First slide

Electrical Resonance series circuit

Question

A telephone wire of length 200 km has a capacitance of 0.014 µF per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

Moderate

A

0.35 mH
B

35 mH
C

3.5 mH
D

Zero

Solution

Capacitance of wire
$C = 0.014 \times {10^{ - 6}} \times 200 = 2.8 \times {10^{ - 6}}F = 2.8\mu F$
For impedance of the circuit to be minimum
${X_L} = {X_C} \Rightarrow 2\pi \nu L = \frac{1}{{2\pi \nu C}}$
$\Rightarrow L = \frac{1}{{4{\pi ^2}{\nu ^2}C}}$ $= \frac{1}{\4{(3.14)^2}\times({5}\times10^3)^2\times2.8\times10^-6}$
$= 0.35 \times {10^{ - 3}}H = 0.35\,mH$

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