GATE 2026 EC Q.62: solution and answer

GATE 2026 · ECQ.62 · 2 marks · NAT

Consider an ideal OP-AMP circuit as shown in the Figure.

The resistances $R_{1} = R_{2} = R_{3} = R_{4} = 50 k Ω$ .

The magnitude of the closed loop gain is ____.

(rounded off to two decimal places)

GATE 2026 EC question 62: Inverting amplifier with a T-network in the feedback path. — Figure from the GATE 2026 EC paper. It is part of the question.

Answer and solution

ANSWER

The closed-loop gain is $v_{o} / v_{i} = - 3$ , so its magnitude is 3.00.

Official answer: 2.90 to 3.10Source: Official answer key, GATE 2026 EC (IIT Guwahati)

Assume an ideal op-amp: infinite open-loop gain and no input current. The (+) input is grounded, so negative feedback holds the (−) input at virtual ground, $v_{-} = 0$ .
KCL at the inverting input: no current enters the op-amp, so all of the current through $R_{1}$ continues through $R_{2}$ towards the T-node X: $\frac{v _{i} - 0}{R _{1}} = \frac{0 - v _{X}}{R _{2}} \Rightarrow v_{X} = - \frac{R _{2}}{R _{1}} v_{i} = - v_{i}$
KCL at the T-node X. The current arriving through $R_{2}$ leaves through $R_{4}$ to ground and through $R_{3}$ to the output: $\frac{0 - v _{X}}{R _{2}} = \frac{v _{X}}{R _{4}} + \frac{v _{X} - v _{o}}{R _{3}}$
Solve for $v_{o}$ : $\frac{v _{o}}{R _{3}} = v_{X} (\frac{1}{R _{2}} + \frac{1}{R _{3}} + \frac{1}{R _{4}}) \Rightarrow v_{o} = v_{X} R_{3} (\frac{1}{R _{2}} + \frac{1}{R _{3}} + \frac{1}{R _{4}}) = v_{X} \cdot 50 k Ω \cdot \frac{3}{50 k Ω} = 3 v_{X}$
Substitute $v_{X} = - v_{i}$ to get $v_{o} = - 3 v_{i}$ . Check against the standard T-network result: $\frac{v _{o}}{v _{i}} = - \frac{R _{2} + R _{3} + \frac{R _{2} R _{3}}{R _{4}}}{R _{1}} = - \frac{50 + 50 + \frac{50 \cdot 50}{50}}{50} = - \frac{150}{50} = - 3$
The closed-loop gain is $- 3$ , so its magnitude is $∣ v_{o} / v_{i} ∣ = 3.00$ .

KEY IDEA

This is an inverting amplifier whose feedback resistor has been replaced by a T-network. Virtual ground at the (−) input fixes the voltage at the T-node, $v_{X} = - (R_{2} / R_{1}) v_{i}$ . KCL at the T-node then gives $v_{o}$ . The general result is $v_{o} / v_{i} = - (R_{2} + R_{3} + R_{2} R_{3} / R_{4}) / R_{1}$ : the shunt resistor $R_{4}$ adds the term $R_{2} R_{3} / R_{4}$ , which raises the gain above what $R_{2} + R_{3}$ alone would give.

WATCH OUT FOR

One mistake is to treat the feedback as a simple series resistor $R_{2} + R_{3}$ and ignore $R_{4}$ , which gives a gain of magnitude 2. Another is to treat $R_{4}$ as a load that can be dropped. $R_{4}$ takes current away from the T-node, so the output has to swing further to push the same current through $R_{2}$ . It adds the term $R_{2} R_{3} / R_{4}$ and brings the gain to 3. A third error is to take the voltage at the (−) input as $v_{i}$ , when with the (+) input grounded it is at virtual ground (0 V).

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

GATE 2026 EC Q.62: solution and answer

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