Alternating Current

JEE Main level, numerical answer, medium

24 of 190

Question 24 of 190

A telegraph line of length 100km100km has a capacity of 0.01μFkm−10.01\mu F{km}^{-1} and it carries an alternating current at 0.50.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is _____ mHmH. (If π=10\pi =\sqrt{10})

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Answer: 100

The frequency given is 0.50.5 kilo cycle per second ⇒f=500Hz⇒ω=2πf=1000πrads−1\Rightarrow f=500Hz\Rightarrow \omega =2\pi f=1000\pi rad{s}^{-1}.
Total capacitance of the wire is C=0.01×10−6×100FC=0.01\times {10}^{-6}\times 100F.
Impedance is minimum at resonance. Therefore, for minimum impedance
XL=XC{X}_{L}={X}_{C}
⇒ωL=1ωC⇒L=1ω2C=110002×π2×0.01×10−6×100=10−1H=100mH\Rightarrow \omega L=\dfrac{1}{\omega C} \Rightarrow L=\dfrac{1}{{\omega }^{2}C}=\dfrac{1}{{1000}^{2}\times {\pi }^{2}\times 0.01\times {10}^{-6}\times 100}={10}^{-1}H=100mH

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