Electrical Engineering questions for GATE, PSU recruitment and SSC JE draw from network theory, electrical machines, power systems, control systems, measurements and instrumentation, analog and digital electronics, and electromagnetic fields. Numerical answers include the formula used and the unit at each stage, which is where marks are commonly lost even when the approach is correct.
In a three-phase inverter, 180° conduction mode means:
Answer: A
In 180° conduction, each switch conducts for half the fundamental cycle (180°), typically producing a 6-step output. 120° conduction has each switch on for 120° only.
Q.2Hard
Which of the following has the lowest on-state voltage drop among power semiconductor devices?
Answer: B
MOSFETs have the lowest on-state voltage drop (few hundred millivolts) at moderate currents. IGBTs: ~1-2V, SCRs: ~1-2V, Diodes: ~0.7-1V depending on rating.
Q.3Hard
The snubber circuit in power electronics is used to:
Answer: A
Snubber circuits (RC networks) protect switching devices from transient voltage spikes during switching transitions and reduce switching losses.
Q.4Hard
In a soft-start thyristor converter, the firing angle is gradually increased to:
Answer: A
Soft-start increases firing angle gradually from maximum value, reducing dv/dt and inrush current, protecting equipment and supply system from transients.
Q.5Hard
The turn-off time (toff) of a power transistor includes:
Answer: B
Turn-off time (toff) = Storage time (ts) + Fall time (tf). Storage time is delay before current starts decreasing; fall time is the time for current to reach zero.
Q.6Hard
In a boost converter operating in continuous conduction mode (CCM), the average output voltage relationship is:
Answer: B
Boost converter output voltage: Vo = Vin / (1 - D), where D is duty cycle. This always produces Vo > Vin for 0 < D < 1.
Q.7Hard
The dv/dt rating of a power semiconductor is important because it:
Answer: B
High dv/dt can cause displacement currents through parasitic capacitances, potentially triggering devices unintentionally. dv/dt rating specifies the maximum safe rate of voltage change.
Q.8Hard
In a three-level diode-clamped inverter (Neutral Point Clamped), the advantage over two-level inverter is:
Answer: B
Three-level inverter generates three voltage levels, reducing the step size and dv/dt compared to two-level inverter, thus reducing EMI and stress on motor windings.
Q.9Hard
Which control strategy for a DC-DC converter provides the tightest output voltage regulation under varying input voltage and load conditions?
Answer: B
Closed-loop feedback control with error amplifier and compensation networks provides superior regulation by continuously adjusting duty cycle based on output voltage error.
Q.10Hard
In a synchronous buck converter, the advantages of using a low-side MOSFET instead of a diode are:
Answer: D
Synchronous MOSFETs have lower on-state resistance than diode forward voltage drop, reducing losses. Active switching also allows higher frequency operation with better control.
Q.11Hard
In a forward converter, the energy stored in the magnetizing inductance during the ON time is transferred to:
Answer: C
In a forward converter, when the switch turns OFF, the magnetizing inductance energy is transferred back to the input source through the demagnetization winding
Q.12Hard
The average output voltage of a three-phase half-wave uncontrolled rectifier is:
Answer: A
For a three-phase half-wave uncontrolled rectifier, Vdc = (3√23π) × Vm ≈ 0.827 × Vm
Q.13Hard
In a synchronous rectifier design, the gate drive signal for the MOSFET switch should be synchronized with:
Answer: D
Synchronous rectifier gate drive must be timed with the primary switch to replace the body diode conduction, typically during the primary switch OFF period
Q.14Hard
The phenomenon of 'latch-up' in power semiconductor devices occurs due to:
Answer: B
Latch-up in CMOS and power devices occurs when parasitic p-n-p and n-p-n transistors form a regenerative feedback loop, causing high current and potential device destruction
Q.15Hard
A single-phase half-wave rectifier with a firing angle of 45° is connected to a 230V, 50Hz AC source. Calculate the RMS output voltage.
Answer: D
For half-wave rectifier: Vrms = (Vm/2)√[(π - α + sin(2α))/(2π)] where α = 45° = π/4, Vm = 230√2 = 325.3V. Vrms ≈ 108.6V
Q.16Hard
In a controlled rectifier circuit, the commutation overlap angle (μ) depends on:
Answer: B
Commutation overlap occurs due to finite source impedance. The overlap angle μ = sin⁻¹(ωLₛI/√2×Vline), depending on source inductance and current.
Q.17Hard
Which parameter determines the minimum OFF-time in a PWM converter?
Answer: B
Minimum OFF-time must account for semiconductor recovery characteristics: tOFF(min) = ts + tf, allowing complete device turn-off before the next cycle.
Q.18Hard
For an AC voltage controller with RMS output voltage of 230V from a 230V, 50Hz source at 30° firing angle, what is the power factor?
Answer: A
For AC voltage controller: PF = (1/π)√(π² - 4α²) × sin(2α)/(2α) where α = 30° = π/6. This yields PF ≈ 0.87 with leading reactive current.
Q.19Hard
In a soft-switching zero-voltage switching (ZVS) converter, the main advantage is:
Answer: B
ZVS eliminates capacitive discharge losses during switching, significantly reducing switching losses and high-frequency EMI emissions.
Q.20Hard
A MOSFET with on-state resistance RDS(on)=0.5Ω carries an average current of 50A at 100kHz. What is the approximate conduction loss?
Answer: B
Conduction loss = I²rms × RDS(on). With 50A average current (≈35.4A RMS) and 0.5Ω: Loss ≈ (35.4)² × 0.5 ≈ 625W. At high frequency, peak current factor increases loss to ~1250W.
