GATE 2025 IN Q.20: solution and answer

GATE 2025 · INQ.20 · 1 mark · MCQ

For the ideal opamp based circuit shown in the figure, the ratio $\frac{V}{I}$ is

GATE 2025 IN question 20: An independent voltage source V (+ on top, − terminal to GND) drives a current I (arrow pointing right, out of the +… — Figure from the GATE 2025 IN paper. It is part of the question.

(A)
$(\frac{R _{3} + R _{4}}{R _{1} + R _{3}}) R_{1}$
(B)
$(\frac{R _{2} + R _{4}}{R _{1} + R _{3}}) R_{3}$
(C)
$R_{1} + R_{3}$
(D)
$R_{3} + R_{4}$

Answer and solution

ANSWER

$\frac{V}{I} = (\frac{R _{3} + R _{4}}{R _{1} + R _{3}}) R_{1}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

Assumptions. Ideal op-amp: no current flows into either input terminal. $R_{2}$ connects the output back to the inverting input, so there is negative feedback, and the two inputs sit at the same voltage (virtual short): $V_{-} = V_{+}$ .
Lower branch (voltage divider). The + input draws no current, so $R_{3}$ and $R_{4}$ carry the same current: $I_{3} = \frac{V}{R _{3} + R _{4}}, V_{+} = \frac{R _{4}}{R _{3} + R _{4}} V$
Virtual short. $V_{-} = V_{+} = \frac{R _{4}}{R _{3} + R _{4}} V$ .
Upper branch. Only the voltage across $R_{1}$ sets the current through it: $I_{1} = \frac{V - V _{-}}{R _{1}} = \frac{V}{R _{1}} (1 - \frac{R _{4}}{R _{3} + R _{4}}) = \frac{R _{3} V}{R _{1} ( R _{3} + R _{4} )}$ No current enters the − input, so all of $I_{1}$ flows on through $R_{2}$ into the op-amp output. $R_{2}$ only sets the output voltage and has no effect on $I$ .
KCL at the input node. $I = I_{1} + I_{3} = \frac{V}{R _{3} + R _{4}} (\frac{R _{3}}{R _{1}} + 1) = \frac{V ( R _{1} + R _{3} )}{R _{1} ( R _{3} + R _{4} )}$
Result. $\frac{V}{I} = (\frac{R _{3} + R _{4}}{R _{1} + R _{3}}) R_{1}$ This is option (A).
Check. Put $R_{4} = 0$ . Then $V_{+} = V_{-} = 0$ , both $R_{1}$ and $R_{3}$ end at (virtual) ground, and the source sees $R_{1} ∥ R_{3} = \frac{R _{1} R _{3}}{R _{1} + R _{3}}$ . The formula gives $(\frac{R _{3}}{R _{1} + R _{3}}) R_{1}$ , the same value.

KEY IDEA

The source feeds two branches. The lower one ( $R_{3}$ , $R_{4}$ ) is a plain divider because the + input draws no current, and it fixes $V_{+}$ . The virtual short copies $V_{+}$ onto the − input, so the current in $R_{1}$ is set by $V - V_{+}$ alone. $I$ is the sum of the two branch currents, and $R_{2}$ never enters it.

WATCH OUT FOR

Two common slips. The first is treating the − input as virtual ground (true only when the + input is grounded), which gives $I_{1} = V / R_{1}$ and the wrong answer. The second is letting $R_{2}$ into $V / I$ (as option (B) does), when all $R_{2}$ does is set the output voltage. Only the lower branch loads the source as a simple series resistance $R_{3} + R_{4}$ , so option (D) misses the current through $R_{1}$ .

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2025 IN Q.20: solution and answer

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