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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
Power Electronics and Drives Question 3 Detailed Solution
Switching Characteristics of SCR
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?
Answer (Detailed Solution Below)
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 ________.
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
Power Electronics and Drives Question 6 Detailed Solution
Download Solution PDFIn 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 deviceIn a three phase (50Hz) full converter, the ripple frequency in output voltage?
Answer (Detailed Solution Below)
Power Electronics and Drives Question 7 Detailed Solution
Download Solution PDFConcept:
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
∴ f0 = 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?
Answer (Detailed Solution Below)
Power Electronics and Drives Question 8 Detailed Solution
Download Solution PDFFormula:
\(V_o=V_{in}(\frac{T}{T-T_{ON}})\) ---(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.
Answer (Detailed Solution Below)
Power Electronics and Drives Question 9 Detailed Solution
Download Solution PDFA 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 voltage. The 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?
Answer (Detailed Solution Below)
Power Electronics and Drives Question 10 Detailed Solution
Download Solution PDFConstruction:
- 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
Answer (Detailed Solution Below)
Power Electronics and Drives Question 11 Detailed Solution
Download Solution PDFConcept:
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,
\({{\rm{V}}_0} = {{\rm{V}}_{{\rm{dc}}}} = \frac{1}{π }\mathop \smallint \limits_0^π {{\rm{V}}_{\rm{m}}}sin\omega t\;d\omega t\)
\( = \;\frac{{{V_m}}}{π }\left. {\left( { - \cos \omega t} \right)} \right|\begin{array}{*{20}{c}} π \\ 0 \end{array}\)
\( = \frac{{{V_m}}}{π }\left( { - cosπ - \left( { - cos 0^\circ } \right)} \right)\)
\( = \frac{{{V_m}}}{π }\left( { - \left( { - 1} \right) + 1} \right)\)
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 ______.
Answer (Detailed Solution Below)
Power Electronics and Drives Question 12 Detailed Solution
Download Solution PDFFull 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.
\(V_{o(avg)}={2V_m\over \pi}\) 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
Answer (Detailed Solution Below)
Power Electronics and Drives Question 13 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Power Electronics and Drives Question 14 Detailed Solution
Download Solution PDFHolding 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.
Answer (Detailed Solution Below)
Power Electronics and Drives Question 15 Detailed Solution
Download Solution PDF
Device |
Circuit symbol |
Silicon unilateral switch |
|
Silicon controlled rectifier |
|
Silicon controlled switch |
|
Light activated SCR |
|