ISRO EC 2025 Q.58: solution and answer

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

1. The two equal $10\text{k}\Omega$ resistors divide the $12\text{V}$ supply, so the node feeding the inverting input sits at $V_{-}=12\times \frac{10}{10+10}=6\text{V}$.

2. The op-amp has no feedback, so it works as an open-loop comparator. Its output is $+V_{sat}$ when $V_{+}>V_{-}$ and $-V_{sat}$ (or $0\text{V}$) when $V_{+}<V_{-}$.

3. Here $V_{+}=V_i$ (non-inverting input) and $V_{-}=6\text{V}$.

4. The LED has its anode on the op-amp output (through $R$) and its cathode on ground. It conducts only when the output is positive, i.e. $V_o\approx +V_{sat}$.

5. The output is positive when $V_i>6\text{V}$, so the LED is ON for $V_i$ more than $6\text{V}$.

6. For $V_i<6\text{V}$ the output is negative (or zero), so the LED is reverse biased or off. This rules out option (a).

7. The switching point is the divider voltage, $6\text{V}$, not $12\text{V}$. Option (c) would include $V_i<6\text{V}$, where the LED is OFF, and option (d) would wrongly leave out $6\text{V}<V_i<12\text{V}$, where the LED is also ON.