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Multiple choice question for engineering

Set 1

1. Which of the following is used as a harmonic reduction technique in inverters?
a) Amplitude modulation
b) Cycloconverter control
c) Transformer connection
d) Series connection of two inverters

View Answer

Answer: c [Reason:] Two or more inverters are connected together by means of a transformer to get the net output voltage with reduced harmonic content.

2. For harmonic reduction by transformer connection, the output voltages from the two inverters must be
a) similar and in-phase with each other
b) dissimilar but in-phase with each other
c) similar but phase shifted from each other
d) dissimilar and phase shifted from each other

View Answer

Answer: c [Reason:] The essential condition of this scheme is that the output voltages from the two inverters must be similar but phase shifted from each other.

3. The output voltage obtained by connecting two inverters through a transformer is a
a) square wave
b) sine wave
c) quasi-square wave
d) none of the mentioned

View Answer

Answer: c [Reason:] The net output voltage is the addition of the two inverter voltages, as they are phase shifted from each other, the waveform obtained is a quasi-square wave.

4. Pulses of different widths and heights are superimposed in case of __________ harmonic reduction technique.
a) transformer connection
b) pulse width modulation
c) stepped-wave inverter
d) none of the mentioned

View Answer

Answer: c [Reason:] In the steeped wave inverters pulses of different widths and heights are superimposed to produce a resultant stepped wave with reduced harmonic content.

5. In case of stepped wave inverters,
a) both the transformers have 1:1 turns ratio
b) both the transformers have 1:2 turns ratio
c) both the transformers have different transformer ratio
d) none of the mentioned

View Answer

Answer: c [Reason:] The two transformers have different turns ratio, hence, different voltages levels are obtained at the secondary’s which are the added to get the net output voltage with reduced harmonic content.

6. In stepped wave inverters, one of the inverters are gated such as to obtain
a) one level modulation
b) two level modulation
c) zero level modulation
d) none of the mentioned

View Answer

Answer: b [Reason:] In one-level modulation, during the first half cycle, the output voltage is either zero or positive. During the second half cycle the output voltage would be either zero or negative.

7. In three-level modulation
a) the output voltage is zero in the first half cycle.
b) the output voltage either zero or positive in the first half cycle.
c) the output voltage either zero or negative in the first half cycle.
d) the output voltage either zero, positive or negative in the first half cycle.

View Answer

Answer: d [Reason:] In stepped wave inverters, one of the inverters is so gated as to obtain a three level modulation.

8. In single-phase modulation of PWM inverters, the lowest harmonic can be eliminated if the pulse width is made equal to __________
a) 30°
b) 0°
c) 120°
d) 60°

View Answer

Answer: c [Reason:] In single-phase modulation of PWM inverters, the nth harmonic can be eliminated, if the pulse width (2d) is made equal to (2π/n). Lowest harmonic is for n = 3. 2d = 2π/3 = 120°.

9. In single-phase modulation of PWM inverters, the 5th order harmonic can be eliminated if the pulse width is made equal to ___________
a) 30°
b) 36°
c) 72°
d) None of the mentioned

View Answer

Answer: c [Reason:] In single-phase modulation of PWM inverters, the nth harmonic can be eliminated, if the pulse width (2d) is made equal to (2π/n).

10. The waveform obtained by __________ is more near to a sinusoidal wave
a) PWM inverters
b) Stepped wave inverters
c) parallel connected inverters
d) None of the mentioned

View Answer

Answer: b [Reason:] The stepped wave output, is more nearer to a sine wave. More the number of steps, closer is the wave to a sinusoidal wave.

Set 2

1. SCRs are connected in parallel to fulfill the ___________ demand
a) high voltage
b) high current
c) size
d) efficiency

View Answer

Answer: b [Reason:] Number of devices connected in parallel can carry huge amounts of current.

2. The term used to measure the degree of utilization of SCRs connected in series & parallel is
a) tuf
b) string efficiency
c) voltage/current utilization ratio
d) rectification efficiency

View Answer

Answer: b [Reason:] String Efficiency = Rating of the whole string/(rating of one SCR x number of SCRs)

3. To have maximum possible string efficiency
a) SCRs of same rating must be used
b) SCRs with similar V-I characteristics must be used
c) SCRs with the same dimensions must be used
d) SCRs with similar thermal characteristics must be used

View Answer

Answer: b [Reason:] Having similar ratings does not mean they have similar charc.

