Select Page
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. Dual converters provide
d) none of the mentioned

Answer: c [Reason:] Dual converters provide four quadrent operation, which means voltage can be positive or negative and so can be the current. Hence, AC-DC, DC-AC any converter configuration can be used.

2. A dual converters has
a) two full converters in series
b) two half converters in series
c) two full converters in anti-parallel
d) two half converters in anti-parallel

Answer: c [Reason:] Dual converters have two full converters connected in anti-parallel which provides a four quadrant operation.

3. The major advantage of using dual converters is that
a) it is cheaply available
b) it has better pf
c) no mechanical switch is required to change the mode of operation
d) its operating frequency is very high

Answer: c [Reason:] No mechanical arrangement is required to change from inverter to converter and converter to inverter, which was required in earlier methods.

4. The four quadrant operation of dual converters can be obtained by
a) moving the mechanical lever
b) adding inductance to the circuit
c) changing the firing angle value
d) none of the mentioned

Answer: c [Reason:] The four quadrant operation can be obtained simply by adjusting appropriate values of firing angles for both the connected converters.

5. A single full converter alone can given a
d) none of the mentioned

Answer: c [Reason:] A single full converter alone gives two quadrent operation, hence for all four quadrant operation two full converter circuits are connected in anti-parallel.

6. Find the error in the below given dual converter circuit.

i) Load is not connected in the right position
ii) Only 4 SCRs must be used
iii) Voltage source is not connected for one of the converter circuit
iv) Voltage source is not connected in the proper place

a) All 4
b) Both (i) and (iv)
c) Both (iii) and (iv)
d) Both (ii) and (iii)

Answer: c [Reason:] The right connection for single-phase dual converter is shown below.

7. In the below given circuit, the right side converter C2 operates in the ___ and ___ quadrant.

a) second, fourth
b) first, fourth
c) second, third
d) first, third

Answer: c [Reason:] The C2 converter will supply the load current in direction opposite to that supplied by the converter C1. For converter C2, when α> 90 it operates in 2nd quadrant and if α<90 both current and voltage are negative, C2 is in inverter mode and operates in 3rd quadrant.

8. Name the below given circuit.

a) Single-phase dual converter circulating current type
b) Single-phase dual converter non-circulating current type
c) Three-phase dual converter non-circulating current type
d) Three-phase dual converter circulating current type

Answer: b [Reason:] The circuit is a single phase dual converter circuit. As the there is no reactor (inductor) in series, it is a non-circulating type.

9. For a single-phase dual converter, with converters C1 and C2 connected in anti-parallel, which relation among the following is true to keep the average voltages from C1 and C2 equal? C1 and C2 have firing angles α1 and α2 respectively.
a) α1 = α2
b) α1 + α2 = 360°
c) α1 + α2 = 180°
d) none of the mentioned

Answer: c [Reason:] By maintaining α1 + α2 = 180°, one converter can operate as converter and another as an inverter hence, the average output voltages are equal . This can be proved as follows Vom cos α1 = Vom cos α2 Vom = 2Vm/π . . (Vm for both the converters is same) cos α1 = cos α2 cos α1 = cos (180 – α2) α1 + α2 = 180°.

10. In non-circulating current mode dual converters, the circulating current is avoided by
a) connecting a series reactor
b) maintaining α1 + α2 = 180°
c) operating only one converter

Answer: c [Reason:] Reactor is added in circulating current mode not in non-circulating mode. The circulating current is avoided by using only one of the converters.

## Set 2

1. In circulating current mode dual converters, the circulating current is avoided by
a) connecting a series reactor
b) maintaining α1 + α2 = 180°
c) operating only one converter

Answer: a [Reason:] Reactor is added in circulating current mode not in non-circulating mode. The reactor limits the current to a reasonable value.

2. Choose the correct statement
a) Circulating current exists only in circulating current mode
b) Circulating current exists only in non-circulating current mode
c) Circulating current exists in both the circulating and non-circulating current mode
d) none of the mentioned

Answer: c [Reason:] The circulating current does exist in both the converter circuits, but it is avoided by using a reactor in non-circulating type and by operating only one circuit in case of circulating type.

