An inductor of 500 mH is in series with a resistance and a variable capacitor connected to a source of frequency 0.4 kHz. The value of capacitance of the ca pacitor to get a maximum current will be

The correct option is 0.32 µF.

Step-by-Step Explanation:

For a series R-L-C circuit connected to an alternating current (AC) source, the current in the circuit is maximum at the condition of electrical resonance. At resonance, the inductive reactance ($X_L$) of the inductor is equal to the capacitive reactance ($X_C$) of the capacitor.

The resonance condition is given by the formula:
$X_L=X_C$

We can express reactances in terms of angular frequency ($\omega$):
$\omega L=\frac{1}{\omega C}$

Rearranging the equation to solve for the capacitance ($C$):
$C=\frac{1}{{\omega }^{2}L}$

Since the angular frequency is related to the frequency ($f$) by the relation $\omega =2\pi f$, the equation becomes:
$C=\frac{1}{4{\pi }^2{f}^{2}L}$

We are given the following values:
Inductance, $L=500\text{mH}=0.5\text{H}$
Frequency, $f=0.4\text{kHz}=400\text{Hz}$

Substituting these values into the capacitance formula:
$C=\frac{1}{4\times {\left(3.1416\right)}^2\times {\left(400\right)}^2\times 0.5}$

Calculate the terms in the denominator:
$4\times 9.8696\times 160,000\times 0.5\approx 3,158,272$

Now, calculate the capacitance:
$C=\frac{1}{3,158,272}\approx 3.166\times {10}^{-7}\text{F}$

Convert the answer to microfarads (µF):
$C\approx 0.317\text{µF}\approx 0.32\text{µF}$

Thus, the required value of capacitance to obtain the maximum current is approximately 0.32 µF.