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Multiple choice question for engineering

Set 1

1. In a resistance firing circuit the firing angle
a) cannot be greater than 120°
b) cannot be greater than 90°
c) cannot be greater than 180°
d) cannot be greater than 160°

View Answer

Answer: b [Reason:] The R firing circuits cannot be used for alpha greater than 90 degrees.

2. For a R firing circuit, the maximum value of source voltage is 100 V. Find the resistance to be inserted to limit the gate current to 2 A.
a) 5 Ω
b) 50 Ω
c) 500 Ω
d) 0.5 Ω

View Answer

Answer: b [Reason:] R = 100/2 = 50 Ohm.

3. The diode in the R firing circuit
a) ensures that the gate voltage is a half wave DC pulse
b) ensures that the gate voltage is a full wave DC pulse
c) ensures that the gate voltage is a half wave AC pulse
d) ensures that the gate voltage is a full wave AC pulse

View Answer

Answer: b [Reason:] The diode is placed between the resistances and gate which ensures that the current flows in one direction only.

4. In case of an RC half wave triggering circuit, the firing angle can be ideally varied between
a) 0 to 180
b) 0 to 90
c) 0 to 120
d) 0 to 360

View Answer

Answer: a [Reason:] Unlike the R firing circuit, the RC firing circuits can be used to obtain firing angle greater than 180. Although practically 0 and 180 degree is improbable.

5. In the figure given below, the resistance R1 is used to
power-electronics-questions-answers-firing-circuits-2-q5
a) keep the gate circuit voltage drop minimum
b) limit the gate current to a safe value when R2 = 0
c) limit the gate current to a safe value when R2 is very large
d) allow the gate power dissipation

View Answer

Answer: b [Reason:] As R2 is the variable, R1 makes sure that the current does not exceed the maximum value when R2 is kept at zero position.

6. In case of a R firing with R2 as the variable resistance, Vgp (peak of gate voltage) and Vgt (gate triggering voltage) the value of R2 is so adjusted such that
a) Vgp = Vgt
b) Vgp > Vgt
c) Vgp < Vgt
d) Vgp = Vgt = 0

View Answer

Answer: a [Reason:] For turning on the device, the peak of gate voltage must be equal to the gate triggering voltage.

7. In case of a R firing circuit with Vgp > Vgt
a) α = 90°
b) α > 90°
c) α < 90°
d) α = 0°

View Answer

Answer: c [Reason:] For the values of Vgp great than the gate triggering voltage the firing angle is less than 90°. And for Vgp = Vgt the firing angle is equal to 90°. Α cannot go beyond 90° in case of a R firing circuit.

8. The figure shown below is that of a
power-electronics-questions-answers-firing-circuits-2-q8
a) R firing circuit
b) RC half-wave firing circuit
c) RC full-wave firing circuit
d) UJT triggering circuit

View Answer

Answer: b [Reason:] The given circuit is a RC half-wave firing circuit.

9. The figure shown below is that of an RC firing circuit.
power-electronics-questions-answers-firing-circuits-2-q8
In case of negative cycle at Vs, the capacitor C
a) charges through D2 with lower plate negative
b) charges through D1 with lower plate negative
c) charges through D2 with lower plate positive
d) charges through D1 with lower plate positive

View Answer

Answer: c [Reason:] The current flows through Vs+ – C – D2 – Load – Vs.

10. Find the value of R in case of an RC firing circuit which is to be turned on with a source voltage of 150 V and the following parameters.
Igt = 2A
Vd = 1.5V
Vgt = 125 V
a) 11.75 Ω
b) 54.25 Ω
c) 96 Ω
d) 5 Ω

View Answer

Answer: a [Reason:] R = (Vs-Vgt-Vd)/Igt.

Set 2

1. In voltage fed thyristor inverters __________ commutation is required.
a) load
b) forced
c) self
d) any commutation technique can be used

View Answer

Answer: b [Reason:] In VSI, the input voltage source Vs keeps the SCRs always forward biased. Hence, forced commutation becomes essential.

2. The McMurray circuit is a
a) commutation circuit
b) force commutated VSI
c) self commutated VSI
d) none of the mentioned

View Answer

Answer: b [Reason:] McMurray is a popularyly used forced commutated VSI circuit.

