purely resistive load, armature current is in quadrature with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in quadrature with the field flux. It will try to increase the net magnetic field. The nature of the armature reaction is cross magnetizing.
13. Calculate the value of the short circuit ratio if Voc=78 V, Isc=15 A with field current = 5 A.
a) 5.2
b) 4.8
c) 3.2
d) 1.8
Answer: a
Explanation: The value of the short circuit ratio is the ratio of open circuit voltage to the short circuit current with same field current. SCR=Voc÷Isc=78÷15=5.2.
14. SCR determines the stability of the synchronous machine.
a) True
b) False
Answer: a
Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. Stability of the synchronous machine depends upon the maximum power transfer capability.
Induction Motors – Current Source Speed Control
1. Calculate the value of the short circuit ratio if the per unit value of synchronous reactance is 1.5 Ω.
a) 0.66
b) 0.33
c) 0.17
d) 0.12
Answer: a
Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. SCR=1÷Xs(p.u)=1÷1.5=.66.
2. A higher value of armature reaction means the lower value of the short circuit ratio.
a) True
b) False
Answer: a
Explanation: SCR is inversely proportional to the per unit value of synchronous reactance. Armature reaction depends upon the value of the synchronous reactance. A higher value of armature reaction means the lower value of the short circuit ratio.
3. The unit of SCR is Ω.
a) True
b) False
Answer: b
Explanation: SCR is the ratio of the field currents required to produce open circuit voltage and short circuit current. It is a ratio. It has no unit.
4. Calculate the power factor if the power angle is 12°.
a) .87
b) .47
c) .97
d) 0
Answer: c
Explanation: Power factor is the ratio of the real power to the apparent power. It measures the useful power contained in the total power cosΦ=cos(12°)=.97.
5. Calculate the reactive power in an 84 MΩ resistor.
a) 1.47 VAR
b) 45.6 VAR
c) 0.59 VAR
d) 0 VAR
Answer: d
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. Q=VIsin0° = 0 VAR.
6. Calculate the value of the resistance of a wire of length = 12 m, area = 2 m2 and Ρ = 6 Ω-m.
a) 36 Ω
b) 6 Ω
c) 26 Ω
d) 10 Ω
Answer: a
Explanation: Resistance is the opposition offered by the body to the flow of the current. It is expressed in terms of Ω. R = ΡL/A = 36 Ω.
7. Calculate the reactive power in a 29.6 Ω resistor.
a) 45.1 VAR
b) 41.1 VAR
c) 46 VAR
d) 0 VAR
Answer: d
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 0°. Q=VIsin0° = 0 VAR.
8. The slope of the I-V curve is 160°. Calculate the value of resistance.
a) .36 Ω
b) .41 Ω
c) .89 Ω
d) 2.74 Ω
Answer: d
Explanation: The slope of the I-V curve is the reciprocal of the resistance. The slope given is 160° so R=1÷tan(160°)=2.74 Ω. The slope of the V-I curve is resistance.
9. Calculate the value of the power if voltage=4.2 V and current=10 A.
a) 42 W
b) 12 W
c) 16 W
d) 18 W
Answer: a
Explanation: Power is the product of the voltage and current. It is generally expressed in term of W. P=vi=4.2×10=42 W. 1 horsepower=746 W.
10. 1 calorie is equal to ________
a) 4.18 J
b) 5.23 J
c) 6.23 J
d) 8 J
Answer: a
Explanation: The calorie is the unit of energy. It is equal to 4.18 Joule. It is the amount of heat energy required to raise the temperature of one 1 g of water by 1° Celsius.
11. 1 Horse-power is equal to _______
a) 745.7 W
b) 741 W
c) 747.7 W
d) 740 W
Answer: a
Explanation: 1 Horse-power is equal to 745.7 W. It is equal to the rate at which work is done. Horse-power is the amount of work done by horse in carrying the weight of 90 kg for 50 meters in 60 seconds.
12. Calculate the value of the time period if the frequency of the signal is 58 sec.
a) .017 sec
b) .014 sec
c) .045 sec
d) .077 sec
Answer: a
Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T=1÷F=1÷58=.017 sec.
13. Calculate the value of the short circuit ratio if the per unit value of synchronous reactance is 4.2 Ω.
a) 0.26
b) 0.23
c) 0.67
d) 0.72
Answer: b
Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. SCR=1÷Xs(p.u)=1÷4.2=.23.
14. Calculate the value of the short circuit ratio if the per unit value of synchronous reactance is 1 Ω. Assume armature resistance is 0 ohm.
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. SCR=1÷Xs(p.u)=1÷1=1.