Alternating Current

JEE Main level, numerical answer, medium

120 of 190

Question 120 of 190

A transmitting station releases waves of wavelength 960m960m. A capacitor of 2.56μF2.56\mu F is used in the resonant circuit. The self-inductance of coil necessary for resonance is x×10−8Hx\times {10}^{-8}H. find xx

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Answer: 10

λ=960m\lambda =960m
C=2.56μF=2.56×10−6FC=2.56\mu F=2.56\times {10}^{-6}F
c=3×108ms−1c=3\times {10}^{8}m{s}^{-1}
L=?L=?
Now at resonance, ω0=1LC{\omega }_{0}=\dfrac{1}{\sqrt{LC}} [Resonant frequency]
2πf0=1LC2\pi {f}_{0}=\dfrac{1}{\sqrt{LC}}
On substituting f0=cλ,{f}_{0}=\dfrac{c}{\lambda }, we have 2πcλ=1LC2\pi \dfrac{c}{\lambda }=\dfrac{1}{\sqrt{LC}}
On substituting f0=cλ,{f}_{0}=\dfrac{c}{\lambda }, we have 2πcλ=1LC2\pi \dfrac{c}{\lambda }=\dfrac{1}{\sqrt{LC}}
Squaring both sides: 4π2c2λ2=1LC4{\pi }^{2}\dfrac{{c}^{2}}{{\lambda }^{2}}=\dfrac{1}{LC}
=4×10×(3×108)2(960)2=1L×2.56×10−6=\dfrac{4\times 10\times {\left(3\times {10}^{8}\right)}^{2}}{{\left(960\right)}^{2}}=\dfrac{1}{L\times 2.56\times {10}^{-6}}
⇒1L=4×10×9×1016×2.56×10−6960×960\Rightarrow \dfrac{1}{L}=\dfrac{4\times 10\times 9\times {10}^{16}\times 2.56\times {10}^{-6}}{960\times 960}
⇒L=10×10−8H\Rightarrow L=10\times {10}^{-8}H

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