First slide

Phasors and Phasor Diagrams

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,
$C = 0 .014 \times 10^{- 6} \times 200 = 2 .8 \times 10^{- 6} F = 2 .8 μF$
For impedance of the circuit to be minimum, $X_{L} = X_{C}$
$\Rightarrow 2 πvL = \frac{1}{2 πvC}$
$\Rightarrow L = \frac{1}{4 π^{2} v^{2} C} = \frac{1}{4 (3 .14)^{2} \times {(5 \times 10^{3})}^{2} \times 2 .8 \times 10^{- 6}}$
$= 0 .35 \times 10^{- 3} H = 0 .35 mH$

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