Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

Multiple choice question for engineering

Set 1

1. Inverters converts
a) dc power to dc power
b) dc power to ac power
c) ac power to ac power
d) ac power to dc power

View Answer

Answer: b [Reason:] Inverter is a dc to ac converter.

2. Line-commutated inverters have
a) AC on the supply side and DC on the load side
b) AC on both supply and load side
c) DC on both supply and load side
d) DC on the supply side and AC on the load side

View Answer

Answer: b [Reason:] Line-commuted inverters are actually phase-controlled converters operated in the inverting mode. They cannot act as standalone inverters as they require a AC supply. It is to be noted that “line commutated inverter” is not the conventional inverter. The conventional inverters are forced or load commutated inverters.

3. In a VSI (Voltage source inverter)
a) the internal impedance of the DC source is negligible
b) the internal impedance of the DC source is very very high
c) the internal impedance of the AC source is negligible
d) the IGBTs are fired at 0 degrees.

View Answer

Answer: a [Reason:] A VSI is the one in which the internal impedance of the source is negligible. It has a stiff DC source at its input.

4. VSIs using GTOs are turned off by
a) load commutation
b) line commutation
c) applying a negative gate pulse
d) removing the base signal

View Answer

Answer: c [Reason:] GTOs are gate turn off SCRs in which turn-off is achieved by applying a negative gate pulse.

5. VSIs using IGBTs are turned off by
a) load commutation
b) line commutation
c) applying a negative gate pulse
d) removing the base signal

View Answer

Answer: d [Reason:] IGBT is a transistor family device. It can be turned off simply by removing the gate signal. All the transistor devices operated in the same way in inverters.

6. __________ based inverters do not require self-commutation.
a) IGBT
b) GTO
c) PMOSFET
d) SCR

View Answer

Answer: d [Reason:] All the devices can be turned off by their gate/base singles expect SCR. SCRs require external commutation circuits.

7. Identify the circuit given below.
power-electronics-questions-answers-single-phase-vsi-1-q7-q9
a) Half wave series inverter
b) Full wave series inverter
c) Half wave bridge inverter
d) Half wave parallel inverter

View Answer

Answer: c [Reason:] It is a half-wave circuit as it has only 2 SCRs are connected. It is a half wave bridge type thyristorised inverter.

8. Single phase half bridge inverters requires
a) two wire ac supply
b) two wire dc supply
c) three wire ac supply
d) three wire dc supply

View Answer

Answer: d [Reason:] They require two voltage sources Vs/2 and Vs/2.

9. What is the voltage across the R load when only T1 is conducting?
power-electronics-questions-answers-single-phase-vsi-1-q7-q9
a) Vs
b) Vs/2
c) 2Vs
d) zero

View Answer

Answer: b [Reason:] Considering T1 as an ideal switch, the load is directly connected to the upper voltage source when T1 is on.

10. In a single-phase half wave inverter ________ SCR(s) are/is gated at a time.
a) one
b) two
c) three
d) none of the mentioned

View Answer

Answer: a [Reason:] Only one SCR is gated at a time, gating the both of them will short the supply.

Set 2

1. What is the voltage across the R load when only T2 is conducting?
power-electronics-questions-answers-single-phase-vsi-2-q1-q5-q6
a) Vs
b) Vs/2
c) 2Vs
d) Zero

View Answer

Answer: b [Reason:] Considering T2 as an ideal switch, the load is directly connected to the lower voltage source when T2 is on.

2. The voltage in a single phase half wave inverter varies between
a) Vs and 0
b) Vs/2 and 0
c) Vs/2 and –Vs/2
d) Vs and –Vs

View Answer

Answer: c [Reason:] It varies between –Vs/2 and Vs/2 as the upper and lower thyristors/IGBTs are fired in sequence.

3. Below given is a
power-electronics-questions-answers-single-phase-vsi-2-q3
a) SCR based inverter
b) MOSFET based inverter
c) IGBT based inverter
d) None of the mentioned

View Answer

Answer: b [Reason:] It is a half wave inverter with MOSFET switches. The diodes are used to bypass the negative current.

4. The output of a single-phase half bridge inverter on R load is ideally
a) a sine wave
b) a square wave
c) a triangular wave
d) constant dc

View Answer

Answer: b [Reason:] Due to rapid switching on and off of the devices, it seems to be a square wave. But practically it is a B-wave.

5. In the below given circuit, T1 is fired at 0,T and so on and T2 at T/2, 3T/2 etc. What is the frequency of the alternating voltage obtained?
power-electronics-questions-answers-single-phase-vsi-2-q1-q5-q6
a) 50 Hz
b) T Hz
c) 1/T Hz
d) T/2 Hz

View Answer

Answer: c [Reason:] T1 conducts from 0 to T/2 and T2 from T/2 to T. Hence, 1 cycle is completed in T time. f = 1/T Hz.

