GATE EE Solved Problems

GATE 2017 Electrical Engineering (EE) Power Electronics (2017)

Solved problems

Author: Prof. Mithun Mondal Subject: Power Electronics Year: 2017 Total Questions: 5
Section 01

1-Mark Questions

QQuestion 1 1 Mark

A thyristor can be turned ON by:

AOptions

  1. Forward voltage breakover
  2. Gate triggering
  3. Temperature rise
  4. All of the above

SSolution

A thyristor can be turned ON by: 1. Exceeding forward breakover voltage 2. Applying positive gate current (normal triggering) 3. Excessive temperature (thermal triggering) 4. High dv/dt (not listed but also possible) Correct answer: D.

Section 02

2-Mark Questions

QQuestion 2 2 Mark

A single-phase full-wave rectifier with center-tapped transformer has an input voltage of 230V RMS. What is the peak inverse voltage (PIV) across each diode?

\begin{figure}[H] \centering

Figure 2.1
Figure 2.1

\end{figure}

AOptions

  1. 230V
  2. 325V
  3. 460V
  4. 650V

SSolution

For a center-tapped transformer, each half has voltage = 230V RMS Peak voltage of each half = \(230 × \sqrt{2} = 325V\) PIV across each diode = \(2 × 325V = 650V\) Correct answer: D.

QQuestion 3 2 Mark

In a PWM inverter, the total harmonic distortion (THD) can be minimized by:

AOptions

  1. Increasing switching frequency
  2. Optimizing switching angles
  3. Using multiple level inverters
  4. All of the above

SSolution

THD in PWM inverters can be reduced by: 1. Higher switching frequency (pushes harmonics to higher frequencies) 2. Optimized switching angles (selective harmonic elimination) 3. Multiple level inverters (more levels = lower THD) All methods are effective for THD reduction. Correct answer: D.

QQuestion 4 2 Mark

The conduction angle of a thyristor in a single-phase half-wave controlled rectifier with resistive load, when the firing angle is 60°, is: \begin{figure}[H] \centering

Figure 4.1
Figure 4.1

\end{figure}

AOptions

  1. 60°
  2. 120°
  3. 180°
  4. 240°

SSolution

For a resistive load, the thyristor conducts from firing angle (60°) until the current naturally becomes zero at 180°. Conduction angle = 180° - 60° = 120° Correct answer: B.

QQuestion 5 2 Mark

In a DC-DC buck converter operating in continuous conduction mode, the relationship between output voltage and input voltage is: \begin{figure}[H] \centering

Figure 5.1
Figure 5.1

\end{figure}

AOptions

  1. \(V_o = V_{in} × D\)
  2. \(V_o = \frac{V_{in}}{D}\)
  3. \(V_o = \frac{V_{in}}{1-D}\)
  4. \(V_o = V_{in} × (1-D)\)

SSolution

In a buck converter, the output voltage is: \(V_o = V_{in} × D\) where D is the duty cycle (ON time/Total time) This relationship holds for continuous conduction mode. Correct answer: A.