Power Electronics and Drives MCQ Quiz - Objective Question with Answer for Power Electronics and Drives - Download Free PDF

Last updated on May 14, 2025

Latest Power Electronics and Drives MCQ Objective Questions

Power Electronics and Drives Question 1:

Which of the following is a characteristic of a group drive system?

  1. Minimal energy loss
  2. Low capital cost 
  3. High flexibility in operation
  4. High maintenance cost

Answer (Detailed Solution Below)

Option 2 : Low capital cost 

Power Electronics and Drives Question 1 Detailed Solution

Electrical Drives

Based on their assembly, electric drives can be of the following three types.

Group Drive:

  • A group drive system uses a high-powered motor to drive a common shaft. When several machines are organized on a single shaft and are driven by a single large motor, this system is known as a group drive.
  • A key characteristic of a group drive system is its lower capital cost compared to other drives. This is because a single, larger motor and associated control gear can power multiple machines connected to a common shaft, rather than requiring a separate motor for each machine. 
  • Examples: Food grinding mills, paper mills, etc.
     

Multi-motor Drive:

  • A multi-motor drive system uses multiple electric motors to power different parts of a machine, each motor operating its own mechanism.
  • This allows for independent control and operation of various machine functions, as seen in traveling cranes where separate motors handle hoisting, long travel, and cross-travel. Example: Cranes
     

Individual Drive:

  • In an electric drive system, if an individual machine is fitted with its motor and each operator has complete control over their machine, then it is termed as an individual electric drive.
  • Examples: Drill machine, lath machine, etc.

Power Electronics and Drives Question 2:

Which control method enables four-quadrant operation of induction motors?

  1. Variable Frequency Drive (VFD)
  2. Rotor resistance control 
  3. Stator voltage control 
  4. Direct On-Line (DOL) starting

Answer (Detailed Solution Below)

Option 1 : Variable Frequency Drive (VFD)

Power Electronics and Drives Question 2 Detailed Solution

Variable Frequency Drive (VFD) enables four-quadrant operation of induction motors.

Rotor resistance control, Stator voltage control, and Direct On-Line (DOL) starting are the starting methods of induction motors.

Variable Frequency Drive (VFD)

  • Variable Frequency Drive (VFD) can enable four-quadrant operation of an induction motor. This means the motor can be controlled in both forward and reverse directions, and it can also generate torque, which is essential for regenerative braking or other applications requiring precise motor control. 
  • Quadrant 1: Forward motoring (positive speed, positive torque). 
  • Quadrant 2: Forward braking (positive speed, negative torque). 
  • Quadrant 3: Reverse motoring (negative speed, negative torque). 
  • Quadrant 4: Reverse braking (negative speed, positive torque). 

Power Electronics and Drives Question 3:

In a thyristor _________ is the time required for the forward blocking voltage to fall from 10% of an anode voltage to the on-state voltage drop of the device.

  1. spread time
  2. turn on time
  3. delay time
  4. fall time

Answer (Detailed Solution Below)

Option 1 : spread time

Power Electronics and Drives Question 3 Detailed Solution

Switching Characteristics of SCR

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SCR or Thyristor Turn-on Time: 

Thyristor turn-on time may be defined as the time required by the SCR to change its state from forward blocking mode to forward conduction mode when a gate pulse is applied.

The total turn-on time of SCR comprises three different time intervals:

Delay Time:

  • It is the time between the instant at which the gate current reaches 90% of its final value and the instant at which the anode current reaches 10% of its final value.

 

Rise Time:

  • Rise time is the time taken by the anode current to rise from 10% of its final value to 90% of its final value. 

 

Spread Time:

  • It is the time taken by the anode current to rise from 90% of its final value to 100%. 
  • It is the time required for the forward blocking voltage to fall from 10% of an anode voltage to the on-state voltage drop of the device.

Power Electronics and Drives Question 4:

If the latching current of a thyristor is 4 mA and it is connected in series with a purely inductive load of 0.2 H across a DC voltage of 200 V. What should be the minimum width of the pulse to properly turn on the thyristor?

  1. 8 ms
  2. 4 μs
  3. 8 μs
  4. 4 ms

Answer (Detailed Solution Below)

Option 2 : 4 μs

Power Electronics and Drives Question 4 Detailed Solution

Concept

The current through an inductor is given by:

\(I_L={V_L\over L}\times t\)

\(t={I_L\over V_L}\times L\)

where, t = time period

IL = Latching current

VL = Voltage

L = Inductor

Calculation

Given, IL = 4 mA

VL = 200 V

L = 0.2 H

\(t={4\times 10^{-3}\over 200}\times 0.2\)

t = 4 μs

Power Electronics and Drives Question 5:

A fully controlled single-phase bridge rectifier is feeding a highly inductive load such that the load current is continuous. If the firing angle is set at 120°, then the average load voltage is ________ and the load current is ________.

