A series L-C-R circuit containing a resistanc | vedclass

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(A) At resonance,the impedance of the circuit is equal to the resistance,$Z = R$.Since the circuit is in series,the current $I$ is the same through al

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$0.2$

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(A) At resonance,the impedance of the circuit is equal to the resistance,$Z = R$.Since the circuit is in series,the current $I$ is the same through all components.$I = \frac{V_R}{R} = \frac{60 \ V}{120 \ \Omega} = 0.5 \ A$.At resonance,the inductive reactance $X_L$ is given by $X_L = \frac{V_L}{I}$.$X_L = \frac{40 \ V}{0.5 \ A} = 80 \ \Omega$.We know that $X_L = \omega L$,where $\omega = 4 	imes 10^5 \ rad \ s^{-1}$.$L = \frac{X_L}{\omega} = \frac{80}{4 	imes 10^5} = 20 	imes 10^{-5} \ H$.$L = 2 	imes 10^{-4} \ H = 0.2 	imes 10^{-3} \ H = 0.2 \ mH$.

$A$ series $L-C-R$ circuit containing a resistance of $120 \Omega$ has an angular frequency of $4 \times 10^5 \ rad \ s^{-1}$. At resonance,the voltage across the resistance and the inductor are $60 \ V$ and $40 \ V$ respectively. The value of the inductance is: (in $mH$)

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