4. For a string voltage of 3300 V, let there be six series connected SCRs each of voltage 600V. Then the string efficiency is
a) 99.36 %
b) 91.7 %
c) 98.54 %
d) 96 %

View Answer

Answer: c [Reason:] String efficiency = 3300/(6 x 600) = 98.54.

5. The measure of reliability of string is given by the factor
a) DRF = 1 – String efficiency
b) DRF = 1 + String efficiency
c) DRF = String efficiency – 1
d) DRF = String efficiency x 2

View Answer

Answer: a [Reason:] DRF is de-rating factor given by the above expression.

6. When an extra SCR is connected in series with a string
a) DRF decreases
b) DRF increases
c) DRF remains constant
d) None of the mentioned

View Answer

Answer: b [Reason:] DRF = 1 – String efficiency String Efficiency = Rating of the whole string/(rating of one SCR x number of SCRs) Extra SCR will reduce the string efficiency which in turn increase the DRF.

7. The most practical way of obtaining a uniform distribution of series connected SCRs is to
a) connect a resistor of value R in series with each of the series connected SCRs
b) connect a resistor of value R in parallel with each of the series connected SCRs
c) connect a resistor of value R in series with one of the series connected SCRs
d) connect a resistor of value R in parallel with one of the series connected SCRs

View Answer

Answer: b [Reason:] For uniform distribution of voltage across series connected SCRs, a resistor of value R in parallel with each series connected SCR.

8. 3 SCRs are connected in series. The string efficiency is 91%. SCRs 1, 2 & 3 have leakage currents 4 mA, 8 mA & 12 mA. Which SCR will block more voltage?
a) SCR-1
b) SCR-2
c) SCR-3
d) All the three will block equal voltage

View Answer

Answer: a [Reason:] The SCR with lower leakage current block more voltage.

9. Two parallel connect SCRs have same voltage drop (Vt) having rated current = 2I1. SCR-1 carries a current of I1=2.6 A whereas SCR-2 carries a current of I2=1.4 A. Find the string efficiency.
a) 45 %
b) 77 %
c) 92 %
d) 84 %

View Answer

Answer: b [Reason:] The total current would be I1+I2 & rated current is 2I1 String efficiency = (I1+I2)/2I1.

10. SCRs with a rating of 1000 V & 200 A are available to be used in a string to handle 6 KV & 1 KV. Calculate the number of series & parallel units required in case the de-rating factor is 0.1. (Round off the fraction to the greatest & nearest integer)
a) Series = 7, Parallel = 6
b) Series = 6, Parallel = 7
c) Series = 6, Parallel = 6
d) Series = 7, Parallel = 7

View Answer

Answer: a [Reason:] DRF = 1-S.E Therefore 0.1 = (1-6000/1000Ns) = (1-1000/200Np) Ns = 6.6 = 7(say) Np = 5.5 = 6(say)

Set 3

1. Solid State Relays (SSRs) have a
a) coil and contact arrangement
b) optocoupler
c) scr
d) none of the mentioned

View Answer

Answer: b [Reason:] Coil and contact arrangement is used in mechanical relays, SSRs have a optocoupler which connects the control circuit to the power circuit via light sensitive devices.

2. The converter circuit which employs turn on and turn off when the voltage and/or current through the device is zero at the instant of switching is ____________
a) a conventional converter
b) a resonant converter
c) a zero switching circuit
d) none of the mentioned

View Answer

Answer: b [Reason:] Resonant converters are used to turn on and turn off when the voltage and/or current through the device is zero at the instant of switching.

3. Induction heating is a ___________ type of heating
a) zero frequency
b) high frequency
c) power frequency
d) none of the mentioned

View Answer

Answer: b [Reason:] As eddy current is proportional to the square of the supply frequency, induction heating is a high frequency heating.

4. The factors governing the induction heating are
a) resistivity
b) relative permeability
c) magnetic field intensity
d) all of the mentioned

View Answer

Answer: d [Reason:] Induction heating depends on all of the above given factors.

