Question

A telephone wire of length $200km$ has a capacitance of $0.014\mu km$ per $km$. If it carries an AC of frequency $5kHz$. What should be the value of the inductor required to be connected in series so that the impedance of the circuit is minimum.
A. $0.35mH$
B. $35mH$
C. $3.5mH$
D. $ZERO$

Answer 1

Hint:We are provided with an LC circuit with given capacitance and frequency. We have to find the inductance with the given statement that impedance of the circuit is minimum. We know the condition for impedance to be minimum is ${X_L} = {X_C}$ where the terms are impedance for inductor and capacitor respectively.

Complete step by step answer:
According to the question, given values are:
Length of the wire $l = 200km$, Capacitance per km =$0.014\mu km$
Hence, capacitance of the wire be $C = 0.014 \times {10^{ - 6}} \times 200$
$C = 2.8 \times {10^{ - 6}}F = 2.8\mu F$
And frequency $v = 5kHz = 5 \times {10^3}Hz$
For impedance of the circuit to be minimum, the condition is ${X_L} = {X_C}$
Where ${X_L}$ is the inductive reactance having value ${X_L} = 2\pi vL$ here $v$ is the frequency and $L$ is the inductance
And ${X_C}$ is the capacitive reactance having value ${X_C} = \dfrac{1}{{2\pi vC}}$ here $C$ is the capacitive.Putting such values in the condition,
$ \Rightarrow 2\pi vL = \dfrac{1}{{2\pi vC}}$
We have to find the inductor. So, from above equation $L = \dfrac{1}{{4{\pi ^2}{v^2}C}}$
Substituting the above values
$
L = \dfrac{1}{{4 \times {{\left( {3.14} \right)}^2} \times {{\left( {5 \times {{10}^3}} \right)}^2} \times 2.8 \times {{10}^{ - 6}}}} \\
\Rightarrow L = 0.35 \times {10^{ - 3}}H \\
\therefore L = 0.35mH \\ $
Hence, the value of the inductor is $0.35mH$.

So, the correct option is A.

Note: Inductive reactance is usually related to the magnetic field surrounding a wire or a coil carrying current. Likewise, capacitive reactance is often linked with the electric field that keeps changing between two conducting plates or surfaces that are kept apart from each other by some insulating medium.