GATE 2026 EC Q.39: solution and answer

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Angular frequency: $\omega =2\pi f=2\pi \times 159.15\times 10^3\text{Hz}\approx 9.9997\times 10^5\approx 10^6\text{rad/s}$. So $X_L=\omega L=10^6\times 10^{-6}=1\Omega$ and $X_C=\frac{1}{\omega C}=\frac{1}{10^6\times 10^{-6}}=1\Omega$.

2. RC branch: $I_C=\frac{V}{R_C-j{X}_C}$. The denominator has angle $-{\tan }^{-1}(X_C/R_C)$, so $I_C$ leads $V$ by ${\theta }_1$ with $\tan {\theta }_1=\frac{X_C}{R_C}$, where $0^{\circ }<{\theta }_1<90^{\circ }$.

3. RL branch: $I_L=\frac{V}{R_L+j{X}_L}$. The denominator has angle $+{\tan }^{-1}(X_L/R_L)$, so $I_L$ lags $V$ by ${\theta }_2$ with $\tan {\theta }_2=\frac{X_L}{R_L}$, where $0^{\circ }<{\theta }_2<90^{\circ }$.

4. Quadrature condition: ${\theta }_1+{\theta }_2=90^{\circ }$ gives $\tan {\theta }_1=\tan (90^{\circ }-{\theta }_2)=\cot {\theta }_2$, i.e. $\tan {\theta }_1\tan {\theta }_2=1$.

5. Substitute: $\frac{X_C}{R_C}\cdot \frac{X_L}{R_L}=1\Rightarrow R_C{R}_L=X_L{X}_C=\omega L\cdot \frac{1}{\omega C}=\frac{L}{C}$.

6. $$R_C{R}_L=\frac{L}{C}=\frac{1\times 10^{-6}\text{H}}{1\times 10^{-6}\text{F}}=1{\Omega }^2$$ The frequency cancels, so the rounding of $\omega$ in step 1 does not affect the result: $R_C{R}_L=1$ at any frequency. The nearest integer is 1, option (B).