In the given circuit, the value of capacitor C that makes current I = 0 is _______ μF.

The correct answer is 20.

From the given circuit diagram, the parameters are:
- An AC voltage source with voltage $V=10\text{V}$ and angular frequency $\omega =5\text{k rad/s}=5000\text{rad/s}$.
- A resistor of $10\text{Ω}$ and an inductor with impedance $j5\text{Ω}$ connected in series with the source.
- A parallel network consisting of a vertical shunt inductor with impedance $Z_2=j5\text{Ω}$ and a horizontal branch consisting of an inductor with impedance $Z_3=j5\text{Ω}$ in series with the capacitor $C$.

The equivalent impedance of the parallel combination is given by:
$Z_p=Z_2\parallel \left(Z_3,+,Z_C\right)=\frac{Z_2\left(Z_3,+,Z_C\right)}{Z_2+Z_3+Z_C}$
where $Z_C=-\frac{j}{\omega C}$ is the impedance of the capacitor.

For the input current $I$ to be zero ($I=0$), the total impedance of the circuit must be infinite. This occurs when the parallel combination behaves as an open circuit (i.e., its impedance becomes infinite):
$Z_p\to \infty$
This condition is satisfied when the denominator of $Z_p$ is equal to zero:
$Z_2+Z_3+Z_C=0$

Substitute the impedance values into the equation:
$j5+j5-\frac{j}{\omega C}=0$
Combine the imaginary terms:
$j10=\frac{j}{\omega C}$
$10=\frac{1}{\omega C}$

Rearranging the equation to solve for $C$:
$C=\frac{1}{10\omega }$
Substitute the angular frequency $\omega =5000\text{rad/s}$:
$C=\frac{1}{10\times 5000}=\frac{1}{50000}=2\times {10}^{-5}\text{F}$
Convert the capacitance to microfarads (μF):
$C=20\times {10}^{-6}\text{F}=20\text{μF}$