  1. negative, negative
  2. negative, positive
  3. positive, negative
  4. positive, positive

Answer (Detailed Solution Below)

Option 2 : negative, positive

Power Electronics and Drives Question 5 Detailed Solution

Concept

The average output voltage for a full-bridge rectifier is given by:

\(V_{o(avg)}={2V_m \over \pi}cosα\)

where, Vm = Maximum value of input voltage

cos α = Firing angle

Explanation

Given, α = 120° 

\(V_{o(avg)}={2V_m \over \pi}cos(120)\)

\(V_{o(avg)}={-V_m \over \pi}\)

Since the result is negative, the average load voltage is negative.

Nature of Load Current:

  • The load is highly inductive, meaning the current is continuous (i.e., it does not become zero).
  • Even though the average voltage is negative, the current remains positive due to the energy stored in the inductor.


The correct answer is option 2.

Top Power Electronics and Drives MCQ Objective Questions

For the power semiconductor devices IGBT, MOSFET, Diode and Thyristor, which one of the following statements is TRUE?

  1. All the four are majority carrier devices.
  2. All the four are minority carrier devices.
  3. IGBT and MOSFET are majority carrier devices, whereas Diode and Thyristor are minority carrier devices.
  4. MOSFET is majority carrier device, whereas IGBT, Diode Thyristor are minority carrier devices.

Answer (Detailed Solution Below)

Option 4 : MOSFET is majority carrier device, whereas IGBT, Diode Thyristor are minority carrier devices.

Power Electronics and Drives Question 6 Detailed Solution

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In majority carrier devices conduction is only because of majority carriers whereas in minority carrier devices conduction is due to both majority and minority carriers.

1. MOSFET is a majority carrier device.

2. Diode is both majority and minority carrier device.

3. Thyristor is minority carrier device

4. IGBT is minority carrier device

In a three phase (50Hz) full converter, the ripple frequency in output voltage?

  1. 50 Hz
  2. 100 Hz
  3. 150 Hz
  4. 300 Hz

Answer (Detailed Solution Below)

Option 4 : 300 Hz

Power Electronics and Drives Question 7 Detailed Solution

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Concept:

Ripple frequency at the output = m × supply frequency

fo = m × fs

Where m = types of the pulse converter

Calculation:

A three-phase full-wave AC to DC converter is a 6-pulse converter

Number of pulses (m) = 6

fo = 6 × supply voltage frequency

∴ f= 6 x 50

f0 = 300 Hz

A step-up chopper is fed with 200 V. The conduction time of the thyristor is 200 µs and the required output is 600 V. If the frequency of operation is kept constant and the pulse width is halved, what will be the new output voltage?

  1. 600 volts
  2. 300 volts
  3. 400 volts
  4. 200 volts

Answer (Detailed Solution Below)

Option 2 : 300 volts

Power Electronics and Drives Question 8 Detailed Solution

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Formula:

\(V_o=V_{in}(\frac{T}{T-T_{ON}})\)     ---(1)

Where, Vo is the output voltage

Vin is the input voltage

TON is the pulse width

Application:

Given,

Vin = 200 volts

TON = 200 µs

V0 = 600 V

From equation (1),

 \(\frac{V_o}{V_{in}}=(\frac{T}{T-T_{ON}})\)

or, \(3=\frac{T}{T-200}\)

or, 3T - 600 = T

Hence, T = 300 µs

If the Pulse width is half then, the new value of pulse width (TON') will be,

\(T_{ON'}=\frac{T_{ON}}{2}=\frac{200}{2}=100\ \mu s\)

Hence,

Hence, the new value of output voltage (V0') will be,

\(V_o'=V_{in}(\frac{T}{T-T_{ON'}})=200\times (\frac{300}{300-100})=300\ volts\)

Choppers are _______ converters. 

  1. DC to DC
  2. AC to AC
  3. AC to DC
  4. DC to AC

Answer (Detailed Solution Below)

Option 1 : DC to DC

Power Electronics and Drives Question 9 Detailed Solution

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A chopper is a static device which is used to obtain a variable dc voltage from a constant dc voltage source. Also known as a dc-to-dc converter.

They can step up the DC voltage or step down the DC voltage levels.

Types of Choppers:

  • Type A Chopper or First-Quadrant Chopper
  • Type B Chopper or Second-Quadrant Chopper
  • Type C Chopper or Two-quadrant type-A Chopper
  • Type D Chopper or Two-quadrant type-B Chopper
  • Type E Chopper or fourth-quadrant Chopper

 

Note:

Power electronic circuits can be classified as follows.