6. In the below circuit, if Vs/2 = 50 V. Find, the rms AC voltage that would be ideally obtained.
power-electronics-questions-answers-single-phase-vsi-2-q1-q5-q6
a) 50 V
b) 100 V
c) 0.707 x 50 V
d) 0.707 x 100 V

View Answer

Answer: c [Reason:] Peak value would be = 50 V RMS = Peak/√2 = Peak x 0.707 V.

7. Find the conduction time of the diodes if the SCRs are fired at 0 and T/2 respectively in a single phase half wave inverter with R load.
a) 0
b) T/2
c) 2/T
d) insufficient data

View Answer

Answer: a [Reason:] With R load, the diodes do not come into play.

8. The output current wave of a single-phase full bridge inverter on RL load is
a) a sine wave
b) a square wave
c) a triangular wave
d) constant dc

View Answer

Answer: c [Reason:] On RL load, the SCRs are revised biased due to the voltage drops across the diodes and negative current flows.

9. Single-phase full bridge inverters requires
a) 4 SCRs and 2 diodes
b) 4 SCRs and 4 diodes
c) 2 SCRs and 4 diodes
d) 2 SCRs and 2 diodes

View Answer

Answer: b [Reason:] Full bridge inverters require 4 SCR and 4 diodes along with a two wire dc source.

10. Identify the circuit given below.
power-electronics-questions-answers-single-phase-vsi-2-q10
a) Full wave series inverter
b) Half wave series inverter
c) Full wave bridge inverter
d) Full wave parallel inverter

View Answer

Answer: c [Reason:] It is a full wave bridge inverter using 4 diodes and 4 SCRs.

Set 3

1. The output voltage from a single phase full wave bridge inverter varies from
a) Vs to –Vs
b) Vs to zero
c) Vs/2 to zero
d) –Vs/2 to Vs/2

View Answer

Answer: a [Reason:] The output from a full wave bridge inverter varies from Vs to –Vs.

2. In a single phase full wave bridge inverter, when the output is Vs or –Vs
a) one SCR and one diode are conducting
b) four SCRs are conducting
c) two SCRs are conducting
d) two diodes are conducting

View Answer

Answer: c [Reason:] When two SCRs are conducting, T1-T2 or T3-T4, the output voltage magnitude is Vs. When the diodes are conducting, freewheeling action is taking place and voltage is zero.

3. In the following circuit, the diodes D1 to D4 are used to
power-electronics-questions-answers-single-phase-vsi-3-q3
a) reduce the switching losses
b) send current back to the dc source when SCRs are off
c) send current back to the load when SCRs are off
d) send current back to the dc source when SCRs are conducting

View Answer

Answer: b [Reason:] Diodes are connected in anti-parallel with the SCRs during inductive loads for freewheeling actions.

4. For a full wave bridge inverter, the output voltage (Vo)
a) Vo = Vs/2 for 0 < t < T/2
b) Vo = Vs for 0 < t <T/2
c) Vo = Vs for T/2< t < T
d) Vo = -Vs for T/2< t < 3T/2

View Answer

Answer: b [Reason:] For the first half cycle, Vo = Vs and for the second half cycle Vo = -Vs.

5. For a half wave bridge inverter, the output voltage
a) Vo = – Vs/2 for 0 < t < T/2
b) Vo = – Vs/2 for T/2< t<T
c) Vo = – Vs for 0 < t < T/2
d) Vo = Vs/2 for T/2< t < T

View Answer

Answer: b [Reason:] In case of a half wave inverter, for the first half cycle, Vo = Vs/2 and for the second half cycle Vo = -Vs/2.

6. The fundamental component of output voltage for a half wave bridge inverter is given by
a) (4Vs/π) sinωt
b) (2Vs/π) sinωt
c) (Vs/2π) sinωt
d) (Vs)

View Answer

Answer: b [Reason:] The fourier analysis of half wave inverter gives, power-electronics-questions-answers-single-phase-vsi-3-q6 Put n = 1 for the fundamental component.

7. The fundamental component of output voltage for a full wave bridge inverter is given by
a) (2Vs/π) sinωt
b) (4Vs/π) sinωt
c) (Vs/2π) sinωt
d) (Vs)

View Answer

Answer: b [Reason:] The fourier analysis of full wave inverter gives, Vo = power-electronics-questions-answers-single-phase-vsi-3-q7 Put n = 1 for the fundamental component.

8. A single phase half bridge inverter has a dc voltage source Vs/2 = 115 V. Find the rms value of the fundamental component of output voltage.
a) 510 V
b) 103.5 V
c) 120 V
d) 96 V

View Answer

Answer: b [Reason:] The fundamental component of voltage = (2Vs/π) sinωt. Peak value Vm = 2Vs/π Rms value = 2Vs/π√2 = 103.552 V.

