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

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(C) At resonance,the angular frequency is given by $\omega = \frac{1}{\sqrt{LC}}$,which implies $C = \frac{1}{\omega^2 L}$.In a series $L-C-R$ circuit

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$\frac{V_R}{V_L R \omega}$

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(C) At resonance,the angular frequency is given by $\omega = \frac{1}{\sqrt{LC}}$,which implies $C = \frac{1}{\omega^2 L}$.In a series $L-C-R$ circuit,the current $I$ is the same through all components.The voltage across the resistor is $V_R = IR$,so $I = \frac{V_R}{R}$.The voltage across the inductor is $V_L = I X_L = I \omega L$.Substituting $I$ into the expression for $V_L$,we get $V_L = \left( \frac{V_R}{R} \right) \omega L$.Solving for $L$,we find $L = \frac{V_L R}{V_R \omega}$.Now,substitute $L$ into the expression for $C$:$C = \frac{1}{\omega^2 \left( \frac{V_L R}{V_R \omega} \right)} = \frac{V_R \omega}{V_L R \omega^2} = \frac{V_R}{V_L R \omega}$.

$A$ series $L-C-R$ circuit containing a resistance $R$ has angular frequency $\omega$. At resonance,the voltages across the resistance and the inductor are $V_R$ and $V_L$ respectively. The value of the capacitance is:

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