GATE 2025 EE Q.48: solution and answer

GATE 2025 · EEQ.48 · 2 marks · MCQ

In the following circuit, the average voltage $V_{o} = 400 (1 + \frac{c o s α}{3})$ V, where $α$ is the firing angle. If the power dissipated in the resistor is 64 W, then the closest value of $α$ in degrees is

GATE 2025 EE question 48: A three-phase star-connected AC source (three sinusoidal sources joined at a common star point) feeds a box containing a… — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
35.9
(B)
46.4
(C)
41.4
(D)
0

Answer and solution

ANSWER

The firing angle is $α \approx 35. 9^{\circ}$ , option (A).

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

The load is a resistor $R = 1 Ω$ in series with an inductor $L = 100 mH$ and a DC source $E = 500 V$ , with its positive side toward the $V_{o} +$ terminal. So the converter must supply a voltage above $E$ for current to flow.
The large inductance keeps the load current essentially constant, so $I_{r m s} \approx I_{o}$ and the ripple is neglected.
The power in the resistor is $P = I_{o}^{2} R = 64 W$ . With $R = 1 Ω$ , $I_{o} = 64/1 = 8 A$ .
The average voltage across the inductor is zero in steady state. Averaging KVL around the load gives $V_{o} = E + I_{o} R = 500 + 8 \times 1 = 508 V$ .
Set this equal to the given expression: $400 (1 + \frac{cos α}{3}) = 508$ , so $1 + \frac{cos α}{3} = 1.27$ .
Then $cos α = 3 \times 0.27 = 0.81$ , so $α = cos^{- 1} (0.81) \approx 35. 9^{\circ}$ , which is option (A).

KEY IDEA

With a large inductor the load current is nearly constant, so the resistor power gives the current. Averaging KVL around the load, where the inductor drops no average voltage, then gives the converter's average output voltage as $V_{o} = E + I_{o} R$ .

WATCH OUT FOR

Forgetting the 500 V source and writing $V_{o} = I_{o} R = 8 V$ , which gives an impossible $cos α$ . Another slip is to treat the inductor as dropping average voltage.

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GATE 2025 EE Q.48: solution and answer

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