Question

In the given circuit, the value of capacitor \(C\) that makes current \(I = 0\) is \(\underline{\hspace{1cm}}\) \(\mu\text{F}\). 

Hint:

At resonance, inductive and capacitive reactances cancel each other, creating infinite impedance in that branch.

Answer 1

Correct Answer: 20

For current to be zero, the net impedance must become infinite. This happens when the series impedance resonates with the parallel combination.
The resonance condition for the branch with \(j5\Omega\) and capacitor is:
\[ \frac{1}{j\omega C} = -j5 \] Given \( \omega = 5000 \text{ rad/s} \),
\[ \frac{1}{5000C} = 5 \] \[ C = \frac{1}{25000} = 40 \times 10^{-6} \] Equivalent for parallel resonance becomes half (due to symmetric branch):
\[ C = 20\ \mu F. \] Thus, the required capacitor is \(20.00\ \mu F\).