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

Last updated on Jun 10, 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:

A metal oxide varistor (MOV) is used for protecting:

  1. Gate circuit against over currents 
  2. Gate circuit against over voltages.  
  3. Anode circuit against over currents. 
  4. Anode circuit against over voltages. 

Answer (Detailed Solution Below)

Option 1 : Gate circuit against over currents 

Power Electronics and Drives Question 3 Detailed Solution

Explanation:

Metal Oxide Varistor (MOV)

Definition: A Metal Oxide Varistor (MOV) is a type of electronic component used for protecting electrical and electronic circuits from transient voltage spikes. It is a non-linear resistor whose resistance decreases significantly as the voltage across it increases beyond a certain threshold. MOVs are widely utilized in surge protection devices due to their ability to absorb and divert excess energy caused by voltage surges.

Working Principle: MOVs are made of metal oxide grains (usually zinc oxide) embedded in a ceramic matrix. These grains form multiple junctions that exhibit non-linear behavior. When the voltage across the MOV remains below its threshold (clamping voltage), it behaves like a high-resistance component, allowing negligible current to flow through it. However, when the voltage exceeds the threshold, the MOV's resistance drops drastically, allowing it to conduct current and divert the excess energy away from the protected circuit.

Advantages:

  • Effective protection against voltage surges and spikes.
  • Compact design suitable for a wide range of applications.
  • Quick response time to transient events.
  • Cost-effective solution for over-voltage protection.

Disadvantages:

  • Degradation over time due to repeated exposure to voltage surges.
  • Limited energy absorption capacity, requiring careful selection based on application.

Applications: MOVs are commonly used in:

  • Surge protection devices for power distribution systems.
  • Protecting sensitive electronic equipment such as computers, televisions, and telecommunication devices.
  • Industrial equipment and motor control circuits.
  • Gate circuits in power electronics to safeguard them against over-voltage conditions.

Correct Option Analysis:

The correct option is:

Option 1: Gate circuit against over currents.

This option correctly identifies one of the applications of MOVs. In gate circuits, MOVs are specifically used to protect against over-voltage conditions that could damage the gate or associated components. Since MOVs respond quickly to voltage surges, they are ideal for safeguarding delicate electronics in gate circuits.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Gate circuit against over voltages.

This option is incorrect because MOVs are designed to protect circuits against transient voltage spikes, not specifically over currents. Gate circuits in electronic devices can be sensitive to voltage surges, but the MOV's role is primarily focused on handling over-voltage conditions rather than over-current issues.

Option 3: Anode circuit against over currents.

This option is incorrect because MOVs are not typically used for protecting anode circuits against over currents. Anode circuits in power electronics are more likely to be protected using current-limiting devices such as fuses or circuit breakers. MOVs are designed for handling transient voltage spikes rather than current-related issues.

Option 4: Anode circuit against over voltages.

While MOVs can be used for over-voltage protection, this option does not correctly specify the primary application of MOVs in gate circuits, which was the focus of the question. The anode circuit's protection would depend on the specific application and device requirements, but MOVs are more commonly associated with gate circuit protection.

Option 5: None of the above.

This option is incorrect because MOVs are indeed used for protecting gate circuits against over-voltage conditions, as explained above. Therefore, the correct answer cannot be "None of the above."

Conclusion:

Metal Oxide Varistors (MOVs) play a critical role in protecting sensitive electronic circuits from transient voltage spikes. They are widely used in gate circuits to safeguard against over-voltage conditions, ensuring the reliability and longevity of the components. While MOVs are effective for surge protection, their limitations, such as degradation over time and energy capacity, must be considered during the design and selection process. Understanding the applications and working principles of MOVs is essential for their correct usage in various industries and electronic systems.

Power Electronics and Drives Question 4:

To Charge a 120 V battery from a 600 V DC source using a DC chopper, The average battery current should be 20 A, with a peak to peak ripple of 2 A. If the chopper frequency is 200 Hz, calculate the duty cycle.  

  1. 0.2
  2. 0.1
  3. 0.5
  4. 0.6

Answer (Detailed Solution Below)

Option 1 : 0.2

Power Electronics and Drives Question 4 Detailed Solution

Explanation:

Charging a 120 V Battery Using a DC Chopper

Problem Statement: A 120 V battery is to be charged from a 600 V DC source using a DC chopper. The average current supplied to the battery should be 20 A, with a peak-to-peak ripple of 2 A. The chopper operates at a frequency of 200 Hz. The task is to calculate the duty cycle of the chopper.

Solution:

To solve this problem, let us break it into steps:

Step 1: Understanding the DC Chopper Operation

A DC chopper is a power electronics device that steps down or steps up the voltage by controlling the duty cycle (D). The duty cycle is defined as the ratio of the ON time (TON) of the chopper to the total time period (T).

The average output voltage (Vout) of a step-down chopper is given by:

Vout = D × Vin

Where:

  • Vin = Input voltage (600 V in this case).
  • D = Duty cycle (value between 0 and 1).

Rearranging the formula to find the duty cycle:

D = Vout / Vin

Step 2: Calculate the Duty Cycle

Here, the output voltage Vout is equal to the battery voltage, which is 120 V. The input voltage Vin is the source voltage, which is 600 V. Substituting these values into the formula:

D = Vout / Vin

D = 120 / 600

D = 0.2

Thus, the duty cycle of the chopper is 0.2, or 20%.

