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JEE Main 2022 (28 Jun, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2022 (28 Jun, Shift 1)

Q82JEE Main 2022 · 28 JUN · SHIFT 1Difficult

Resonance, Bandwidth & Q Factor

A telegraph line of length 100 km has a capacity of 0.01 $\mu$ F/km and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is _____________ mH. (if $\pi = \sqrt{10}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 100

Total capacitance $C = 0.01\ \mu\text{F/km}\times100\text{ km} = 1\ \mu\text{F}$ .
Minimum impedance means resonance: $L = \frac{1}{\omega^2 C} = \frac{1}{4\pi^2 f^2 C}$
$L = \frac{1}{4\pi^2(500)^2(10^{-6})} = \frac{1}{\pi^2} = \frac{1}{10}\text{ H}$
$L = 100\text{ mH}$

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