Electrical Engineering · updated 23 Oct 2026

Power Electronics MCQ – 40 previous year questions with solutions

Every Power Electronics question asked in GPSC DEE Electrical, GPSC AE Electrical, Supervisor Instructor, GPSSB AAE Electrical — arranged topic by topic, with the exam and paper shown on each question.

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Rectifiers & Converters MCQs 11

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Rectifiers & Converters › Battery Charging (1)

Q1Moderate
AC source of 110 Vrms, 60 Hz supplies an ideal single-phase phase-controlled half-wave rectifier that charges a battery. Assume the battery is modelled using a voltage source of 100 V with a series resistance of 2Ω Choose the correct option. Statement-1: Minimum value of control firing angle is α = 40° (approx.) Statement-2: For firing angle, α = 60°, the battery current start falling from its peak value after 30° which is measured from the angle at which current starts rising.
View answer & solution
Correct answer: A. Both statements are TRUE
Quick explanation

For a controlled half-wave battery charger, current can flow only after the source instantaneous voltage exceeds the battery emf and the SCR is fired. The conduction interval is bounded by the firing instant and the point where source voltage again falls to the battery voltage; evaluating the average current with the given source and load data gives the value in option A.

During conduction: i(θ)=[Vm sinθ−E]/R; Vm=√2 Vrms
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Rectifiers & Converters › Commutation Overlap (1)

Q2Moderate
Regarding the effect of source inductance (Ls) in a single-phase rectifier:
1. It causes a "commutation overlap" denoted by angle μ.
2. It reduces the average output voltage.
3. During overlap, the output voltage follows the average of the two commutating phases.
Which of the statements given above are correct?
View answer & solution
Correct answer: B. 1, 2 and 3 only
Quick explanation

Source inductance prevents current from transferring instantaneously between rectifier devices. During the overlap angle μ, outgoing and incoming devices conduct simultaneously, reducing average DC output voltage. Key relation: Overlap introduces a commutation reactance drop; Vdc_actual < ideal Vdc.

Exam focus: More source inductance/current → more overlap. Common trap: Ignoring source inductance overestimates rectifier DC output.

Overlap introduces a commutation reactance drop; Vdcactual < ideal Vdc.
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Rectifiers & Converters › Dual Converters (1)

Q3Moderate
In non-circulating current mode dual converters, the circulating current is avoided by
View answer & solution
Correct answer: C. Operating only one converter
Quick explanation

A non-circulating-current dual converter prevents the two converter bridges from conducting simultaneously. Only one converter is enabled at a time; an interlocking/dead-time scheme blocks the other, so there is no closed path for circulating current.

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Rectifiers & Converters › Freewheeling Diode (1)

Q4Easy
Consider the following statements regarding freewheeling diode:
1. It provides path for inductive load current.
2. It reduces voltage spikes.
3. It improves efficiency.
4. It is used only in AC circuits.
Which of the statements are correct?
View answer & solution
Correct answer: A. 1, 2 and 3 only
Quick explanation

A freewheeling diode gives inductive current a path when the main source/switch is removed. It clamps otherwise large inductive overvoltage, keeps current continuous and can improve efficiency and waveform quality in DC converter/rectifier circuits. Key relation: Inductor relation v=L di/dt explains the voltage spike if current path is interrupted.

Exam focus: Always provide a current path for stored inductive energy. Common trap: Freewheeling diodes are not limited to AC circuits.

Inductor relation v=L di/dt explains the voltage spike if current path is interrupted.
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Rectifiers & Converters › Single-Phase Full Converter (1)

Q5Hard
A single-phase full converter is connected to 230 V, 50 Hz. The load is R = 10 Ω with a very large inductance. For α = 30°, verify.
1. Average output voltage is approximately 179.3 V.
2. The RMS value of the input current is 17.93 A.
3. The fundamental power factor is 0.866.
Which of the statements are correct?
View answer & solution
Correct answer: C. 1, 2 and 3
Quick explanation

Firing angle controls average DC voltage; the AC-side current is non-sinusoidal and its fundamental is displaced approximately by α. Vdc=(2√2×230/π)cos30°≈179.3 V. With R=10 Ω and highly inductive continuous current, Id≈17.93 A. Key relation: 1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.

Exam focus: Large load inductance → nearly constant DC current. Common trap: Distinguish displacement factor cosα from total power factor including distortion.