3. What causes circulating current in dual converters?
a) Temperature issues
c) Out of phase voltages from both the converters
d) none of the mentioned

Answer: c [Reason:] In case of practical dual converters, the voltages from both the converter circuits though equal in magnitude is out of phase. This indifference in voltages causes circulating currents to flow.

4. Name the below given circuit.

a) Single-phase dual converter with circulating current type
b) Single-phase dual converter with non-circulating current type
c) Three-phase dual converter with non-circulating current type
d) Three-phase dual converter with circulating current type

Answer: c [Reason:] The circuit is a three phase dual converter circuit. As the there is no reactor (inductor) in series, it is a non-circulating type.

5. In case of three-phase dual converter, one of the converter circuits is fired at an angle of 60°. For both the converter circuits to have equal average output voltage, what is the value of the firing angle for the other converter circuit?
a) 60°
b) 120°
c) 100°
d) Insufficient data

Answer: b [Reason:] For equal average output voltage, α1 + α2 = 180°.

6. In case of circulating current type dual converters, the reactor is inserted between
a) supply and converter
c) between the converters
d) no reactor is used in case of circulating type dual converter

Answer: c [Reason:] Reactor (inductance) is added in circulating current mode between both the converters. The reactor limits the current to a reasonable value.

7. Choose the correct statement.
a) Circulating current type is faster in operation
b) Non-circulating current type is faster in operation
c) Both the types have the same speed of operation
d) Circulating current improves power factor

Answer: a [Reason:] In case of non-circulating type, to shift the operation from one mode to another mode a delay of 10 to 20msec is required to let the current decay to zero value and let the outgoing SCRs safely turn off. This delay is not required in case of circulating current mode.

8. Circulating current flows
b) from one converter to another converter
c) in the whole circuit
d) none of the mentioned

Answer: b [Reason:] The circulating current flows only between the converters and not to through the load.

9. The reactor in circulating current type dual converters
a) increases losses
b) reduces power factor
c) increase the weight of the circuit
d) all of the above

Answer: d [Reason:] All of the above mentioned are the major drawbacks of using reactors to reduce circulation current.

10. If V1 and V2 are the instantaneous voltages of the two converter circuits in the dual convert, then the output voltage is
a) V1 + V2
b) (V1 + V2)/2
c) V1 – V2
d) 2(V1 + V2)

Answer: b [Reason:] The load voltage is the average value of the instantaneous converter outputs.

## Set 3

1. For the circuit shown in the figure below,

C = 4 pF
L = 16 μH
Vs = 200V
The capacitor voltage after the SCR is self commutated is
a) -100 V
b) -200 V
c) -400 V
d) 0 V

Answer: b [Reason:] It is simply the negative of Vs as it is a series circuit & after turn off the whole voltage appears across the thyristor.

2. For the circuit shown below,

The maximum value of current through thyristors T1 & TA can be given by
a) Vs/R, Vs√C/L
b) Vs √C/L, Vs/R
c) Vs/R + LC, Vs
d) Vs/LC, Vs/R

Answer: a [Reason:] I1 = (Vs/R) T2 = (Vs√C/L).

3. For the circuit shown in the figure below,

C = 4 pF
L = 16 μH
Vs = 200 V
The peak thyristor current is
a) 800 A
b) 400 A
c) 100 A
d) 50 A

Answer: c [Reason:] The peak SCR current is Vs/2.

4. For the circuit shown in the figure below, ___ type of commutation would take place for the SCR (THY1)

a) line
c) forced
d) external-pulse

Answer: c [Reason:] The given figure is that of a class B commutation or resonant-pulse commutation, which is a type of forced commutation technique.

5. For the circuit shown in the figure below,

R1 = 50 Ω, R2 = 100 Ω, Vs = 100 V. At the time of turn-on of the SCR, initial thyristor current is
a) 0 A
b) 10 A
c) 5 A
d) 3.5 A

Answer: c [Reason:] Io = Vs x [(2/R1)+(1/R2)].