1. Diode rectifiers:

  • A diode rectifier circuit converts AC input voltage into a fixed DC voltage.
  • The input voltage may be single phase or three phase.
  • They find use in electric traction, battery charging, electroplating, electrochemical processing, power supplies, welding and UPS systems.

 

2. AC to DC converters (Phase controlled rectifiers):

  • These convert AC voltage to variable DC output voltage.
  • They may be fed from single phase or three phase.
  • These are used in dc drives, metallurgical and chemical industries, excitation systems for synchronous machines.

 

3. DC to DC converters (DC Choppers):

  • A dc chopper converts DC input voltage to a controllable DC output voltage.
  • For lower power circuits, thyristors are replaced by power transistors.
  • Choppers find wide applications in dc drives, subway cars, trolley trucks, battery driven vehicles, etc.

 

4. DC to AC converters (Inverters):

  • An inverter converts fixed DC voltage to a variable AC voltageThe output may be a variable voltage and variable frequency.
  • In inverter circuits, we would like the inverter output to be sinusoidal with magnitude and frequency controllable. In order to produce a sinusoidal output voltage waveform at a desired frequency, a sinusoidal control signal at the desired frequency is compared with a triangular waveform.
  • These find wide use in induction motor and synchronous motor drives, induction heating, UPS, HVDC transmission etc.

 

5. AC to AC converters: These convert fixed AC input voltage into variable AC output voltage. These are two types as given below.

i. AC voltage controllers:

  • These converter circuits convert fixed AC voltage directly to a variable AC voltage at the same frequency.
  • These are widely used for lighting control, speed control of fans, pumps, etc.

 

ii. Cycloconverters:

  • These circuits convert input power at one frequency to output power at a different frequency through a one stage conversion.
  • These are primarily used for slow speed large ac drives like rotary kiln etc.

 

6. Static switches:

  • The power semiconductor devices can operate as static switches or contactors.
  • Depending upon the input supply, the static switches are called ac static switches or dc static switches.

In an SCR with four layers represented as shown in the given figure, which are the layers that are equally doped?

F1 U.B Madhu 29.01.20 D10

  1. P1 and N2
  2. P2 and N2
  3. P1 and N1
  4. N1 and N2

Answer (Detailed Solution Below)

Option 1 : P1 and N2

Power Electronics and Drives Question 10 Detailed Solution

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F1 U.B 20.6.20 Pallavi D10

Construction:

  • The SCR is a four-layer and three-terminal device.
  • The four layers made of P and N layers are arranged alternately such that they form three junctions J1, J2, and J3.
  • These junctions are either alloyed or diffused based on the type of construction.

 

Doping level:

  • The level of doping varies between the different layers of the thyristor.
  • Out of these four layers, the first layer (P1 or P+) and Last layer (N2 or N+) are heavily doped layers.
  • The second layer (N1 or N-) is a lightly doped layer and the third layer (P2 or P+) is a moderately doped layer.
  • The junction J1 is formed by the P+ layer and N- layer.
  • Junction J2 is formed by the N- layer and P+ layer
  • Junction J3 is formed by P+ layer and N+ layer.
  • Thinner layers would mean that the device would break down at lower voltages.

A full-wave rectifier uses 2 diodes. The internal resistance of each diode is 20 Ω. The transformer RMS secondary voltage from centre tap to each end of the secondary is 50 V and the load resistance is 980 Ω. Mean load current will be

  1. 45 A
  2. 4.5 A
  3. 45 mA
  4. 45 μA

Answer (Detailed Solution Below)

Option 3 : 45 mA

Power Electronics and Drives Question 11 Detailed Solution

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Concept:

Center tapped full wave rectifier:

  • The Center tapped full-wave rectifier is a device used to convert the AC input voltage into DC voltage at the output terminals.
  • It employs a transformer with the secondary winding tapped at the center point. And it uses only two diodes, which are connected to the opposite ends of a center-tapped transformer as shown in the figure below.
  • The center tap is usually considered as the ground point or the zero voltage reference point.

     Diagram DMRC

Analysis:

The DC output voltage or average output voltage can be calculated as follows,

\({{\rm{V}}_0} = {{\rm{V}}_{{\rm{dc}}}} = \frac{1}{π }\mathop \smallint \limits_0^π {{\rm{V}}_{\rm{m}}}sin\omega t\;d\omega t\)

\( = \;\frac{{{V_m}}}{π }\left. {\left( { - \cos \omega t} \right)} \right|\begin{array}{*{20}{c}} π \\ 0 \end{array}\)

\( = \frac{{{V_m}}}{π }\left( { - cosπ - \left( { - cos 0^\circ } \right)} \right)\)

\( = \frac{{{V_m}}}{π }\left( { - \left( { - 1} \right) + 1} \right)\)

V0 = 2Vm / π 

Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance RL. Therefore mean load current is given by

I0 = V0 / RL

If the internal resistance of the diode is given in that case mean load current I0 = V0 / (RL + r)

Where r = internal resistance of the diode.