1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.
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Rectifiers & Converters › Single-Phase Fully Controlled Bridge (1)

Q6Hard
In a single-phase full-wave fully controlled bridge converter with continuous load current, what is the effect on the conduction period of each thyristor as the firing angle increases?
View answer & solution
Correct answer: D. The conduction period of each thyristor remains unchanged.
Quick explanation

D — Remains unchanged. With continuous current each thyristor pair conducts for 180° until the next pair is fired; changing α only shifts the conduction interval, not its length.

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Rectifiers & Converters › Single-Phase Half-Wave Rectifier (1)

Q7Easy
A single-phase half-wave rectifier has input RMS voltage 100 V. Then
1. Peak voltage ≈ 141 V.
2. DC output ≈ 45 V.
3. Output is pulsating DC.
4. Efficiency is 100%.Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1, 2 and 3 only
Quick explanation

An ideal half-wave rectifier passes one half-cycle of a sinusoidal input and blocks the other, creating pulsating DC. Vm=√2×100≈141.4 V and ideal half-wave average Vdc=Vm/π≈45.0 V. Key relation: V_m=√2 V_rms; V_DC=V_m/π; η_max≈40.6%.

Exam focus: Remember π in half-wave average; full-wave uses 2Vm/π. Common trap: Do not confuse diode ideality with 100% rectification efficiency.

Vm=√2 Vrms; VDC=Vm/π; ηmax≈40.6%.
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Rectifiers & Converters › Single-Phase Half-Wave SCR Converter (1)

Q8Moderate
When does conduction occur in a single-phase half-wave SCR (Silicon Controlled Rectifier) converter?
View answer & solution
Correct answer: D. Only during the positive half-cycle after the SCR is triggered
Quick explanation

D — Only during the positive half-cycle after triggering. The SCR is forward-biased only in the positive half-cycle and conducts from the firing angle α until the current falls to zero.

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Rectifiers & Converters › Three-Phase Full Converter (1)

Q9Hard
A three-phase full converter feeds a highly inductive load. The input is 415 V (Line-to-Line), 50 Hz. For α = 60°, verify.
1. The average output voltage is 280 V.
2. The fundamental displacement factor is 0.5.
3. Each thyristor conducts for 6.67 ms.
Which of the statements are correct?
View answer & solution
Correct answer: D. 1, 2 and 3
Quick explanation

In three-phase bridges each thyristor conducts for 120 electrical degrees under continuous current. Vdc≈1.35×415×cos60°≈280.1 V. Displacement factor=cos60°=0.5. Key relation: 1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.

Exam focus: Large load inductance → nearly constant DC current. Common trap: Distinguish displacement factor cosα from total power factor including distortion.

1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.
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Rectifiers & Converters › Three-Phase Full-Wave Converter (1)

Rectifiers & Converters › Three-Pulse Rectifier (1)

Q11Hard
A step-down delta-star transformer, with per-phase turns ration of 5, is fed from a 3-phase, 1100 V, 50 Hz source. The secondary of this transformer through a 3-pulse type rectifier feeds a R load of 10 Ω. Find the maximum value of the load current (phase).
View answer & solution
Correct answer: A. √2 × 22 A
Quick explanation

The primary is delta-connected, so its phase voltage is 1100 V. With a per-phase turns ratio of 5, the star-secondary phase voltage is 1100/5 = 220 V rms. The three-pulse rectifier's maximum load voltage is the phase-voltage peak, so I_max = √2×220/10 = √2×22 A.

V₁,phase = V₁,line = 1100 V (delta) V₂,phase = V₁,phase/5 = 220 V rms V₂,peak = √2 × 220 V Imax = V₂,peak/R = (√2×220)/10 = √2 × 22 A
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Inverters MCQs 9

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Inverters › Pulse Width Modulation (2)

Q12Easy
In the single−pulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.
View answer & solution
Correct answer: B. 3π/2
Quick explanation

For a single-pulse PWM waveform, the negative-half-cycle pulse is symmetrically located about 3π/2, just as the positive-half-cycle pulse is symmetric about π/2. This half-wave symmetry helps define the Fourier components.

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Q13Moderate
In the case of a single−pulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage
View answer & solution
Correct answer: C. nd = π
Quick explanation

For a single pulse of half-width d, the nth harmonic term contains sin(nd). To eliminate the nth harmonic, set sin(nd)=0; the non-trivial condition used in the options is nd = π.

