Question

A telephone wire of length $200\text{km}$ has a capacitance of $0.014\mu \text{F}$ per $\text{km}$. If it carries an $\text{AC}$ of frequency $5\text{kHz},$ what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum.

• A. $0.36\text{mH}$
• B. $36\text{mH}$
• C. $3.6\text{mH}$
• D. $0$

Answer 1

The correct option is A $0.36\text{mH}$

Let the inductance of inductor be $L$

Length of the wire, $l=200\text{km}$

Capacitance per unit length of the wire, $C_0=0.014\frac{\mu \text{F}}{\text{km}}$

Frequency of the source, $f=5\text{kHz}$

The capacitance of the whole wire is

$C=C_{0}l$

$C=0.014\times {10}^{-6}\times 200$

$\therefore C=2.8\mu \text{F}$

For minimum impedance

$X_L=X_C$

$\omega L=\frac{1}{\omega C}$

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

$L=\frac{1}{(2\pi \times 5000{)}^2\times 2.8\times {10}^{-6}}$

$\therefore L=0.36\text{mH}$

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