Question

A wire of length $10\text{km}$ has a capacitance of $0.014\mu \text{F}$ per meter. If it carries an AC of frequency $70\text{Hz}$, then what should be the value of the inductance of the inductor required to be connected in series with it, so that the impedance of the circuit is minimum?

Take ${\pi }^2\approx 10$

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

Answer 1

The correct option is B $36\text{mH}$

Capacitance of the wire,

$C=0.014\times {10}^{-6}\times 10\times {10}^3=1.4\times {10}^{-4}\text{F}$

The impedance of a series LCR circuit is minimum at resonance.

Now, resonant frequency,

$f_0=\frac{1}{2\pi \sqrt{LC}}$

$\Rightarrow L=\frac{1}{4{\pi }^2{f}_0^{2}C}$

$\Rightarrow L=\frac{1}{4\times 10\times {70}^2\times 1.4\times {10}^{-4}}$

$\Rightarrow L\approx 0.036\text{H}=36\text{mH}$

Hence, option $\left(B\right)$ is the correct answer.