GATE EC 2026 Set 1 — Question 39

The circuit consists of two parallel branches connected to a common voltage $V$. Let's assume $V$ is the reference phasor, $V = |V| \angle 0^\circ$.Branch 1: $R_C$ in series with $C$.
The impedance of this branch is $Z_C = R_C - jX_C$, where $X_C = \frac{1}{\omega C}$.
The current through this branch is $I_C = \frac{V}{Z_C} = \frac{V}{R_C - jX_C}$.
The phase angle of $I_C$ with respect to $V$ is $\theta_1 = \arctan\left(\frac{X_C}{R_C}\right)$. This is a leading phase angle, consistent with the phasor diagram.Branch 2: $R_L$ in series with $L$.
The impedance of this branch is $Z_L = R_L + jX_L$, where $X_L = \omega L$.
The current through this branch is $I_L = \frac{V}{Z_L} = \frac{V}{R_L + jX_L}$.
The phase angle of $I_L$ with respect to $V$ is $-\theta_2 = -\arctan\left(\frac{X_L}{R_L}\right)$. The phasor diagram shows $\theta_2$ as the magnitude of this lagging phase angle, so $\theta_2 = \arctan\left(\frac{X_L}{R_L}\right)$.We are given that the sum of these phase angles is $90^\circ$: $\theta_1 + \theta_2 = 90^\circ$.
$\arctan\left(\frac{X_C}{R_C}\right) + \arctan\left(\frac{X_L}{R_L}\right) = 90^\circ$.This implies:
$\arctan\left(\frac{X_C}{R_C}\right) = 90^\circ - \arctan\left(\frac{X_L}{R_L}\right)$.
Taking the tangent of both sides:
$\frac{X_C}{R_C} = \tan\left(90^\circ - \arctan\left(\frac{X_L}{R_L}\right)\right)$.
Using the identity $\tan(90^\circ - \phi) = \cot(\phi) = \frac{1}{\tan(\phi)}$:
$\frac{X_C}{R_C} = \frac{1}{\tan\left(\arctan\left(\frac{X_L}{R_L}\right)\right)} = \frac{1}{\frac{X_L}{R_L}} = \frac{R_L}{X_L}$.So, we have the relationship: $\frac{X_C}{R_C} = \frac{R_L}{X_L}$.
Rearranging this, we get $R_C R_L = X_C X_L$.Now, let's calculate $X_C$ and $X_L$ using the given values:
$L = 1 \text{ µH} = 1 \times 10^{-6} \text{ H}$
$C = 1 \text{ µF} = 1 \times 10^{-6} \text{ F}$
Frequency $f = 159.15 \text{ kHz} = 159.15 \times 10^3 \text{ Hz}$Angular frequency $\omega = 2\pi f = 2\pi \times 159.15 \times 10^3 \text{ rad/s}$.Inductive reactance $X_L = \omega L = (2\pi \times 159.15 \times 10^3) \times (1 \times 10^{-6})$
$X_L = 2\pi \times 159.15 \times 10^{-3} = 2\pi \times 0.15915 \approx 1.000 \text{ } \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{(2\pi \times 159.15 \times 10^3) \times (1 \times 10^{-6})}$
$X_C = \frac{1}{2\pi \times 0.15915} \approx \frac{1}{1.000} = 1.000 \text{ } \Omega$.Now, substitute these values into the derived relationship:
$R_C R_L = X_C X_L \approx (1.000) \times (1.000) = 1.000$.The nearest integer value of $R_C \times R_L$ is 1.The final answer is $\boxed{\text{B}}$