GATE 2026 EE Q.37: solution and answer

GATE 2026 · EEQ.37 · 2 marks · MCQ

In the circuit shown, the open loop gain of the operational amplifier is $A_{0} = 105$ .

What is the voltage gain of the circuit?

(Round off to two decimal places)

GATE 2026 EE question 37: Inverting amplifier. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$- 16.67$
(B)
$- 20.00$
(C)
$- 21.00$
(D)
$- 12.67$

Answer and solution

ANSWER

The voltage gain is $V_{o u t} / V_{in} = - 20/ (1 + 21/105) = - 16.67$ , so the answer is option (A).

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

Model. The op-amp has a finite open-loop gain $A_{0} = 105$ (printed as 105, not $1 0^{5}$ ), with $R_{in} = \infty$ and $R_{o u t} = 0$ . So no current flows into either input, and $V_{o u t} = A_{0} (v_{+} - v_{-})$ exactly. Because $A_{0}$ is finite, there is no virtual short: $v_{-}$ is small but not zero.
Input relation. The non-inverting input is grounded, so $v_{+} = 0$ . Call the inverting-input voltage $v_{x}$ . Then $V_{o u t} = A_{0} (0 - v_{x})$ , which gives $v_{x} = - \frac{V _{o u t}}{A _{0}}$ .
KCL at the inverting node. $R_{in} = \infty$ , so all the current through the 5 kΩ resistor also flows through the 100 kΩ resistor: $\frac{V _{in} - v _{x}}{5 k Ω} = \frac{v _{x} - V _{o u t}}{100 k Ω}$
Simplify. Multiply both sides by $100 k Ω$ : $20 (V_{in} - v_{x}) = v_{x} - V_{o u t}$ , so $20 V_{in} = 21 v_{x} - V_{o u t}$ . Substitute $v_{x} = - V_{o u t} / A_{0}$ : $20 V_{in} = - V_{o u t} (1 + \frac{21}{A _{0}})$
Closed-loop gain. $\frac{V _{o u t}}{V _{in}} = \frac{- 20}{1 + \frac{21}{A _{0}}} = \frac{- 20}{1 + \frac{21}{105}} = \frac{- 20}{1.2} = - 16.67$
Cross-check with feedback theory. The feedback factor is $β = \frac{5}{5 + 100} = \frac{1}{21}$ , so the loop gain is $A_{0} β = 105/21 = 5$ . The ideal gain is $- R_{f} / R_{1} = - 20$ . Scaling it by $\frac{A _{0} β}{1 + A _{0} β} = \frac{5}{6}$ gives $- 20 \times \frac{5}{6} = - 16.67$ , the same result.
Saturation check. $V_{o u t} = - 16.67 \times 0.1 V = - 1.667 V$ . That is well inside the $\pm 15 V$ rails, so the op-amp stays linear and the gain holds. (The inverting input sits at $v_{x} = 1.667/105 \approx 15.9 mV$ , which is clearly not a virtual ground.) Answer: (A).

KEY IDEA

When the open-loop gain is finite, drop the virtual-ground assumption and use $V_{o u t} = A_{0} (v_{+} - v_{-})$ together with KCL at the inverting node. For an inverting amplifier this gives $A_{C L} = \frac{- R _{f} / R _{1}}{1 + ( 1 + R _{f} / R _{1} ) / A _{0}}$ . The noise gain $1 + R_{f} / R_{1} = 21$ (that is, $1/ β$ ) is what gets divided by $A_{0}$ .

WATCH OUT FOR

Assuming an ideal op-amp (virtual ground) and writing $- R_{f} / R_{1} = - 20.00$ , which is option (B). This is the answer you also get if you read the printed 105 as $1 0^{5}$ , since $- 20/ (1 + 21 \times 1 0^{- 5}) \approx - 20.00$ . The paper prints 105, and with a loop gain of only $A_{0} β = 5$ the finite gain matters. Another slip is using the non-inverting formula $- (1 + R_{f} / R_{1}) = - 21.00$ (option C), which mixes up the noise gain with the signal gain.

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

GATE 2026 EE Q.37: solution and answer

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