Alternating Current

JEE Main level, numerical answer, medium

85 of 190

Question 85 of 190

An LCR series circuit of capacitance 62.5nF62.5nF and resistance of 50Ω50\Omega, is connected to an A.C. source of frequency 2.0kHz2.0kHz. For maximum value of amplitude of current in circuit, the value of inductance is ____mHmH.(Takeπ2=10)\left.(\text{Take}{\pi }^{2}=10\right)

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Answer: 100

Given here, f=2kHz=2000Hz,C=62.5nF=62.5×10−9FandR=50Ωf=2kHz=2000Hz,C=62.5nF=62.5\times {10}^{-9}F\text{and}R=50\Omega
Resonance frequency (amplitude of the current will be maximum in this case) is given by f=12πLCf=\dfrac{1}{2\pi \sqrt{LC}}, here, LL is the inductance.
Putting the values,
2000Hz=12πL×62.5×10−92000Hz=\dfrac{1}{2\pi \sqrt{L\times 62.5\times {10}^{-9}}}
L=14π2×20002×62.5×10−9=0.1H=100mHL=\dfrac{1}{4{\pi }^{2}\times {2000}^{2}\times 62.5\times {10}^{-9}}=0.1H=100mH

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