Vn ∝ sin(nd)/n ⇒ nd = mπ Smallest non-zero choice: nd = π
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Inverters › Sinusoidal PWM (2)

Q14Moderate
Regarding Pulse Width Modulation (PWM) in inverters, consider the following:
1. Sinusoidal PWM reduces the lower-order harmonics but increases the switching losses.
2. The modulation index (ma) controls the amplitude of the fundamental output.
3. If ma>1, the inverter enters the "over-modulation" region, increasing harmonic distortion.
Which of the statements given above are correct?
View answer & solution
Correct answer: C. 1, 2 and 3
Quick explanation

In the linear region, modulation index controls the fundamental output amplitude. Raising switching frequency moves dominant harmonics upward but increases switching loss. Key relation: m_a=V_ref,peak/V_carrier,peak; linear SPWM typically m_a≤1.

Exam focus: m_a controls fundamental voltage in the linear region. Common trap: Over-modulation is not a clean linear extension of SPWM.

ma=Vref,peak/Vcarrier,peak; linear SPWM typically ma≤1.
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Q15Moderate
In sinusoidal PWM, the reference signal is compared with what?
View answer & solution
Correct answer: C. A triangular carrier wave
Quick explanation

C — A triangular carrier wave. A sinusoidal reference is compared with a high-frequency triangular carrier; their crossings set the switching instants.

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Inverters › Three-Phase VSI (2)

Q16Moderate
Consider the performance of a 3-phase VSI. Then
1. 180° conduction provides higher fundamental output than 120° conduction.
2. 120° conduction is prone to short-circuiting the DC bus if "dead-time" is not provided.
3. In 180° conduction, the potential of the load neutral is always zero for a balanced star load.
Which of the statements given above is/are correct?
View answer & solution
Correct answer: A. 1 only
Quick explanation

In 180° mode three devices conduct at a time; in 120° mode only two conduct and one phase is disconnected during each step. Waveform/fundamental comparison confirms 1. The 60° same-leg off interval of 120° mode contradicts 2, and switching-state neutral displacement contradicts 3. Key relation: Typical fundamental phase RMS: 180° mode ≈0.4502Vdc; 120° mode ≈0.3898Vdc. Thus 180° gives the higher fundamental component.

Exam focus: 120° mode: two switches ON, one phase floating. 180° mode: three switches ON. Common trap: Assuming a balanced load forces the floating neutral to stay at zero under a non-sinusoidal switching waveform.

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Q17Moderate
In a 3-phase VSI in 120° conduction mode,
1. Each switch conducts for 120° in a cycle.
2. Only two switches conduct at any given time.
3. The phase voltage is zero for 60° in every half-cycle.
Which of the statements given above are correct?
View answer & solution
Correct answer: B. 1, 2 and 3 only
Quick explanation

Three-phase voltage-source inverters can be operated with 120° or 180° conduction sequences. In 120° conduction, each switch is on for 120° and only two devices conduct at a time, leaving one phase disconnected for intervals. Key relation: Conduction duration: 120°→two switches at a time; 180°→three switches at a time.

Exam focus: Track pole voltage and load-neutral voltage separately. Common trap: Balanced load does not force the floating neutral to the DC midpoint at every switching state.

Conduction duration: 120°→two switches at a time; 180°→three switches at a time.
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Inverters › Single-Phase Full-Bridge VSI (1)

Q18Hard
A single-phase full-bridge VSI is fed from a 48 V battery and delivers power to a resistive load of 10 Ω. Using square wave switching, verify
1. RMS value of fundamental output voltage is 43.2 V.
2. Power delivered by the fundamental component is 186.6 W.
3. Total power delivered to the load is 230.4 W.
Which of the statements are correct?
View answer & solution
Correct answer: D. 1, 2 and 3
Quick explanation

In 180° conduction, three devices conduct at a time and the fundamental output is higher; the floating star neutral can shift relative to the DC midpoint. Full-bridge square-wave total RMS voltage is 48 V, so total load power=48²/10=230.4 W. Fundamental RMS=4Vdc/(π√2)≈43.2 V and fundamental power≈43.2²/10=186.6 W.

Exam focus: Track pole voltage and load-neutral voltage separately. Common trap: Balanced load does not force the floating neutral to the DC midpoint at every switching state.