5. The reverse recovery time of a diode is trr = 3 μs and the rate of fall of the diode current (di/dt) = 30 A/μs. Determine the storage charge.
a) 145 μs
b) 135 μs
c) 0
d) none of the mentioned

View Answer

Answer: b [Reason:] Storage charge = (1/2) x (di/dt) x (trr)2 = 135 μs.

6. For a SCR, conduction angle is 120° when average on-state current is 20 A. When the conduction angle is halved the earlier value, the on-state average current will be?
a) 5 A
b) 40 A
c) 10 A
d) 20 A

View Answer

Answer: b [Reason:] When the conduction angle is halved, the device will conduct twice then it was conducting earlier. Hence, I = 2x 20 = 40 A.

7. A single-phase full bridge diode rectifier delivers power to a constant load current of 10 A. The average and rms values of the source currents will be respectively.
a) 5 A, 10 A
b) 10 A, 10 A
c) 5 A, 5 A
d) 10 A, 5A

View Answer

Answer: b [Reason:] As the load current is continuous, Iavg = Irms = 10 A.

8. TRIAC is used in
a) chopper
b) speed control of induction machine
c) speed control of universal motor
d) none of the mentioned

View Answer

Answer: c [Reason:] TRIAC is used in speed control of universal motor.

9. The ratio Vrms/ Vdc is known as
a) Form factor
b) Ripple factor
c) Utilization factor
d) None of the mentioned

View Answer

Answer: a [Reason:] Vrms/ Vdc = FF.

10. Determine the loss in the snubber circuit, if C = 0.545 μF and supply is 200 V, 10 kHz.
a) 233 W
b) 133 W
c) 333 W
d) 233 W

View Answer

Answer: b [Reason:] Snubber loss Ps = (1/2) x C x V2 x f = 133.1 W.

Set 4

1. _________ device from the thyristor family has its gate terminal connected to the n-type material near the anode.
a) SCR
b) RCT
c) PUT
d) SUT

View Answer

Answer: c [Reason:] PUT is Programmable Unijunction Transistor which is a p-n-p-n device just like the SCR with its gate connected to the n-type material.

2. The Programmable Unijunction Transistor (PUT) turns on & starts conducting when the
a) gate voltage exceeds anode voltage by a certain value
b) anode voltage exceeds gate voltage by a certain value
c) gate voltage equals the anode voltage
d) gate is given negative pulse w.r.t to cathode

View Answer

Answer: b [Reason:] The device only starts to conduct when the forward anode to cathode voltage exceeds the applied gate to cathode voltage.

3. The equivalent circuit of SUS (Silicon Unilateral Switch) consists of
a) a diode in series with a PUT
b) a diode in parallel with a PUT
c) a diode in anti-parallel with a PUT
d) two diodes

View Answer

Answer: c [Reason:] It is a diode connected in anti-parallel with a PUT.

4. From the following list of devices, choose the device that only turns-on for a fixed-value of anode-cathode voltage
a) PUT
b) SCR
c) SUS
d) BJT

View Answer

Answer: c [Reason:] Unlike the other devices the SUS only turns-on for a fixed value of anode to cathode voltage.

5. The SCS (Silicon Controlled Switch) is a
a) two terminal device
b) three terminal device
c) four terminal device
d) five terminal device

View Answer

Answer: c [Reason:] The SCS is a four terminal device A,K,KG & AG.

6. The SCS is a four layer, four terminal thyristor. Can be turned on by
a) the anode gate
b) the cathode gate
c) either of the gates
d) gating both the gates together

View Answer

Answer: c [Reason:] The SCS has two gates, anode-gate and cathode-gate. Either of the gates could be used to turn on the device.

7. The SCS (Silicon Controlled Switch) can be turned on by two methods, by applying __________ and __________
a) positive pulse to the anode gate, positive pulse to the cathode gate
b) positive pulse to the anode gate, negative pulse to the cathode gate
c) negative pulse to the anode gate, positive pulse to the cathode gate
d) negative pulse to the anode gate, negative pulse to the cathode gate

View Answer

Answer: c [Reason:] Either of the gates could be used to turn on the device.