6. Natural commutation of an SCR takes place when
a) voltage across the device becomes negative
b) voltage across the device becomes positive
c) gate current becomes zero
d) anode current becomes zero

Answer: d [Reason:] Anode current (load current) becomes zero and turns off the device, hence the name line commutation.

7. ___________ commutation is usually used in phase-controlled rectifiers
a) line
c) forced
d) external-pulse

Answer: a [Reason:] Line commuataion is used in converters.

8. Parallel-capacitor commutation is
a) line commutation
c) forced commutation
d) external-pulse commutation

Answer: c [Reason:] Parallel capacitor is another name for forced commutation.

9. Class E commutation is a/an
a) line commutation technique
c) forced commutation technique
d) external-pulse commutation technique

Answer: d [Reason:] As an external source is used it is a external-pulse commutation technique.

10. Below is the circuit of a resonant-pulse commutation, the maximum voltage to which the capacitor C charges is

a) Vs
b) Vs/2
c) 2
d) √2Vs

Answer: a [Reason:] The capacitor charges from 0 to Vs with the right side plate positive.

## Set 4

1. A motor load is connected to a single-phase full converter B-2 type controlled rectifier. The net energy is transferred from ac source to the motor (dc load) when
a) π+α > 90
b) π-α > α
c) π+α > α
d) π-α > 90

Answer: b [Reason:] Converter will work as a line commuted inverter when π-α > α.

2. The below shown rectifier configuration has continues load current, find the expression for the average value of output voltage when the supply Vs = Vm sinωt is connected.

a) (Vm/π)cosα
b) (2Vm/π)cosα
c) (Vm/π) (1+cosα)
d) (2Vm/π) (1+cosα)

Answer: b [Reason:] Vo = 1/π x [∫ Vm sinωt d(ωt)] Where the integration runs from α to π+α.

3. Find the expression of the rms value of output voltage for a single-phase M-2 type rectifier with RL load and continues load current. Transformer ratio is 1:1 with supply voltage Vs = Vm sinωt
a) Vm/√2
b) Vs
c) 2Vs
d) Vs/√2

Answer: b [Reason:] Vor2 = 1/π x [∫ Vm2 sin2ωt d(ωt)], Where the integration runs from α to π+α Vor = Vm2/2 = Vs.

4. A single-phase full controlled converted with RLE load will act like a line-commutated inverter when the firing angle α
a) α > 180°
b) α > 90°
c) α < 90°
d) α = 90°

Answer: b [Reason:] It will act like a inverter i.e. most of the current would flow from the battery or back emf E to the source. For α=90° it will not act like a converter nor an inverter.

5. In converter operation, with output voltage = Vo and RLE load.
a) Vo < E
b) Vo = E
c) Vo > E
d) None of the mentioned

Answer: c [Reason:] The output voltage obtained by a converter is always greater than the counter or back emf E. Vo = IR + E.

6. In inverter operation, with output voltage = Vo and a RLE load connected
a) Vo < E
b) Vo = E
c) Vo > E
d) None of the mentioned

Answer: a [Reason:] In inverter operation Vo < E.

7. A motor load is connected to a single-phase full converter B-2 type controlled rectifier, the net energy is transferred from ac source to the motor (dc load) when
a) π+α > 90
b) π-α < α
c) π+α < α
d) π-α > α

Answer: b [Reason:] It works as a converter when π-α < α.

8. The below shown rectifier configuration has continues load current, find the expression of the RMS value of output voltage when the supply Vs = Vm sinωt
a) Vs2/2
b) 2Vs/π
c) Vs
d) √Vs/2

Answer: c [Reason:] Vor2 = 1/π x [∫Vm2 sin2ωt d(ωt)] Where the integration runs from α to π+α Vor = Vm2/2 = Vs.