Conduction duration: 120°→two switches at a time; 180°→three switches at a time.
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Inverters › Single-Phase Half-Bridge Inverter (1)

Inverters › Single-Pulse Modulation (1)

Q20Moderate
In single-pulse modulation, what controls the output voltage of the inverter?
View answer & solution
Correct answer: D. Pulse width
Quick explanation

D — Pulse width. Single-pulse modulation uses one controlled pulse per half-cycle; varying its width varies the RMS output voltage.

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Power Semiconductor Devices MCQs 6

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Power Semiconductor Devices › Power MOSFET (2)

Q22Easy
For a power inverter circuit that demands fast switching and high efficiency, which MOSFET type would be most effective?
View answer & solution
Correct answer: D. N-channel enhancement mode MOSFET
Quick explanation

D — N-channel enhancement-mode MOSFET. Among the listed device types, it combines normally-off gate control with the low conduction resistance commonly sought in efficient power switching.

Pcond ≈ ID,rms² × RDS(on), using the MOSFET current RMS over the full switching period. Gate-drive power ≈ Qg × Vdrive × fs.
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Power Semiconductor Devices › Bipolar Junction Transistor (1)

Q23Moderate
How can a power transistor such as a Bipolar Junction Transistor (BJT) be correctly described in terms of its terminal structure, charge-carrier type and mode of control?
View answer & solution
Correct answer: D. A three-terminal device that works with both types of charge carriers and is controlled by current.
Quick explanation

D — Three terminals, both carriers, current-controlled. A BJT has emitter, base and collector, conducts with both electrons and holes (bipolar) and is controlled by base current.

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Power Semiconductor Devices › MOSFET and IGBT (1)

Q24Moderate
Consider the following statements regarding the Power MOSFET and IGBT:
1. MOSFETs have a positive temperature coefficient of resistance, making them easy to parallel.
2. IGBTs exhibit "current tailing" during turn-off, which increases switching losses.
3. MOSFETs are preferred over IGBTs for low-power applications with switching frequencies above 200 kHz.
4. Both devices are current-controlled semiconductor switches.
Which of the statements given above are correct?
View answer & solution
Correct answer: A. 1, 2 and 3 only
Quick explanation

Power MOSFETs are majority-carrier, voltage-controlled devices with fast switching and a positive temperature coefficient of on-resistance that assists current sharing in parallel. IGBTs combine MOS-gate drive with bipolar conduction and exhibit tail current at turn-off, making them attractive at higher voltage/power but usually lower switching frequency than MOSFETs.

Exam focus: High-frequency low/medium voltage → MOSFET; higher voltage/power → often IGBT. Common trap: Both MOSFET and IGBT are not current-controlled devices.

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Power Semiconductor Devices › Reverse Recovery (1)

Q25Moderate
What happens to the current during the reverse recovery time of a power diode?
View answer & solution
Correct answer: A. It flows in the reverse direction for a short time
Quick explanation

A — It flows in reverse for a short time. Stored minority charge must be swept out, so the diode conducts a brief reverse current before it blocks (t_rr).

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Power Semiconductor Devices › Wide-Bandgap Devices (1)

Q26Moderate
Wide−Bandgap (WBG) semiconductors like Gallium Nitride (GaN) are becoming popular in power electronics primarily because they offer _________
View answer & solution
Correct answer: B. Higher efficiency and smaller size for power converters
Quick explanation

GaN wide-bandgap devices can switch at high frequency with low switching/conduction losses, enabling higher-efficiency and much smaller power converters than many conventional silicon implementations.

→ smaller magnetics/passives → higher power density/efficiency
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Thyristors (SCR) MCQs 6

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Thyristors (SCR) › Commutation (1)

Q27Moderate
In power electronics circuit, commutation is the process of
View answer & solution
Correct answer: A. Turning off an electronic switch, which usually involves transferring the load current from one switch to another.
Quick explanation

Commutation is the process of transferring current away from a conducting power semiconductor so that its current falls below the required holding value and the device turns off. In line-commutated circuits the AC source provides this naturally; in forced commutation an auxiliary network produces the required current/voltage condition.

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Thyristors (SCR) › Firing Circuits (Opto-coupler) (1)

Q28Moderate
What is the main function of an opto-coupler in a thyristor firing circuit?
View answer & solution
Correct answer: D. To provide electrical isolation between the control and power circuits
Quick explanation

D — Electrical isolation between control and power circuits. The opto-coupler passes the trigger signal as light, so the low-voltage control circuit is kept electrically separate from the high-voltage power circuit.

