Question

An inductor of inductance $100\text{mH}$ is connected in series with a resistance, a variable capacitance and an $\text{AC}$ source of frequency $2.0\text{kHz}$. What should be the value of the capacitance so that maximum current may be drawn into the circuit ?

• A. $10\text{nF}$
• B. $28\text{nF}$
• C. $47\text{nF}$
• D. $63\text{nF}$

Answer 1

The correct option is D $63\text{nF}$

Given:-

$L=100\text{mH}$

$f=2\text{kHz}=2\times {10}^3\text{Hz}$

The given circuit is a series $\text{LCR}$ circuit. The current will be maximum when the net reactance of the circuit is zero.

i.e., $X_L=X_C$

$\omega L=\frac{1}{\omega C}\Rightarrow C=\frac{1}{{\omega }^{2}L}$

$C=\frac{1}{(2\pi f{)}^{2}L}$

$C=\frac{1}{(2\pi \times 2\times {10}^3{)}^2\times 100\times {10}^{-3}}$

$\therefore C=63\text{nF}$

Hence, option $\left(D\right)$ is correct.