ISRO EC 2025 Q.79: solution and answer

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

1. The $4\text{pF}$ capacitor, the $10\text{mH}$ inductor and the $5\text{M}\Omega$ resistor are in parallel across the output. They are fed from $V_i$ through the series $1.25\text{M}\Omega$ resistor, so the circuit is a parallel resonant circuit.

2. Resonant frequency: $${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10\times 10^{-3}\times 4\times 10^{-12}}}=\frac{1}{2\times 10^{-7}}=5\text{Mrad/s}.$$

3. The damping is set by the total resistance seen by the $L$-$C$ pair. Looking back from those terminals, the ideal source $V_i$ is replaced by a short, so the series resistor appears in parallel with the $5\text{M}\Omega$ resistor: $$R_{eq}=\frac{1.25\times 5}{1.25+5}\text{M}\Omega =1\text{M}\Omega .$$

4. Quality factor of a parallel RLC circuit: $$Q=R_{eq}{\omega }_{0}C=10^6\times 5\times 10^6\times 4\times 10^{-12}=20.$$

5. Bandwidth: $$\beta =\frac{{\omega }_0}{Q}=\frac{5\times 10^6}{20}=250\text{krad/s},$$ which agrees with $\frac{1}{R_{eq}C}=\frac{1}{10^6\times 4\times 10^{-12}}=250\text{krad/s}$.

6. Hence ${\omega }_0=5\text{Mrad/s}$, $\beta =250\text{krad/s}$ and $Q=20$, which is option (b). Using only the $5\text{M}\Omega$ resistor would wrongly give $Q=100$ and $\beta =50\text{krad/s}$.