Home Subjects Electrical Engg (EEE) Power Electronics

Electrical Engg (EEE)
Power Electronics

Electrical machines, power systems, circuits

100 Q 7 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 1–10 of 100
Topics in Electrical Engg (EEE)
Q.1 Medium Power Electronics
In a boost converter operating in continuous conduction mode (CCM), if the input voltage is 24V, output voltage is 48V, and switching frequency is 100kHz, what is the duty cycle of the converter?
A 0.25
B 0.5
C 0.75
D 0.9
Correct Answer:  B. 0.5
EXPLANATION

For a boost converter, Vo/Vin = 1/(1-D), where D is duty cycle. 48/24 = 1/(1-D) → 2 = 1/(1-D) → 1-D = 0.5 → D = 0.5 or 50%

Test
In isolated DC-DC converters for EV fast-charging stations, which topology provides the best isolation and efficiency above 1kW?
A Flyback converter
B Push-pull converter
C Full-bridge phase-shift PWM converter
D Half-bridge converter
Correct Answer:  C. Full-bridge phase-shift PWM converter
EXPLANATION

Full-bridge phase-shift topology offers high efficiency (>97%), ZVS operation, and high isolation ratio suitable for high-power fast-charging applications.

Test
Q.3 Medium Power Electronics
A three-phase PWM rectifier with active front-end reduces grid distortion to THD < 5%. This is primarily achieved by:
A Using passive LC filters only
B Real-time current injection control and modulation
C Increasing switching frequency alone
D Reducing the load current
Correct Answer:  B. Real-time current injection control and modulation
EXPLANATION

Active front-end rectifiers use closed-loop current control with PWM modulation to inject harmonics that cancel grid distortion.

Test
For an industrial induction motor variable frequency drive (VFD) using vector control, what is the typical output frequency range in 2024-25 applications?
A 5-50Hz
B 0-400Hz
C 0.1-500Hz
D 1-1000Hz
Correct Answer:  C. 0.1-500Hz
EXPLANATION

Modern VFDs support 0.1Hz to 500Hz+ output, enabling ultra-low speed precision applications and extended-speed motor operation.

Test
A resonant DC-link inverter operates at 200kHz. The resonant frequency must be:
A Equal to switching frequency
B Half the switching frequency
C Twice the switching frequency
D Independent of switching frequency
Correct Answer:  A. Equal to switching frequency
EXPLANATION

In resonant DC-link inverters, the resonant frequency is designed to equal or track the switching frequency for zero-voltage switching.

Test
Q.6 Medium Power Electronics
In a matrix converter (direct AC-to-AC), how many bi-directional switches are required for three-phase input and three-phase output?
A 6
B 9
C 12
D 18
Correct Answer:  B. 9
EXPLANATION

Matrix converter requires m×n switches where m=output phases and n=input phases. For 3×3: 9 bi-directional switches.

Test
Q.7 Medium Power Electronics
Which semiconductor device combination is used in a cycloconverter to convert AC-to-AC with frequency reduction?
A Diodes and resistors
B Back-to-back thyristor pairs
C IGBTs with freewheeling diodes
D MOSFETs and capacitors
Correct Answer:  B. Back-to-back thyristor pairs
EXPLANATION

Cycloconverters use back-to-back (antiparallel) thyristor pairs to enable bidirectional power flow for frequency conversion.

Test
A single-phase AC voltage controller with a resistive load has a firing angle of 45°. What is the RMS output voltage as a fraction of input RMS voltage?
A 0.571
B 0.642
C 0.707
D 0.866
Correct Answer:  B. 0.642
EXPLANATION

For AC controller: Vo/Vi = √[(π-α+sin(2α)/2)/π] with α=45°≈0.785rad gives ≈0.642.

Test
A diode-clamped five-level inverter reduces the voltage stress on individual switches compared to a two-level inverter by a factor of:
A 2
B 2.5
C 4
D 5
Correct Answer:  A. 2
EXPLANATION

In a five-level inverter, maximum voltage across each switch is Vdc/4, compared to Vdc/2 in a two-level inverter, reducing stress by half.

Test
In a synchronous buck converter, if the duty cycle is 0.7 and the switching frequency is 500kHz, what is the approximate inductor ripple current (ΔIL) with input voltage 48V and output voltage 24V?
A 1.2A
B 2.4A
C 3.6A
D 4.8A
Correct Answer:  B. 2.4A
EXPLANATION

ΔIL = (Vin × D × (1-D))/(L × f). Assuming standard inductor ~10μH: ΔIL ≈ (48 × 0.7 × 0.3)/(10e-6 × 500e3) ≈ 2.4A.

Test
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