GATE 2024 IN Q.39: solution and answer

Official answer: CSource: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Identify the arms. The detector D is connected between the top and bottom nodes, and the source between the left and right nodes. So the opposite-arm pairs are (left-top, bottom-right) and (top-right, left-bottom). From $v=5\sin (5000t)\text{V}$, the angular frequency is $\omega =5000\text{rad/s}$.

2. Write the arm impedances: $Z_1=500+j\omega L=500+j(5000)(1)=500+j5000\Omega$ (left-top), $Z_2=1000\Omega$ (top-right), $Z_3=1000\Omega$ (left-bottom), and $Z_4=R-\frac{j}{\omega C}$ (bottom-right, $R$ and $C$ in series).

3. At balance the products of opposite arms are equal: $Z_1{Z}_4=Z_2{Z}_3$. So $Z_4=\frac{Z_2{Z}_3}{Z_1}=\frac{10^6}{500+j5000}$.

4. Rationalise the denominator: $$Z_4=\frac{10^6(500-j5000)}{500^2+5000^2}=\frac{10^6(500-j5000)}{25.25\times 10^6}=19.802-j198.02\Omega$$

5. Equate the real parts: $R=19.802\Omega$.

6. Equate the imaginary parts: $\frac{1}{\omega C}=198.02\Omega$, so $C=\frac{1}{5000\times 198.02}\text{F}=1.01\times 10^{-6}\text{F}=1.01\mu \text{F}$.

7. Check: $Z_1$ is inductive (positive angle), so the arm opposite it must be capacitive (negative angle) for $Z_1{Z}_4$ to be a real number, $10^6$. The series RC arm is capacitive, and $R$ comes out positive, so the balance can actually be reached. $R=19.802\Omega$ and $C=1.01\mu \text{F}$ is option (C).