GATE 2026 IN Q.34: solution and answer

GATE 2026 · INQ.34 · 1 mark · NAT

In the circuit below the maximum value of $v_{o u t}$ is ________ V. (rounded off to the nearest integer)

GATE 2026 IN question 34: An AC source 15 cos ω t (sinusoidal source symbol) on the left; — Figure from the GATE 2026 IN paper. It is part of the question.

Answer and solution

ANSWER

The maximum value of $v_{o u t}$ is $11 V$ . It occurs at the source peak of $15 V$ , when the Zener is in breakdown and $1 A$ flows through the $12 Ω$ loop: $v_{o u t} = 3 + 8 (1) = 11 V$ .

Official answer: 11Source: Official answer key, GATE 2026 IN (IIT Guwahati)

Model: take the Zener as ideal. It is an open circuit when reverse biased below $V_{Z} = 3 V$ . In reverse breakdown it holds exactly $3 V$ , and forward biased it is a short (zero forward drop). Its cathode is at the output node and its anode goes to the $8 Ω$ resistor, so it is reverse biased whenever $v_{o u t} > 0$ . Let $v_{s} = 15 cos ω t$ . The source swings through both signs, so both half cycles occur whichever terminal is taken as positive.
Positive half cycle, $0 < v_{s} < 3 V$ : the Zener has not broken down, so the Zener and $8 Ω$ branch carries no current. No current flows in the $4 Ω$ resistor, so it drops nothing and $v_{o u t} = v_{s}$ . The $8 Ω$ resistor also drops nothing, so the whole of $v_{o u t}$ appears across the Zener, and breakdown starts when $v_{s}$ reaches $3 V$ .
Positive half cycle, $v_{s} > 3 V$ : the Zener is in breakdown and acts as a $3 V$ battery in series with the two resistors. KVL around the loop gives the current flowing down through the branch: $i = \frac{v _{s} - 3}{4 + 8} = \frac{v _{s} - 3}{12}$ This is positive for $v_{s} > 3 V$ , which confirms the Zener really is in reverse breakdown.
The output is taken across the Zener and the $8 Ω$ resistor together: $v_{o u t} = 3 + 8 i = 3 + \frac{2}{3} (v_{s} - 3) = \frac{2 v _{s}}{3} + 1$
This rises with $v_{s}$ , so it is largest at the source peak $v_{s} = 15 V$ : $i = \frac{15 - 3}{12} = 1 A, v_{o u t} = 3 + 8 (1) = 11 V$
Negative half cycle: the Zener is forward biased and acts as a short, so the two resistors form a voltage divider: $v_{o u t} = \frac{8}{12} v_{s}$ . Its lowest value is $- 10 V$ , at $v_{s} = - 15 V$ . This half cycle never gives a positive output, and a small forward drop would not change that. So $v_{o u t, m a x} = 11 V$

KEY IDEA

The Zener is in series with the $8 Ω$ resistor and does not sit directly across the output, so it does not clamp $v_{o u t}$ at $3 V$ . Once it breaks down it behaves like a $3 V$ battery. The output is then that $3 V$ plus the drop across the $8 Ω$ resistor, which keeps rising with the source: $v_{o u t} = \frac{2 v _{s}}{3} + 1$ .

WATCH OUT FOR

One mistake is to treat the Zener as a clamp across the output and answer $3 V$ , which ignores the $8 Ω$ drop in series with it. Another is to apply the voltage divider to the whole peak, $\frac{8}{12} \times 15 = 10 V$ , which leaves the Zener's $3 V$ out of the loop. That value is actually the size of the negative-half-cycle extreme, not the positive maximum.

From the GATE 2026 IN paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2026 IN Q.34: solution and answer

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