GATE 2025 Electrical Engineering (EE) Power Electronics (2025)

SSolution

Given:

• \(v(t) = 300 \sin(\omega t)\) V
• \(i(t) = 10 \sin(\omega t - \frac{\pi}{6}) + 2 \sin(3\omega t + \frac{\pi}{6}) + \sin(5\omega t + \frac{\pi}{2})\) A
• \(\omega = 2\pi \times 50\) rad/s

Solution:

The power factor is defined as:

Step 1: Identify voltage and current components

Voltage (only fundamental):

Current (fundamental + harmonics): \begin{align*} i(t) &= 10 \sin(\omega t - \frac{\pi}{6}) \text{(fundamental)}\\ &+ 2 \sin(3\omega t + \frac{\pi}{6}) \text{(3rd harmonic)}\\ &+ \sin(5\omega t + \frac{\pi}{2}) \text{(5th harmonic)} \end{align*}

Step 2: Calculate RMS current

where: \begin{align*} I_{1,rms} &= \frac{10}{\sqrt{2}} = 7.071 \text{ A}\\ I_{3,rms} &= \frac{2}{\sqrt{2}} = 1.414 \text{ A}\\ I_{5,rms} &= \frac{1}{\sqrt{2}} = 0.707 \text{ A} \end{align*}

Step 3: Calculate real power

Only the fundamental components contribute to real power:

where \(\phi_1\) is the phase angle between fundamental voltage and fundamental current.

Voltage phase: \(0°\)\\ Fundamental current phase: \(-30°\) (or \(-\frac{\pi}{6}\))

Step 4: Calculate apparent power

Step 5: Calculate power factor

Alternative approach:

Power factor can also be expressed as:

This is called displacement power factor × distortion factor.

Correct answer: A (0.845)

SSolution

Given:

• Duty cycle: \(D = 0.4\)
• Current source: \(I = 10\) A
• DC source: \(V = 15\) V
• All devices are ideal

Circuit Analysis:

The current source drives 10 A into the upper node. From that node there are exactly two paths: down through the MOSFET, or to the right through the diode into the 15 V source. Since the source current is constant, one of the two carries all 10 A at every instant.

MOSFET ON (\(DT = 0.4T\))

The MOSFET shorts the upper node to the return rail, so the diode sees \(-15\) V and blocks. The full 10 A circulates through the MOSFET and the voltage across the current source is

MOSFET OFF (\((1 - D)T = 0.6T\))

The 10 A must now flow through the diode into the 15 V source, which clamps the upper node:

Average power delivered by the current source

The ideal MOSFET and diode dissipate nothing, so all 90 W is absorbed by the 15 V source.

Answer: 90 W

SSolution

Given:

• Modulating waveforms: \(v_a(t) = 0.8 \sin(\omega t)\) V (and similar for phases b, c)
• Fundamental frequency: \(f = 40\) Hz
• Carrier frequency: \(f_c = 10\) kHz
• Carrier magnitude: \(\pm 1\) V
• DC link voltage: \(V_{dc} = 600\) V
• Load: Star connected motor

Solution:

Step 1: Determine modulation index

The modulation index (amplitude modulation index) is:

Peak of modulating signal: \(V_{m,peak} = 0.8\) V\\ Peak of carrier signal: \(V_{c,peak} = 1\) V

Step 2: Output voltage for sinusoidal PWM

For a 3-phase VSI with sinusoidal PWM, the peak fundamental phase voltage (line-to-neutral) is:

This is for linear modulation region where \(m_a \leq 1\).

Step 3: Calculate RMS phase voltage

The RMS value of the fundamental component:

Verification:

For a 3-phase VSI with sinusoidal PWM: - DC link voltage: \(V_{dc} = 600\) V - Modulation index: \(m_a = 0.8\) - RMS phase voltage: \(V_{ph,rms} = \frac{m_a \times V_{dc}}{2\sqrt{2}} = \frac{0.8 \times 600}{2\sqrt{2}} = \frac{480}{2.828} = 169.71\) V

Note on other options:

• 300.00 V would be \(V_{dc}/2\)
• 424.26 V would be \(V_{dc}/\sqrt{2}\) (line-to-line RMS)
• 212.13 V would be \(V_{dc}/(2\sqrt{2})\) without modulation index

Correct answer: A (169.71 V)

SSolution

Given:

• Voltage source: \(v(t) = 230\sqrt{2} \sin(2\pi \times 50t)\) V
• Inductance: \(L = 100\) mH \(= 0.1\) H
• Firing angle: \(\alpha = 0°\)
• Ideal SCR and ideal inductor

Solution:

Step 1: Identify voltage parameters

where: - Peak voltage: \(V_m = 230\sqrt{2} = 325.27\) V - Angular frequency: \(\omega = 2\pi \times 50 = 314.16\) rad/s - Frequency: \(f = 50\) Hz

Step 2: Current through inductor

For an inductor:

Integrating:

Step 3: Apply initial conditions

With SCR firing at \(\alpha = 0°\): - SCR turns ON at \(\omega t = 0\) - At \(t = 0\): \(i(0) = 0\) (current through inductor cannot change instantaneously)

Therefore:

Step 4: Find peak current

The maximum value of \(i(t)\) occurs when \(\cos(\omega t) = -1\) (at \(\omega t = \pi\)):

Alternative approach using inductive reactance:

Peak current would be:

But this gives the steady-state AC amplitude, not the peak with DC offset.

With firing angle \(\alpha = 0°\) and pure inductive load, the current will have a DC offset, and the peak reaches:

Correct answer: A (20.71 A)

SSolution

Given:

• Average output voltage: \(V_o = 400\left(1 + \frac{\cos\alpha}{3}\right)\) V
• Power dissipated in resistor: \(P = 64\) W
• Circuit: Controlled rectifier with resistive load

Solution:

Step 1: Load current from the resistor loss

The load is \(1\) \(\Omega\) in series with 100 mH and a 500 V battery. Since \(\omega L = 2\pi \times 50 \times 0.1 = 31.4\) \(\Omega \gg R = 1\) \(\Omega\), the load current is essentially ripple free, so

Step 2: Average KVL around the load

Averaged over a cycle the inductor voltage is zero, so

Step 3: Solve for the firing angle

The larger angles are ruled out: they give \(V_o < 500\) V, which would need current to flow out of the battery back through the converter, and thyristors cannot conduct in that direction.

Correct answer: A (35.9°)

SSolution

Given:

• DC-DC Buck converter in CCM (Continuous Conduction Mode)
• Input voltage: \(V_{in} = 30\) V
• Capacitor current waveform is provided (steady state)
• Need to find: Inductance \(L\) in mH

Step 1: Read the waveform

The capacitor current is the ac part of the inductor current, \(i_C = i_L - I_o\). It ramps up from \(-0.1\) A at \(t = 0\) to \(+0.1\) A at \(t = 30\) μs and back down to \(-0.1\) A at \(t = 50\) μs; the zero crossings at 15 μs and 40 μs are the midpoints of the two ramps. Hence

Step 2: Output voltage

Step 3: Inductance from the ripple

During the on-time the inductor sees \(V_{in} - V_o\):

The off-time gives the same figure: \(L = V_o t_{off}/\Delta i_L = 18 \times 20 \times 10^{-6}/0.2 = 1.8\) mH.

Answer: 1.8 mH