First slide

Electrical Resonance series circuit

Question

A telephone wire of length 200 km has a capacitance of 0.014 $μ$ F 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
$C = 0 .014 \times 10^{- 6} \times 200 = 2 .8 \times 10^{- 6} F = 2 .8 μF$
For impedance of the circuit to be minimum $X_{L} = X_{C} \Rightarrow 2 πνL = \frac{1}{2 πνC}$
$\Rightarrow L = \frac{1}{4 π^{2} ν^{2} C} = \frac{1}{4 {(3 .14)}^{2} \times {(5 \times 10^{3})}^{2} \times 2 .8 \times 10^{- 6}} = 0 .35 \times 10^{- 3} H = 0 .35 mH$

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