8. Which of the following devices provide complete isolation between triggering circuit and power circuit?
a) PUT
b) LASCR
c) SUS
d) DIAC

View Answer

Answer: b [Reason:] Complete Isolation between triggering circuit & power circuit is the major advantage of using LASCR as they are light activated or light trigged.

9. The DIAC can be represented by
a) two SCRs in anti-parallel
b) two SCRs in parallel
c) two diodes in anti-parallel
d) two diodes in parallel

View Answer

Answer: c [Reason:] The DIAC is nothing but a bi-directional diode.

10. The TRIAC can be represented by
a) two SCRs in anti-parallel
b) two SCRs in parallel
c) two diodes in anti-parallel
d) two diodes in parallel

View Answer

Answer: a [Reason:] The TRIAC is a bidirectional SCR.

Set 5

1. In the single-pulse width modulation method, the output voltage waveform is symmetrical about __________
a) π
b) 2π
c) π/2
d) π/4

View Answer

Answer: c [Reason:] The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.

2. In the single-pulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.
a) 2π
b) 3π/2
c) π/2
d) 3π/4

View Answer

Answer: b [Reason:] In the negative half the wave is symmetrical about 3π/2.

3. The shape of the output voltage waveform in a single PWM is
a) square wave
b) triangular wave
c) quasi-square wave
d) sine wave

View Answer

Answer: c [Reason:] Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasi-square wave.

4. In the single-pulse width modulation method, the Fourier coefficient bn is given by
a) (Vs/π) [ sin(nπ/2) sin(nd) ].
b) 0
c) (4Vs/nπ) [sin(nπ/2) sin(nd)].
d) (2Vs/nπ) [sin(nπ/2) sin(nd)].

View Answer

Answer: c [Reason:] The Fourier analysis is as under: bn = (2/π) ∫ Vs sin⁡ nωt .d(ωt) , Where the integration would run from (π/2 + d) to (π/2 – d) 2d is the width of the pulse.

5. In the single-pulse width modulation method, the Fourier coefficient an is given by
a) (Vs/π) [ cos(nπ/2) cos(nd) ].
b) 0
c) (4Vs/nπ) [sin(nπ/2) sin(nd)].
d) (2Vs/nπ) [sin(nπ/2) sin(nd)].

View Answer

Answer: b [Reason:] As the positive and the negative half cycles are identical the coefficient an = 0.

6. In the single-pulse width modulation method, when the pulse width of 2d is equal to its maximum value of π radians, then the fundamental component of output voltage is given by
a) Vs
b) 4Vs/π
c) 0
d) 2Vs/π

View Answer

Answer: b [Reason:] The Fourier representation of the output voltage is given by power-electronics-questions-answers-pwm-inverters-1-q6 Put 2d = π & n = 1.

7. In case of a single-pulse width modulation with the pulse width = 2d, the peak value of the fundamental component of voltage is given by the expression
a) 4Vs/π
b) Vs
c) (4Vs/π) sin 2d
d) (4Vs/π) sin d

View Answer

Answer: d [Reason:] For the fundamental component put n = 1. power-electronics-questions-answers-pwm-inverters-1-q6 Vo = (4Vs/π) sin (d) sin (ωt) Hence the peak value is (4Vs/π) sin d.

8. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage
a) d = π
b) 2d = π
c) nd = π
d) nd = 2π

View Answer

Answer: c [Reason:] To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n. From the below expression, power-electronics-questions-answers-pwm-inverters-1-q6 when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.

9. Find the peak value of the fundamental component of voltage with a pulse width of 2d = 90 and Vs = 240 V for single-pulse modulation in a full wave bridge inverter.
a) 305 V
b) 216 V
c) 0 V
d) 610 V

View Answer

Answer: b [Reason:] The peak value of the fundamental component of voltage is given by (4Vs/π) sin d.

10. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 3rd harmonic from the output voltage waveform, the value of the pulse width (2d) must be
a) 0°
b) 60°
c) 120°
d) 180°

View Answer

Answer: c [Reason:] To eliminate the nth harmonic, nd = π. Therefore, d = π/n = π/3 = 60° Hence, 2d = 120°.

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