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(A) In an $LCR$ series circuit,the frequencies at which the current amplitude becomes $\frac{1}{\sqrt{2}}$ times its maximum value are known as half-p

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) In an $LCR$ series circuit,the frequencies at which the current amplitude becomes $\frac{1}{\sqrt{2}}$ times its maximum value are known as half-power frequencies,denoted as $\omega_1$ and $\omega_2$.The bandwidth of the circuit is given by $\Delta\omega = \omega_2 - \omega_1$.Given $\omega_1 = 212\,rad\,s^{-1}$ and $\omega_2 = 232\,rad\,s^{-1}$,the bandwidth is $\Delta\omega = 232 - 212 = 20\,rad\,s^{-1}$.The formula for bandwidth in an $LCR$ series circuit is $\Delta\omega = \frac{R}{L}$.Substituting the given values: $20 = \frac{5}{L}$.Solving for $L$: $L = \frac{5}{20} = 0.25\,H$.Converting to millihenries $(mH)$: $L = 0.25 	imes 1000 = 250\,mH$.

The frequencies at which the current amplitude in an $LCR$ series circuit becomes $\frac{1}{\sqrt{2}}$ times its maximum value are $212\,rad\,s^{-1}$ and $232\,rad\,s^{-1}$. The value of resistance in the circuit is $R = 5\,\Omega$. The self-inductance in the circuit is $.........\,mH$.

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