Calculation:

Given that 

Rms value of supply voltage V = 50 V

The internal resistance of diode r = 20 Ω 

The load resistance RL = 980 Ω 

Maximum voltage on the secondary side Vm = √2 V = √2 × 50 = 70.7 V

Average or DC output voltage V0 = \(\frac{(2 × 70.7) }{π} = 45\; V\)

Average or mean load current is

I0 = V0 / (RL + r) = \(\frac{45 }{(980 + 20)} \) = 45 mA

The centre-tapped full wave rectifier's efficiency is ______.

  1. 50%
  2. 40.6%
  3. 81.2%
  4. 45.3%

Answer (Detailed Solution Below)

Option 3 : 81.2%

Power Electronics and Drives Question 12 Detailed Solution

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Full wave rectifier

F3 Vilas Engineering 8.12.2022 D3

Case 1: During +ve half

Do ON and D1 OFF

Vo = Vs

Case 2: During -ve half

Do OFF and D1 ON

Vo = -Vs

The output waveform is:

F3 Vilas Engineering 8.12.2022 D4

The rectification efficiency is the ratio of the DC output power to the AC input power. 

\(V_{o(avg)}={2V_m\over \pi}\) and \(I_{o(avg)}={2V_m\over \pi R}\)

\(V_{o(rms)}={V_m\over \sqrt{2}}\) and \(I_{o(rms)}={V_m\over \sqrt{2}R}\)

% η = \({V_{o(avg)}\times I_{o(avg)}\over V_{o(rms) \times I_{o(rms)}}}\)

% η = \({{2V_m\over \pi}\times{2V_m\over \pi R}\over {V_m\over \sqrt{2}}\times {V_m\over \sqrt{2}R}}\)

% η = 81.2%

Mistake Points The rectification efficiency of half wave rectifier is 40.6%

In a 3 phase semi-converter, firing angle = 120° and extinction angle = 110°. Each SCR and freewheeling diode conducts respectively for

  1. 60°, 50°
  2. 30°, 50°
  3. 60°, 10°
  4. 30°, 40°

Answer (Detailed Solution Below)

Option 1 : 60°, 50°

Power Electronics and Drives Question 13 Detailed Solution

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Concept:

In a three-phase semi-converter,

The conduction period of each thyristor = π – α

The conduction period of freewheeling diode = β – 60°

Where α is the firing angle

β is the extinction angle

Calculation:

Given that, firing angle (α) = 120°

Extinction angle (β) = 110°

The conduction period of each thyristor = π – α = 180 – 120 = 60°

The conduction period of freewheeling diode = β – 60° = 110 – 60 = 50°

What is the relation between latching current and holding current in a thyristor?

  1. Latching current = Holding current
  2. Latching current > Holding current
  3. Latching current < Holding current
  4. Latching current ≥ Holding current

Answer (Detailed Solution Below)

Option 2 : Latching current > Holding current

Power Electronics and Drives Question 14 Detailed Solution

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Key Points Latching Current: It is the minimum anode current required to maintain the thyristor in the ON state immediately after a thyristor has been turned on and the gate signal has been removed.

Holding Current: It is the minimum anode current to maintain the thyristor in the on-state. 

Latching current is always greater than holding current.

Additional Information The thyristor or SCR is a power semiconductor device which is used in power electronic circuits.

They work like a bistable switch and it operates from nonconducting to conducting.

The designing of thyristors can be done with 3-PN junctions and 4 layers.

It includes three terminals namely anode, gate, and cathode. 

F1 Vinanti Engineering 18-11-22 D5

F1 U.B Madhu 29.01.20 D7

Identify the device shown in the given figure.

  1. Silicon unilateral switch
  2. Silicon controlled rectifier
  3. Silicon controlled switch
  4. Light activated SCR

Answer (Detailed Solution Below)

Option 3 : Silicon controlled switch

Power Electronics and Drives Question 15 Detailed Solution

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Device

Circuit symbol

Silicon unilateral switch

F1 U.B 20.6.20 Pallavi D7

 

Silicon controlled rectifier

F1 U.B 20.6.20 Pallavi D8

Silicon controlled switch

F1 U.B Madhu 29.01.20 D7

Light activated SCR

F1 U.B 20.6.20 Pallavi D9

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