A series LCR circuit containing a resistance | vedclass

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(C) At resonance,the impedance of the circuit is purely resistive,so $X_L = X_C$. The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \fra

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$0.2\,mH$ and $0.03125\,\mu F$

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(C) At resonance,the impedance of the circuit is purely resistive,so $X_L = X_C$. The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \frac{60}{120} = 0.5\,A$.The voltage across the inductor is $V_L = I X_L = I \omega_0 L$. Substituting the known values:$40 = 0.5 	imes (4 	imes 10^5) 	imes L$$L = \frac{40}{0.5 	imes 4 	imes 10^5} = \frac{40}{2 	imes 10^5} = 2 	imes 10^{-4}\,H = 0.2\,mH$.At resonance,$V_C = V_L = 40\,V$. Since $V_C = I X_C = \frac{I}{\omega_0 C}$,we have:$C = \frac{I}{\omega_0 V_C} = \frac{0.5}{(4 	imes 10^5) 	imes 40} = \frac{0.5}{160 	imes 10^5} = \frac{0.5}{1.6 	imes 10^7} = 0.3125 	imes 10^{-7}\,F = 0.03125 	imes 10^{-6}\,F = 0.03125\,\mu F$.

$A$ series $LCR$ circuit containing a resistance of $120\,\Omega$ has a resonance frequency of $4 \times 10^5\, rad\, s^{-1}$. The voltages,at resonance,across the resistance and inductance are $60\,V$ and $40\,V$ respectively. The values of $L$ and $C$ respectively are:

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