GATE 2026 IN Q.16: solution and answer

Official answer: DSource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. Reading the circuit: $R$ is the series arm. $V_{out}$ is taken, with no load, across the shunt branch, which is $L$ in series with $C$. The circuit is therefore a voltage divider with $Z_{sh}=j\omega L+\frac{1}{j\omega C}$.

2. Voltage divider. Multiplying the numerator and the denominator by $j\omega C$: $$H(j\omega )=\frac{V_{out}}{V_{in}}=\frac{j\omega L+\frac{1}{j\omega C}}{R+j\omega L+\frac{1}{j\omega C}}=\frac{1-{\omega }^{2}LC}{1-{\omega }^{2}LC+j\omega RC}$$ so $$|H(j\omega )|=\frac{\left|1-{\omega }^{2}LC\right|}{\sqrt{{\left(1-{\omega }^{2}LC\right)}^2+{\left(\omega RC\right)}^2}}$$

3. Low-frequency limit, $\omega \to 0$: $C$ behaves as an open circuit, so the shunt branch carries no current. No current flows through $R$, it drops no voltage, and $V_{out}=V_{in}$. The formula gives the same result: $|H|\to \frac{1}{1}=1$.

4. High-frequency limit, $\omega \to \infty$: $L$ behaves as an open circuit, so again no current flows through $R$ and $V_{out}=V_{in}$. From the formula, the ${\omega }^{2}LC$ terms dominate the numerator and the denominator, so $|H|\to 1$.

5. At series resonance, ${\omega }_0=\frac{1}{\sqrt{LC}}$: $j{\omega }_{0}L+\frac{1}{j{\omega }_{0}C}=0$, so the $L$-$C$ branch is a short circuit across the output and $|H(j{\omega }_0)|=0$.

6. The gain is 1 at both ends of the frequency range and falls to 0 in a band around ${\omega }_0$, which is a notch. The circuit is therefore a band stop filter, option (D).