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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsEasy2 minPYQ_2020

PhysicsEasysingle choice

A telephone wire of length $200 km$ has a capacitance of $0.014 μ F$ per $km$ . If it carries an AC 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?

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Answer:

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

$X_{L} = X_{C}$

$\Rightarrow 2 π f L = \frac{1}{2 π f C}$

$L = \frac{1}{4 π^{2} f^{2} C}$

$= \frac{1}{4 {(3.14)}^{2} \times {(5 \times 10^{3})}^{2} \times 2.8 \times 10^{- 6}}$

$= 0.35 \times 10^{- 3} H = 0.35 m H$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2020

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