3. Forced commutation requires
a) a precharged inductor
b) a precharged capacitor
c) an overdamped RLC load
d) a very high frequency ac source

View Answer

Answer: b [Reason:] Forced commutation requires a pre charged capacitor of correct polarity to turn-off the earlier conducting SCR.

4. Which elements are not present in the original McMurray inverter but are present in the modified McMurray inverters?
a) Two auxiliary thyristors and the di/dt inductor
b) Two auxiliary diodes and the damping resistor
c) One auxiliary SCR and one auxiliary diode
d) None of the mentioned

View Answer

Answer: b [Reason:] In the original inverter circuit given by McMurray the elements DA1, DA2 and the damping resistor Rd were not present.

5. Let Im be the maximum load current and Vm be the minimum supply voltage value. Than the expression for the design of commutation circuit parameters in a single-phase modified McMurray half-bridge inverter is given by
power-electronics-questions-answers-force-commutated-inverters-q5
a) (1.5 x Im)/Vm
b) Im/(1.5 x Vm)
c) Vm/(1.5 x Im)
d) (1.5 x Vm)/Im

View Answer

Answer: c [Reason:] The design is carried out on the basis of the worst operating conditions which consist of the minimum supply voltage Vm and the maximum load current. (c) gives the empirical formula for calculation of the L and C commutation circuit parameters.

6. The number of diodes, SCRs and other components in full-bridge inverter McMurray inverter is ____________ of those in half-bridge McMurray inverter.
a) same
b) double
c) three times
d) none of the mentioned

View Answer

Answer: b [Reason:] The modified McMurray full-bridge requires twice the components that required in the half-wave type.

7. In the single-phase modified McMurray full-bridge inverter, for commutating the main SCRs T1 and T2
a) The capacitor is charged
b) TA1 is triggered
c) The commutation circuit is switched on
d) TA1 and TA2 are triggered

View Answer

Answer: d [Reason:] TA1 and TA2 are triggered together, which then turn-off the main SCRs. TA1 and TA2 are the auxiliary SCRs in the commutation circuit of the McMurray inverter.

8. How many diodes are there in total in the single-phase, modified McMurray full-bridge inverter?
a) 4
b) 6
c) 8
d) 10

View Answer

Answer: c [Reason:] 4 belong to the main full-bridge circuit and 4 in the commutation circuit (2 on both the sides).

9. The McMurray-Bedford half-bridge inverter requires
a) 4 SCRs, 2 diodes, 2 capacitors and 2 inductors
b) 4 SCRs, 4 diodes, 2 capacitors and 2 inductors
c) 2 SCRs, 4 diodes, 2 capacitors and 1 inductor
d) 2 SCRs, 2 diodes, 2 capacitors and 2 inductors

View Answer

Answer: d [Reason:] McMurray-Bedford type requires less number of SCRs and diodes as compared to the McMurray type, however the number of capacitors and inductors required is the same.

10. The single-phase McMurray-Bedford type bridge inverter is a/an
a) auxiliary-commutated inverter
b) complementary-commutated inverter
c) supplementary-commutated inverter
d) none of the mentioned

View Answer

Answer: b [Reason:] If one SCR gets turned on, the other conducting SCR gets turned off. This type of commutation is called as complementary commutation.

Set 3

1. Choose the correct statement with respect to the below given circuit.
power-electronics-questions-answers-freshers-q1
a) The load current can be negative
b) The load voltage can never be negative
c) The load voltage can never be zero
d) The load voltage can never be positive

View Answer

Answer: b [Reason:] The voltage cannot be negative due to the FD (freewheeling diode or commutating diode so connected).

2. In the below given circuit, the FD (Freewheeling diode) is forward biased at ωt =
power-electronics-questions-answers-freshers-q1
a) 0
b) α
c) π
d) 2π

View Answer

Answer: c [Reason:] It is forward biased at π by the conducting SCR & the current starts to through the FD & Load.

3. In the below given circuit, when the commutating diode or FD is conducting than the
power-electronics-questions-answers-freshers-q1
a) SCR has reverse bias voltage and the load current is zero
b) SCR has reverse bias voltage and the load current is positive
c) SCR has forward bias voltage and the load current is zero
d) SCR has forward bias voltage and the load current is positive

View Answer

Answer: b [Reason:] When the FD is forward biased at π by the conducting SCR & the current starts to through the FD & load.

