An LCR series circuit of capacitance 62.5 , n | vedclass

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(D) For maximum current amplitude in an $LCR$ series circuit,the circuit must be in resonance.At resonance,the inductive reactance equals the capaciti

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$100$

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(D) For maximum current amplitude in an $LCR$ series circuit,the circuit must be in resonance.At resonance,the inductive reactance equals the capacitive reactance,i.e.,$X_L = X_C$.The resonant frequency is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.Given: $f = 2.0 \, kHz = 2000 \, Hz$,$C = 62.5 \, nF = 62.5 	imes 10^{-9} \, F$,and $\pi^2 = 10$.Rearranging the formula for $L$: $L = \frac{1}{4 \pi^2 f^2 C}$.Substituting the values: $L = \frac{1}{4 	imes 10 	imes (2000)^2 	imes 62.5 	imes 10^{-9}}$.$L = \frac{1}{40 	imes 4 	imes 10^6 	imes 62.5 	imes 10^{-9}}$.$L = \frac{1}{160 	imes 62.5 	imes 10^{-3}} = \frac{1}{10000 	imes 10^{-3}} = \frac{1}{10} = 0.1 \, H$.Converting to $mH$: $0.1 \, H = 100 \, mH$.

An $LCR$ series circuit of capacitance $62.5 \, nF$ and resistance of $50 \, \Omega$ is connected to an $A.C.$ source of frequency $2.0 \, kHz$. For the maximum value of the amplitude of current in the circuit,the value of inductance is $.......... \, mH$. (take $\pi^2 = 10$)

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