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Rectifiers & Converters MCQ [Free PDF] – 11 Objective Questions with Answers

Every Rectifiers & Converters question asked in GPSC DEE Electrical, GPSC AE Electrical, Supervisor Instructor — solve, check instantly, then read the formula and exam tip.

11 PYQs11 subtopics3 examsUpdated 23 Oct 2026
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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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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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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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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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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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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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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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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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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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Three-Phase Full-Wave Converter (1)

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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