4. The output voltage waveform of the below given circuit would be the same that obtained from a
power-electronics-questions-answers-freshers-q1
a) full-wave R load circuit
b) half-wave R load circuit
c) half-wave RL load circuit
d) full-wave RL load with freewheeling diode

View Answer

Answer: b [Reason:] The wave from will be like a half wave diode rectifier circuit. Which is the same as that obtained from a half-wave R load circuit and the above given circuit.

5. In a single-phase half-wave circuit with RL load and a freewheeling diode, the load voltage during the freewheeling period will be
a) zero
b) positive
c) negative
d) positive than negative

View Answer

Answer: a [Reason:] The FD short circuits the load and voltage across a short circuit would be = 0.

6. In a single-phase half-wave circuit with RL load and a freewheeling diode, the freewheeling period is
a) 0 to π
b) α to π+α
c) π to 2π+α
d) π/2 to 2π-α

View Answer

Answer: c [Reason:] Freewheeling period is the one in which the FD diode conducts.

7. A single-phase half wave rectifier with a FD is supplied by Vs = 240 V, AC with a load R = 10 Ω, L = 0.5 mH and a firing angle α = 30°. Find the average value of the load voltage.
a) 50 V
b) 100 V
c) 150 V
d) 200 V

View Answer

Answer: b [Reason:] Vo = (Vm/2π) x (1+cosα) Vm = √2Vs.

8. A single-phase HW rectifier with a FD is supplied by Vs = 240 V, 50 Hz, AC with a load R = 10 Ω, L = 0.5 mH and a firing angle α = 30°. Find the average value of the load current.
a) 10 A
b) 0.063 A
c) 6.3 A
d) 0.1 A

View Answer

Answer: a [Reason:] Vo = (Vm/2π) x (1+cosα) Vm = √2Vs Io = Vm/R (Due to the FD).

9. A single phase half-wave controlled rectifier has 400 sin314t as the input voltage and R as the load. For a firing angle of 60°,the average output voltage is
a) 200/π
b) 300/π
c) 100/π
d) 400/π

View Answer

Answer: b [Reason:] Vo = (Vm/2π) x (1+cosα) = 400/2π x (1+cos60) = 300/π.

10. Choose the incorrect statement with respect to the use of FD in half-wave circuits.
a) Input pf is improved
b) Load current waveform is improved
c) It prevents the load voltage from becoming negative
d) Reduces the reverse voltage (PIV) faced by the SCR

View Answer

Answer: d [Reason:] PIV is unaffected with the use of FD (freewheeling diodes).

Set 4

1. If the RC firing circuit used for firing an SCR is to be used to fire a TRIAC then
a) the capacitor should be removed
b) the diode should be replaced by a diac
c) the diode should be replaced by a bjt
d) the diode should be shorted using a resistor

View Answer

Answer: b [Reason:] The TRIAC is a bidirectional SCR, hence it will need gating in both the directions. This can be achieved by replacing the diode by a DIAC (bidirectional diode).

2. In the thyristor gating circuit, the supply to the pulse amplifier is provided by the
a) zcd
b) isolation transformer
c) synchronizing transformer
d) control signal generator

View Answer

Answer: b [Reason:] Isolation transformer provides the supply to the amplifier and also provides the necessary isolation for the load and triggering circuit.

3. In the thyristor gating circuit, the ZCD is used to
a) amplify the voltage
b) produce a train of pulses
c) convert AC input the ramp voltage
d) used to step-down the voltage

View Answer

Answer: c [Reason:] It is used to convert the AC synchronizing input voltage into ramp voltage & synchronizes it with the zero crossing of the AC supply.

4. The firing-angle delay is
a) inversely proportional to the synchronizing transformer voltage
b) inversely proportional to the control signal voltage
c) directly proportional to the synchronizing transformer voltage
d) directly proportional to the control signal voltage

View Answer

Answer: d [Reason:] If Ec is lowered the firing angle decreases & vice-verse.

5. The pulse gating is not suitable of
a) R loads
b) RC loads
c) RL loads
d) It is suitable of every type of load

View Answer

Answer: c [Reason:] It is not suitable of RL load because initiation of SCR conduction is not well defined in these types of loads.

