GATE 2026 IN Q.45: solution and answer

GATE 2026 · INQ.45 · 2 marks · MCQ

In the circuit shown below, $R$ is $1 k Ω$ and $R_{f}$ is $10 k Ω$ . If $V_{in}$ is $100 mV$ and the op-amps have supply voltages of $\pm 15 V$ , then $V_{o u t}$ is ________ V.

GATE 2026 IN question 45: Two op-amps, one above the other: the input stage of an instrumentation amplifier. — Figure from the GATE 2026 IN paper. It is part of the question.

(A)
$- 15$
(B)
$15$
(C)
$2.1$
(D)
$1.1$

Answer and solution

ANSWER

$V_{o u t} = V_{in} (1 + \frac{2 R _{f}}{R}) = 21 \times 100 mV = 2.1 V$ , option (C).

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

Assumptions: both op-amps are ideal (infinite open-loop gain, zero input current). Each has negative feedback through its own $R_{f}$ to its inverting input, so each works in its linear region with a virtual short between its inputs. Call the input terminal voltages $V_{1}$ (top, +) and $V_{2}$ (bottom, −), so $V_{in} = V_{1} - V_{2}$ . Call the outputs $V_{o 1}$ (top) and $V_{o 2}$ (bottom), so $V_{o u t} = V_{o 1} - V_{o 2}$ .
Virtual short at each op-amp: the top end of $R$ is at $V_{1}$ and the bottom end of $R$ is at $V_{2}$ . So the whole input difference appears across $R$ .
Ohm's law on $R$ : $I = \frac{V _{1} - V _{2}}{R} = \frac{V _{in}}{R}$ flowing downward through $R$ .
No current enters either op-amp input, so the same current $I$ flows through both $R_{f}$ resistors. It comes out of the top output through the upper $R_{f}$ , passes down through $R$ , and flows through the lower $R_{f}$ into the bottom output.
Add up the voltage drops from the top output to the bottom output through $R_{f}$ , $R$ and $R_{f}$ in series: $V_{o u t} = V_{o 1} - V_{o 2} = I (R_{f} + R + R_{f}) = V_{in} (1 + \frac{2 R _{f}}{R})$
Differential gain: $1 + \frac{2 ( 10 k Ω )}{1 k Ω} = 21$ , so $V_{o u t} = 21 \times 100 mV = 2.1 V$ .
Saturation check: $I = 100 mV /1 k Ω = 0.1 mA$ , so $I R_{f} = 1 V$ . This gives $V_{o 1} = V_{1} + 1 V$ and $V_{o 2} = V_{2} - 1 V$ . Each output sits only about $1 V$ away from its own input terminal. The question gives no input common-mode level, so take the usual reading that it is well inside the rails. Then both outputs stay far from $\pm 15 V$ , both op-amps stay linear, and $V_{o u t} = 2.1 V$ , option (C). Options (A) and (B) would need an op-amp to saturate, and that does not happen here.

KEY IDEA

This is the input stage of the three-op-amp instrumentation amplifier. The two virtual shorts force the full input difference across $R$ . Since no current enters either op-amp input, the same current $V_{in} / R$ flows through the string $R_{f}$ – $R$ – $R_{f}$ between the two outputs. That makes the differential gain $1 + 2 R_{f} / R$ .

WATCH OUT FOR

Using the single non-inverting gain $1 + R_{f} / R = 11$ gives $1.1 V$ , option (D). That forgets the second $R_{f}$ , which carries the same current. Another mistake is to see the $\pm 15 V$ supply and assume the output saturates, which leads to (A) or (B). In fact each output moves only about $1 V$ from its input, so both op-amps stay linear.

From the GATE 2026 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2026 IN Q.45: solution and answer

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