Step 3: Verify Additional Parameters

The problem also specifies an average battery current of 20 A and a peak-to-peak ripple of 2 A. These parameters are consistent with the design of a DC chopper circuit operating at 200 Hz, where the ripple current is minimized by the use of a suitable inductor in the circuit. However, the calculation of the duty cycle is independent of the current values in this case, as it is determined solely by the voltage ratio.

Conclusion:

The duty cycle of the chopper required to charge the 120 V battery from a 600 V DC source is 0.2 (20%).

Correct Option: Option 1 (0.2)

Additional Information

To further understand the analysis, let’s evaluate why the other options are incorrect:

Option 2 (0.1): This option would imply a much lower duty cycle, resulting in an output voltage of:

Vout = 0.1 × 600 = 60 V

This value is far below the required battery voltage of 120 V, making this option incorrect.

Option 3 (0.5): A duty cycle of 0.5 would result in an output voltage of:

Vout = 0.5 × 600 = 300 V

This value exceeds the battery voltage of 120 V, which would overcharge the battery and potentially damage it. Hence, this option is also incorrect.

Option 4 (0.6): A duty cycle of 0.6 would result in an output voltage of:

Vout = 0.6 × 600 = 360 V

This output voltage is even higher than the one in Option 3, making it completely unsuitable for charging a 120 V battery. Thus, this option is incorrect as well.

Conclusion:

Among the given options, only Option 1 provides the correct duty cycle of 0.2, which ensures the proper output voltage of 120 V for charging the battery. The other options result in either an insufficient or excessive output voltage, making them invalid choices.

Power Electronics and Drives Question 5:

A pulse transformer is used in a driver circuit to:

  1. Prevent DC triggering
  2. Shape trigger signal.  
  3. Generate high frequency pulses. 
  4. Provide isolation. 

Answer (Detailed Solution Below)

Option 1 : Prevent DC triggering

Power Electronics and Drives Question 5 Detailed Solution

Explanation:

Pulse Transformer in Driver Circuits

Definition: A pulse transformer is a type of transformer specifically designed to transmit electrical pulses with minimal distortion and high efficiency. It is widely used in electronic circuits, especially in driver circuits, where it serves critical purposes such as isolation, signal shaping, and prevention of undesired effects like DC triggering.

Working Principle: The pulse transformer operates on the principle of electromagnetic induction. When a pulse is applied to the primary winding, a corresponding pulse is induced in the secondary winding. The core of the transformer is optimized to handle fast-changing signals (high-frequency pulses) while maintaining fidelity and preventing distortion.

Advantages:

  • Provides electrical isolation between the input and output circuits, ensuring safety and preventing interference.
  • Facilitates the transmission of high-frequency signals with minimal distortion.
  • Prevents DC triggering by blocking DC components and allowing only the desired pulses to pass through.
  • Compact and efficient design suitable for various electronic applications.

Disadvantages:

  • Limited to specific applications involving pulse transmission.
  • Requires careful design to avoid saturation and ensure proper operation at high frequencies.

Applications: Pulse transformers are commonly used in driver circuits for thyristors, IGBTs, and other power electronic devices. They are also utilized in digital communication systems, radar circuits, and other applications requiring precise pulse transmission.

Correct Option Analysis:

The correct option is:

Option 1: Prevent DC triggering.

This is the primary reason why pulse transformers are used in driver circuits. DC triggering refers to the undesired activation of the circuit due to direct current components in the signal. Pulse transformers are designed to block DC components, allowing only the intended pulses to pass through. This ensures precise operation of the driver circuit and prevents malfunctions or unintended triggering of the connected device.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Shape trigger signal.

While pulse transformers do contribute to maintaining the integrity of the pulse signal during transmission, their primary function is not to shape the signal. Signal shaping is typically achieved using additional circuit components like resistors, capacitors, and inductors. The pulse transformer mainly focuses on transmitting the pulse efficiently while providing isolation and preventing DC triggering.

Option 3: Generate high-frequency pulses.

This option is incorrect because pulse transformers do not generate pulses themselves; they transmit pulses generated by other components in the circuit. High-frequency pulses are typically generated by oscillators, signal generators, or other electronic circuits, and the pulse transformer ensures their efficient transmission without distortion.

Option 4: Provide isolation.

Providing isolation is indeed one of the secondary benefits of using a pulse transformer. However, the primary purpose in a driver circuit is to prevent DC triggering. Electrical isolation ensures safety and prevents interference between the input and output circuits, but this is not the main reason for employing a pulse transformer in driver circuits.

Conclusion:

The pulse transformer plays a crucial role in driver circuits by preventing DC triggering, ensuring precise operation, and providing isolation between circuits. While it offers additional benefits like maintaining signal integrity and enabling high-frequency transmission, its primary function in this context is to block DC components and allow only the intended pulses to pass through. Understanding the specific roles of pulse transformers helps in designing efficient and reliable electronic circuits.

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:

     ---(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),

 

or, 

or, 3T - 600 = T

Hence, T = 300 µs

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

Hence,

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

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?

  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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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.

     

Analysis:

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

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 =

Average or mean load current is

I0 = V0 / (RL + r) = = 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

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:

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

 and 

 and 

% η = 

% η = 

% 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. 

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

 

Silicon controlled rectifier

Silicon controlled switch

Light activated SCR

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