6. In case of a cosine firing scheme, __________ is used to get a cosine wave
a) ic 555
b) a comparator
c) an integrator circuit
d) a differentiator circuit

View Answer

Answer: c [Reason:] The Sync. Transformer is connected to a integrator to obtain a cosine-wave.

7. If the gating circuits generator negative pulses, then those can be removed by using
a) schmit triggers
b) clippers
c) clampers
d) zener diodes

View Answer

Answer: b [Reason:] The clippers can be used to clip the negative part.

8. The improved version of the UJT oscillator triggering circuit is the
a) ramp & pedal triggering
b) rc triggering
c) cosine-pulse triggering
d) ramp triggering

View Answer

Answer: a [Reason:] The ramp & pedal triggering is the improved version of the UJT oscillator triggering circuit.

9. RB1 = 3 kΩ & RB2 = 6 kΩ. Find the intrinsic stand-off ratio (η) of the UJT.
a) 9
b) 1/3
c) 2/3
d) 3

View Answer

Answer: b [Reason:] η = RB1/(RB1+RB1).

10. The decaying factor in the wave shape of the output pulses from the pulse transformer is its
a) transformer ratio
b) inductance
c) capacitance
d) resistance

View Answer

Answer: b L is the decaying factor in the waveform which emerge from the PT.

Set 5

1. The GTO (gate turn-off thyristor) is a
a) p-n-p-n device
b) p-n-p device
c) p-metal-n device
d) p-n single junction device

View Answer

Answer: a [Reason:] Just like a SCR, the GTO is a four layer p-n-p-n device.

2. The GTO can be turned off
a) by a positive gate pulse
b) by a negative gate pulse
c) by a negative anode-cathode voltage
d) by removing the gate pulse

View Answer

Answer: b [Reason:] The GTO can be turned off by applying a negative gate pulse to the gate terminal.

3. The anode current is ideally limited by the
a) gate pulse amplitude
b) internal impedance of the device
c) load Impedance
d) gate circuit impedance

View Answer

Answer: c [Reason:] The SCR or any device is connected through the load, hence the magnitude of the anode current (same as load current) will depend on the supply voltage and load impedance.

4. In a GTO the n+ layer forms the
a) anode & gate
b) cathode & gate
c) cathode
d) gate

View Answer

Answer: c [Reason:] The bottom n+ layer forms the cathode.

5. The turn-off gain βoff of the GTO is given by
a) Ig/Ia
b) Ia/Ig
c) Vg/Va
d) Vg/Va

View Answer

Answer: b [Reason:] βoff = (anode current/gate current).

6. A GTO can be represented by two transistors T1 & T2. The current gain of both transistors are α1 and α2 respectively. A low value of gate current requires
a) low value of α1 and α2
b) low value of α1 and high value of α2
c) high value of α1 and low value of α2
d) high values of α1 and α2

View Answer

Answer: b [Reason:] In order that the gate current for turning-off the device is low, α2 should be made as nearer to unity as possible whereas α1 should be small.

7. Gold doped GTOs have _____________ as compared to the conventional GTOs
a) high turn-off time
b) low negative gate current requirement
c) low reverse voltage blocking capabilities
d) lower positive gate current requirement

View Answer

Answer: b [Reason:] Gold doping reduces the negative gate current requirements, different kinds of dopings have different advantages over the others.

8. Latching current for the GTOs is ________ as compared to CTs (Conventional thyristors).
a) more
b) less
c) constant
d) cannot be said

View Answer

Answer: a [Reason:] Latching current of GTOs is 2-4A as compared to 200 to 400 mA in case of CT’s.

9. In case of the two-transistor model (T1 & T2) of GTO with anode-short, the anode-short is placed between the
a) emitter of T1 & T2
b) emitter of T1 & base of T2
c) emitter of T1 & base of T1
d) emitter of T1 & collector of T2

View Answer

Answer: c [Reason:] Draw the model. The anode-short resistor is connected between emitter (Anode A of GTO) with base of T1 transistor.

10. Choose the correct statement:
GTOs have _________ as compared to the CTs.
a) less on-state voltage drop
b) less gate drive losses
c) higher reverse blocking capabilities
d) faster switching speed

View Answer

Answer: d [Reason:] GTOs have less turn-on and turn-off time, making it efficient for high frequency applications.