GATE 2025 EE Q.17: solution and answer

Official answer: DSource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Assumptions: both op-amps are ideal (infinite open-loop gain, infinite input resistance, zero output resistance). The circuit has no resistors, capacitors or diodes, only wires.

2. Op-amp 2: its output is wired straight back to its own inverting input, which is 100% negative feedback. It is therefore a unity-gain voltage follower. By the virtual short $v_{-}=v_{+}$, its output is $v_{o2}=v_{+2}=V_{out}$.

3. Loop polarity: $V_{out}$ passes through the follower (gain $+1$) and arrives at the inverting input of op-amp 1. The overall loop is therefore negative feedback, so the virtual-short rule applies to op-amp 1 as well.

4. Op-amp 1 inputs: $v_{+1}=V_{in}$ and $v_{-1}=v_{o2}=V_{out}$.

5. Virtual short at op-amp 1's inputs: $v_{-1}=v_{+1}$, so $$V_{out}=V_{in}\Rightarrow \frac{V_{out}}{V_{in}}=1$$

6. The gain is unity with no phase inversion. The input resistance is very high, because the input drives only the $+$ terminal of an ideal op-amp, which draws no current. The output resistance is zero (the ideal op-amp output). A unity-gain stage with these properties is a buffer. Answer (D).