ISRO EC 2025 Q.55: solution and answer

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

1. The op-amp is ideal and the feedback path returns to the inverting input, so the $+$ input being grounded makes the inverting input a virtual ground: $V_{-}=0$.

2. $I_1=\frac{1\text{V}-0}{10\text{k}\Omega }=100\mu \text{A}$.

3. An ideal op-amp draws no input current, so all of $I_1$ continues through the $10\text{k}\Omega$ feedback resistor: $I_2=I_1=100\mu \text{A}$.

4. The voltage at the junction after the feedback resistor is $V_x=0-I_2\times 10\text{k}\Omega =-100\mu \text{A}\times 10\text{k}\Omega =-1\text{V}$.

5. The $100\Omega$ resistor connects this junction to ground. The arrow of $I_3$ points up, from ground into the junction, so $I_3=\frac{0-(-1\text{V})}{100\Omega }=10\text{mA}$.

6. Check: at the junction, $I_2+I_3=10.1\text{mA}$ leaves through the $1\text{k}\Omega$ resistor toward the output, so $V_o=V_x-10.1\text{mA}\times 1\text{k}\Omega =-1-10.1=-11.1\text{V}$. This is a finite output, so the ideal op-amp can supply it, and the result is consistent.

7. So $I_1=100\mu \text{A}$, $I_2=100\mu \text{A}$, $I_3=10\text{mA}$, which is option (a).