GATE 2026 EE Q.27: solution and answer

GATE 2026 · EEQ.27 · 1 mark · MCQ

The figure shows a straight-line approximation for the forward characteristics of a power diode. A continuous on-state current of 15 A is flowing through the diode. What is the power loss in the diode?

GATE 2026 EE question 27: Plot with horizontal axis 'On-state voltage (V)' and vertical axis 'On-state current (A)', origin 0. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
32.8 W
(B)
21.2 W
(C)
18.6 W
(D)
23.1 W

Answer and solution

ANSWER

The power loss in the diode is 18.6 W, option (C).

Official answer: CSource: Official answer key, GATE 2026 EE (IIT Guwahati)

The straight-line model of a power diode is a threshold voltage in series with a resistance: $v = V_{0} + r_{d} i$ .
From the graph the line leaves the voltage axis at $V_{0} = 1.0 V$ (where the current is zero).
The line passes through $(1.8 V, 50 A)$ (the dashed lines meet on it), so $r_{d} = \frac{1.8 - 1.0}{50} = 0.016 Ω$ .
At $I = 15 A$ the on-state voltage is $v = 1.0 + 0.016 \times 15 = 1.24 V$ .
The current is continuous (dc), so the instantaneous and average loss are the same: $P = v I = 1.24 \times 15 = 18.6 W$ .
Check with the split form: $P = V_{0} I + r_{d} I^{2} = 1.0 \times 15 + 0.016 \times 225 = 15 + 3.6 = 18.6 W$ . This matches option (C).

KEY IDEA

Replace the diode by a threshold voltage $V_{0}$ in series with a slope resistance $r_{d}$ read from the straight-line characteristic. For continuous current $I$ , the loss is $P = V_{0} I + r_{d} I^{2}$ .

WATCH OUT FOR

Using the 1.8 V point as the on-state voltage at 15 A (giving 27 W), or taking $r_{d} = 1.8/50$ with no threshold voltage. The line does not pass through the origin. The slope is measured from the 1.0 V intercept, not from zero.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 9 Switches questions, with notes

GATE 2026 EE Q.27: solution and answer

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