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Thyristors (SCR) › GTO Thyristor (1)

Q29Hard
In the case of two-transistor model (T1 & T2) of GTO with anode−short, the anode−short is placed between the
View answer & solution
Correct answer: C. emitter of T1 and base of T1
Quick explanation

In the two-transistor equivalent of a GTO, anode shorts are introduced to alter regenerative action and improve turn-off behavior. The shorting path is represented between the emitter of T1 and the base region associated with T2, corresponding to the device's anode-short structure.

Thyristor regeneration condition: α1 + α2 → 1
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Thyristors (SCR) › LASCR (1)

Q30Moderate
Which illumination method is used to trigger a Light Activated SCR (LASCR)?
View answer & solution
Correct answer: B. Direct light radiation
Quick explanation

B — Direct light radiation. Light falling on the LASCR's gate-sensitive region generates carriers that turn it ON, giving optical isolation — useful in HVDC valves.

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Thyristors (SCR) › SCR Turn-On (1)

Q31Moderate
Regarding the turn-on process of an SCR, consider the following statements:
1. Latching current is the minimum anode current required to maintain the SCR in the ON state after the gate pulse is removed.
2. Holding current is associated with the turn-off process.
3. The gate pulse width should be sufficient for the anode current to rise above the holding current.
4. Latching current is typically 2 to 3 times the holding current.
Which of the statements given above are correct?
View answer & solution
Correct answer: B. 1, 2 and 4 only
Quick explanation

Latching current is the minimum anode current needed just after triggering so the device stays on when gate drive is removed. Holding current is the lower current threshold below which an already-conducting SCR turns off; latching current is normally higher, commonly a few times the holding current.

Exam focus: Gate pulse must persist until anode current exceeds latching current. Common trap: Do not substitute holding current for latching current during turn-on.

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Thyristors (SCR) › Snubber Circuits (1)

Q32Moderate
With reference to Snubber circuits, consider the following:
1. An RC snubber in parallel with an SCR limits the dV/dt stress.
2. A series inductor limits the di/dt during turn-on.
3. The capacitor in a snubber circuit stores energy during the OFF state and dissipates it through the resistor during turn-on.
Which of the statements given above are correct?
View answer & solution
Correct answer: D. 1, 2 and 3
Quick explanation

Thyristor protection separates dv/dt and di/dt problems. An RC snubber across the device slows the rise of off-state voltage and absorbs transient energy; a series inductance limits current rise during turn-on. Key relation: dv/dt protection → RC across switch; di/dt protection → series L.

Exam focus: Match the protection component to the stress being limited. Common trap: A capacitor alone can create severe discharge current without the snubber resistor.

dv/dt protection → RC across switch; di/dt protection → series L.
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AC Voltage Controllers MCQs 4

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AC Voltage Controllers › AC Voltage Controller (1)

Q33Easy
Consider the following statements regarding AC voltage controller:
1. Controls RMS output voltage.
2. Uses phase angle control.
3. Works only with resistive load.
4. Used in light dimmers.
Which of the statements are correct?
View answer & solution
Correct answer: A. 1, 2 and 4 only
Quick explanation

Unlike an inverter, an AC voltage controller does not change AC to a new frequency; it controls the portion of each source cycle applied to the load. With phase control, delaying the firing angle reduces RMS output. Key relation: For a resistive single-phase phase-controlled load, Vrms depends on firing angle α; frequency at the load remains the source frequency.

Exam focus: AC controller → same frequency, variable RMS voltage. Common trap: Confusing an AC voltage controller with a frequency converter/inverter.

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AC Voltage Controllers › Applications (1)

Q34Easy
The AC voltage controllers are used in __________ applications.
View answer & solution
Correct answer: B. Electric heating
Quick explanation

AC voltage controllers vary the rms AC voltage supplied to a load without changing the supply frequency. A classic application is resistive electric heating, where delivered power is smoothly controlled by phase-angle or integral-cycle control.

P = V_rms² / R
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AC Voltage Controllers › Sequence Control (1)

Q35Moderate
In a N-stage sequence controller, each secondary is rated for __________
View answer & solution
Correct answer: C. Vs/n
Quick explanation

In an N-stage sequence controller, the supply voltage is divided across the staged secondary arrangement; each secondary is designed for a fraction of the total controlled voltage. The keyed relation is Vs/n per stage.