9. A single phase half bridge inverter has load R = 2 Ω and a dc voltage source Vs/2 = 115 V. Find the rms value of the fundamental load current.
a) 10.25 A
b) 51.7 A
c) 86 A
d) 24.8 A

View Answer

Answer: b [Reason:] Peak value Vm = 2Vs/π Rms value = 2Vs/π√2 = 103.552 V Io = Vo/R = 103.552/2 = 51.776 A.

10. A single phase half bridge inverter has load R = 2 Ω and a dc voltage source Vs/2 = 115 V. Find the power delivered to the load due to the fundamental component.
a) 536 kW
b) 53.61 kW
c) 5.361 kW
d) 536 W

View Answer

Answer: c [Reason:] Peak value Vm = 2Vs/π Rms value = 2Vs/π√2 = 103.552 V Io = Vo/R = 103.552/2 = 51.776 A P = Io2 x R = 5361.5 Watts.

Set 4

1. A single phase full bridge inverter has a dc voltage source Vs = 230 V. Find the rms value of the fundamental component of output voltage.
a) 90 V
b) 207 V
c) 350 V
d) 196 V

View Answer

Answer: b [Reason:] The fundamental component of voltage = (4Vs/π) sinωt. Peak value Vm = 4Vs/π Rms value = 4Vs/π√2.

2. A single phase full bridge inverter has load R = 2 Ω, and dc voltage source Vs = 230 V. Find the rms value of the fundamental load current.
a) 96 A
b) 0 A
c) 103 A
d) none of the mentioned

View Answer

Answer: c [Reason:] The fundamental component of voltage = (4Vs/π) sinωt. Peak value Vm = 4Vs/π Rms voltage = 4Vs/π√2 = 207 V Current = 207/2 = 103.5 A.

3. A certain full bridge type inverter circuit has its rms value of fundamental load current component given by W. The fundamental frequency component of the load current would be given by
a) W sin ωt
b) (W/√2) sin ωt
c) √2 W sin ωt
d) sin ωt

View Answer

Answer: c [Reason:] The fundamental frequency component of the load current is given by Im sin ωt As W = Irms . . . (Given) Im = √2 Irms = √2 W.

4. In a half wave bridge inverter circuit, the power delivered to the load by each source is given by
a) Vs x Is
b) (Vs x Is)/2
c) 2(Vs x Is)
d) None of the mentioned

View Answer

Answer: b [Reason:] Power delivered by each source (Vs/2) each is (Vs/2) x Is.

5. In a half wave circuit, forced commutation is essential when the
a) load is inductive
b) load is resistive
c) source voltage is below 150 V
d) none of the mentioned

View Answer

Answer: b [Reason:] When the load is resistive (R load) , the diodes do not conduct, hence they cannot help stop the conduction of the SCRs. Hence, forced commutation in such cases becomes essential.

6. A single phase full bridge inverter circuit, has load R = 2 Ω and dc source Vs = 230 V. Find the value of power delivered to the load in watts only due to the fundamental component of the load current.
a) 5361.5 W
b) 2142.5 W
c) 21424.5 W
d) 214.2 W

View Answer

Answer: c [Reason:] The fundamental component of voltage = (4Vs/π) sinωt Peak value Vm = 4Vs/π Rms voltage = 4Vs/π√2 = 207 V RMS Current (Irms) = 207/2 = 103.5 A P = (Irms)2 x R = 21424.5 W.

7. A single phase full bridge inverter is fed from a dc source such that the fundamental component of output voltage = 230 V. Find the rms value of SCR and diode current respectively, for a R load of 2 Ω.
a) 115 A, 80 A
b) 81.33 A, 36.2 A
c) 36.2 A, 0 A
d) 81.33 A, 0 A

View Answer

Answer: d [Reason:] Fundamental component of load current = V/R = 230/2 = 115 A. SCR current = 115/2 = 81.33 A Diode current = 0 as the diodes do not come into picture for R loads.

8. For a full bridge inverter with the following load: R = 2 Ω, XL = 8 Ω and XC = 6 Ω.
a) The output voltage lags the current by 45°
b) The output current lags the voltage by 45°
c) The output current lags the voltage by 90°
d) The output current lags the voltage by more than 90°

View Answer

Answer: b [Reason:] As the inductive effect is more than the capacitive effect, of course the current will lag the voltage by an angle P. P = tan-1 (XL – XC)/R = tan-1 (1) = 45°.

