GATE 2026 EC Q.53: solution and answer

GATE 2026 · ECQ.53 · 2 marks · MCQ

A circuit using an ideal OP-AMP is shown in the Figure.

Which of the following options gives the correct value of the current $I_{X}$ ?

GATE 2026 EC question 53: Ideal op-amp with its non-inverting (+) input grounded. — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
2.0 mA
(B)
1.5 mA
(C)
3.0 mA
(D)
0 mA

Answer and solution

ANSWER

$I_{X} = 3.0 mA$ , flowing from node X into the op-amp output. The answer is (C).

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

Assume an ideal op-amp with negative feedback. The output drives node X through the $2 k Ω$ resistor, and node X feeds back to the inverting input through the $1 k Ω$ resistor. If $V_{o}$ rises, $V_{X}$ rises and so does $V_{-}$ , which pulls $V_{o}$ back down, so the feedback is negative. That gives a virtual short, $V_{-} = V_{+} = 0 V$ , and no current flows into either op-amp input.
Current in the input resistor (Ohm's law): $I_{1} = \frac{V _{i} - V _{-}}{1 k Ω} = \frac{2 - 0}{1 k Ω} = 2 mA$ , flowing toward the inverting node.
No current enters the op-amp input, so the same $2 mA$ flows on through the feedback $1 k Ω$ resistor, from the inverting node into node X. Therefore $V_{X} = V_{-} - (2 mA) (1 k Ω) = 0 - 2 = - 2 V$ .
Current in the grounded $2 k Ω$ resistor: $\frac{0 - V _{X}}{2 k Ω} = \frac{0 - ( - 2 )}{2 k Ω} = 1 mA$ , flowing from ground into node X.
KCL at node X. The arrow shows $I_{X}$ leaving X toward the op-amp output, so the two currents entering X must both leave as $I_{X}$ : $I_{X} = 2 mA + 1 mA = 3 mA$
Check: $V_{o} = V_{X} - I_{X} (2 k Ω) = - 2 - (3 mA) (2 k Ω) = - 8 V$ . That is a finite output, so the ideal op-amp stays in its linear region and sinks $3 mA$ . The answer is (C).

KEY IDEA

With a virtual ground at the inverting input, the input current $V_{i} /1 k Ω = 2 mA$ has to pass through the feedback $1 k Ω$ resistor, which puts node X at $- 2 V$ . The op-amp output then has to sink this $2 mA$ as well as the $1 mA$ drawn from ground through the $2 k Ω$ load (KCL at X).

WATCH OUT FOR

Picking 2.0 mA by assuming the output resistor carries only the feedback current, which ignores the $1 mA$ in the grounded $2 k Ω$ resistor at node X. Another mistake is to treat the circuit as a plain inverting amplifier with gain $- 1$ , so that $V_{X} = - 2 V$ is taken to be the output voltage. The op-amp output is actually on the far side of the $2 k Ω$ resistor, at $- 8 V$ .

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.53: solution and answer

Answer and solution

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