ISRO EC 2025 Q.9: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. The virtual short means $v_{+}\approx v_{-}$. Since $v_{+}-v_{-}=v_o/A_{OL}$, this difference is nearly zero only when $v_o$ stays finite (within the supply limits) and $A_{OL}$ is very large.

2. $v_o$ stays within the supply limits only if the output is not saturated. That needs negative feedback, which feeds part of the output back to the inverting input and drives the differential input toward zero, and it needs the op-amp to operate in its linear region.

3. So the condition is negative feedback together with linear-region operation, which is option (c).

4. Option (a) describes an ideal op-amp (infinite input impedance, zero output impedance). It does not guarantee a virtual short: in open loop an ideal op-amp still saturates and $v_{+}\ne v_{-}$.

5. Option (b) is wrong. In open loop the tiny differential input is multiplied by $A_{OL}$ and drives the output into saturation, so there is no virtual ground.

6. Option (d) is wrong. A very low gain makes $v_o/A_{OL}$ non-negligible, so the virtual short fails. A very high gain is needed.