9. A single phase full bridge inverter has RLC load. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the expression for the load voltage up to the fifth harmonic.
a) 292 sin 314t + 97.62 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t
b) 292 sin 314t + 97.62 sin (3 x 314t) + 58.57 sin (5 x 318t)
c) 292 sin 314t + 97.62 sin (2 x 314t) + 58.57 sin (3 x 318t) + 28.31 sin (4 x 318t) + 3.686 sin (5 x 318t)
d) 292 sin 512t + 25.62 sin 249t + 6.74 sin 508t

View Answer

Answer: b [Reason:] In a single phase full bridge inverter only odd harmonics are present. i.e. 1,3,5 etc. Vo = (4Vs/π) sin ωt + (4Vs/3π) sin 3ωt + (4Vs/5π) sin 5ωt (4Vs/π) = 292 V ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.

10. A single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the rms value of the fundamental load current.
a) 28.31 A
b) 20.02 A
c) 16.69 A
d) 26.90 A

View Answer

Answer: b [Reason:] Average value of fundamental output voltage = 4Vs/π = 292.85 V XL = 2 x 3.14 x 50 x 0.035 = 10.99 Ω XC = 1/(2 x 3.14 x 50 x 155 x 10-6) = 20.54 Ω Z = 10.345 Ω I = V/Z = 28.31 I(rms) = V/Z√2 = (292.85)/(1.414 x 10.345) = 20.02 A.

Set 5

1. If a step up chopper’s switch is always kept off then (ideally)
a) Vo = 0
b) Vo = ∞
c) Vo = Vs
d) Vo > Vs

View Answer

Answer: c [Reason:]; If it is said the a chopper is always kept off, that means the switch is always open. As such, Ton = 0 Duty cycle = 0 Vo = Vs/1-duty cycle . . . (for a step-up chopper).

2. If a step up chopper’s switch is always kept open then (ideally)
a) Vo = 0
b) Vo = ∞
c) Vo = Vs
d) Vo > Vs

View Answer

Answer: b [Reason:] If it is always in then, Ton = T. Duty cycle = 1. Vo = Vs/1-duty cycle = Vs/0 Therefore, Vo = ∞.

3. Find the average value of output voltage of a basic step-down chopper, with duty cycle = α and load = R Ω.
power-electronics-questions-answers-test-q3-q4
a) I = Vs x α
b) I = (Vs x α)/R
c) I = 0
d) I = Vs/R

View Answer

Answer: b [Reason:] Vo = Vs x α I = Vo/R.

4. For the below given circuit, find the output current at the instant of commutation.
power-electronics-questions-answers-test-q3-q4
a) Vo/R
b) Vs/R
c) 0
d) α/2

View Answer

Answer: b [Reason:] The circuit is that of a step-down chopper. The output current is commutated by the switch(IGBT) at the instant t = Ton. Therefore, output current at the instant of commutation is Vs/R.

5. For a step-down chopper, find the rms value of output voltage. Let α be the duty cycle and Vs be the input voltage.
a) α x Vs
b) Vs/α
c) √α x Vs
d) Vs/2

View Answer

Answer: c [Reason:] [(Ton/T).Vs2]1/2 = √α x Vs.

6. A step down chopper is operated at 240V at duty cycle of 75%. Find the value of RMS switch (IGBT/MOSFET) current. Take R = 10 Ω.
a) 2.07 A
b) 200 mA
c) 1.58 A
d) 2.4 A

View Answer

Answer: a [Reason:] The switch could be anything, a IGBT, a SCR, a GTO etc. The current through the switch is the same as the current through the load for a step down chopper. Vo = Vs/1-α α = 0.75 I = √α x Vs/R.

7. Find the expression for effective input resistance of a step down chopper. With R load and duty cycle = α.
a) R x α
b) R/2
c) 0
d) R/α

View Answer

Answer: d [Reason:] Effective input resistance of the chopper circuit = (dc source voltage)/(average source current) = (Vs.R)/(Vs.α) = R/α.

8. A step-up chopper has input voltage of 220 V and output voltage of 660 V. If the conducting time of the IGBT based chopper is 100 μs, compute Toff width of the output voltage pulse.
a) 100 μs
b) 150 μs
c) 50 μs
d) Insufficient data

View Answer

Answer: c [Reason:] 660 = 220/1-α α = 2/3 = Ton/T Ton = 2T/3 = 100μs. This gives chopping period T = 100 x 3/2 = 150 μs Toff = T – Ton = 150 – 100 = 50 μs.

9. For a step-up chopper, when the duty cycle is increased the average value of the output voltage
a) increases
b) decreases
c) remains the same
d) none of the mentioned

View Answer

Answer: a [Reason:] Vo = Vs/1-duty cycle Hence, as duty cycle increases the output voltage increases.

10. For a step-down chopper, when the duty cycle is increased the average value of the output voltage
a) increases
b) decreases
c) remains the same
d) none of the mentioned

View Answer

Answer: a [Reason:]; Vo = Duty cycle x Vs. Hence, output voltage is directly proportional to the duty cycle.