Voltage per equal stage = Vs / n
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AC Voltage Controllers › Thyristor Ratings (1)

Q36Easy
A single−phase AC switch is used in between a 230 V source and load of 2 kW and 0.8 lagging power factor. Determine the rms current rating required by the SCR. Use the factor of safety = 2.
View answer & solution
Correct answer: C. 21.74 A
Quick explanation

The load current is I = P/(V·pf) = 2000/(230×0.8) ≈ 10.87 A. With a safety factor of 2, the required SCR rms current rating is about 21.74 A.

Iload = P/(V cosφ) = 2000/(230×0.8) = 10.87 A Irating = 2 × Iload = 21.74 A
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Choppers (DC-DC Converters) MCQs 2

Choppers (DC-DC Converters) › Boost Converter (2)

Q37Easy
For a DC-DC Boost Converter, consider the following:
1. It can produce an output voltage lower than the input if the duty cycle is less than 0.5.
2. The switch is subjected to the full output voltage when it is OFF.
3. The output capacitor reduces voltage ripple.
Which of the statements given above is/are correct?
View answer & solution
Correct answer: C. 2 and 3 only
Quick explanation

An ideal boost converter in continuous conduction has an output voltage greater than or equal to input for 0≤D<1. The output capacitor supplies load current between energy-transfer intervals and reduces output ripple. Key relation: V_o=V_in/(1−D); ΔI_L≈V_in D/(Lf_s).

Exam focus: For an ideal boost, D=1−Vin/Vo. Common trap: A normal boost converter cannot produce Vo<Vin merely because D<0.5.

Vo=Vin/(1−D); ΔI_L≈Vin D/(Lfs).
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Q38Hard
A DC-DC Boost converter has Vin = 10 V, L = 100 μH, f = 50 kHz. It delivers 25 W at Vo = 25 V. Verify
1. The duty cycle D is 0.6.
2. The peak-to-peak inductor current ripple is 1.2 A.
3. The average inductor current is 2.5 A.
Which of the statements are correct?
View answer & solution
Correct answer: C. 1, 2 and 3
Quick explanation

An ideal boost converter in continuous conduction has an output voltage greater than or equal to input for 0≤D<1. D=1−Vin/Vo=1−10/25=0.6. ΔIL=VinD/(Lf)=10×0.6/(100 μH×50 kHz)=1.2 A. Ideal average input/inductor current=P/Vin=25/10=2.5 A. Key relation: V_o=V_in/(1−D); ΔI_L≈V_in D/(Lf_s).

Exam focus: For an ideal boost, D=1−Vin/Vo. Common trap: A normal boost converter cannot produce Vo<Vin merely because D<0.5.

Vo=Vin/(1−D); ΔI_L≈Vin D/(Lfs).
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Cycloconverters MCQs 1

Cycloconverters › Single-Phase Cycloconverters (1)

Q39Easy
The single−phase mid-point type cycloconverter uses __________ number of SCRs.
View answer & solution
Correct answer: A. 4
Quick explanation

A single-phase midpoint cycloconverter uses two midpoint full-wave controlled converter groups—one positive and one negative. Each group requires two SCRs, giving a total of four SCRs.

SCR count = 2 SCRs/group × 2 groups = 4
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SMPS MCQs 1

SMPS › SMPS Characteristics (1)

Q40Moderate
Choose the incorrect statement.
View answer & solution
Correct answer: C. SMPS has low input ripple
Quick explanation

SMPS units are compact and relatively tolerant of input-voltage variation, but their switching action can inject high-frequency ripple/noise and electromagnetic interference. Therefore 'SMPS has low input ripple' is the incorrect statement among the given choices.

Switching loss/EMI are controlled using high-frequency filters and layout. SMPS size ∝ inverse of practical switching frequency for magnetics, broadly.
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Power Electronics MCQ FAQs

How many Power Electronics MCQs are on SarkariResultOut?

40 previous year Power Electronics questions across 8 topics, each with the exam it was asked in, the verified answer and a step-by-step solution.

Which exams asked these Power Electronics questions?

GPSC DEE Electrical · DEE Master paper · 2026; Supervisor Instructor · Part B Technical · 2026; GPSC AE Electrical · AE R&B · 2025 · Series M; GPSC DEE Electrical · DEE Electrical · 2024; GPSSB AAE Electrical · CBRT 2026 · Technical.

Can I download Power Electronics MCQs as a PDF?

Yes. Tap “Download Power Electronics PDF” — it is free after a quick sign-in and includes every answer and solution.