9. Choose the correct statement
a) M-2 type connection requires SCRs with higher PIV as compared to those in a B-2 type
b) M-2 type connection requires SCRs with lower PIV as compared to those in a B-2 type
c) The average output voltage in M-2 type is more than that obtained from a B-2 type configuration of the same rating
d) The average output voltage in M-2 type is less than that obtained from a B-2 type configuration of the same rating

Answer: a [Reason:] The PIV of diodes in M-2 is 2Vm, whereas that in B-2 type of connection is Vm.

10. An SCR has the peak forward voltage = 1000 V. Find the maximum voltage that the SCR can handle if employed in a M-2 type full controlled converter circuit. Use factor of safety (FOS) = 2.5
a) 500 V
b) 400 V
c) 200 V
d) 1000 V

Answer: c [Reason:] In M-2 type configuration maximum voltage handled is 2Vm Therefore, 1000/2×2.5 = 200 V.

## Set 5

1. In a single pulse semi-converter using two SCRs, the triggering circuit must produce
a) two firing pulses in each half cycle
b) one firing pulse in each half cycle
c) three firing pulses in each cycle
d) one firing pulse in each cycle

Answer: b [Reason:] A single phase semi-converter has only two SCRs & two diodes. Hence, only two pulses are required in each cycle, one in each half.

2. In a 3-phase full converter using six SCRs, gating circuit must provide
a) one firing pulse every 30°
b) one firing pulse every 90°
c) one firing pulse every 60°
d) three firing pulses per cycle

Answer: c [Reason:] 60° x 6(devices) = 360°.

3. In the complete firing circuit, the driver circuit consists of
a) pulse generator & power supply
b) gate leads & power supply
c) pulse amplifier & pulse transformer
d) pulse detector & pulse amplifier

Answer: c [Reason:] The driver circuit consists of a pulse amplifier to increase the magnitude of the gate pulse to a sufficient value. The pulse transformer then provides pulses to individual SCRs.

4. Find the average gate power dissipation (Pgav) when the maximum allowable gate power dissipation (Pgm) = 10 kW, with a duty cycle = 50 %.
a) 10 KW
b) 5 KW
c) 2.5 KW
d) 7 KW

Answer: b [Reason:] (Pgm) = (Pgav)/Duty Cycle.

5. The magnitude of gate voltage and gate current for triggering an SCR is
a) inversely proportional to the temperature
b) directly proportional to the temperature
c) inversely proportional to the anode current requirement
d) directly proportional to the anode current requirement

Answer: a [Reason:] Higher the temperature lesser will be the gate current required as the temperature must have already excited some of the atoms.

6. Find the amplitude of the gate current pulse, when the gate-cathode curve is given by the relation Vg = [(1+10) x Ig]
The peak gate drive power is 5 Watts.
a) 359mA
b) 659mA
c) 1.359 A
d) 1.659 A

Answer: b [Reason:] (1+10 Ig).Ig = 5 Watts Ig = 0.59 A.

7. The gate-cathode curve for an SCR is given by the relation Vg = (1+10)Ig. The gate voltage source is a rectangular pulse of peak value 15 V and current = 0.659 A. Find the source resistance.

a) 90.2 Ω
b) 11.24 Ω
c) 46.2 Ω
d) 39 Ω

Answer: b [Reason:] Es = Rs.Ig + Vg Vg = 1+10 Ig Therefore Rs = (15-1)/0.659.

8. Find the triggering frequency when the average gate power dissipation = 0.3 W and the peak gate drive power is 5 Watts. The gate source has a pulse width of 20 μsec duration.
a) 3 kHz
b) 0.3 kHz
c) 30 kHz
d) 0.03 mHz

Answer: a [Reason:] (Pgm) = (Pgav)/Duty Cycle Duty Cycle = f x T = (Pgav)/(Pgm) Duty Cycle = 0.3/5 T = 20 μsec. 0.3/5 = f x T f = (0.3)/(5 x 20 x 10-6) = 3000 Hz.

9. The duty cycle can be written as
a) f x T
b) f/T
c) T/f
d) f