In an LCR series circuit,C = 2 mu F,L = 1 mH | vedclass

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(A) In an $LCR$ series circuit,the current is maximum at resonance,where the inductive reactance equals the capacitive reactance $(X_L = X_C)$.At reso

Vedclass · Accepted Answer

$1 : 1$

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(A) In an $LCR$ series circuit,the current is maximum at resonance,where the inductive reactance equals the capacitive reactance $(X_L = X_C)$.At resonance,the impedance $Z = R$,so the maximum current is $I_{\max} = \frac{V}{R}$.The energy stored in the inductor is $W_L = \frac{1}{2} L I_{\max}^2$.The energy stored in the capacitor is $W_C = \frac{1}{2} C V_C^2$,where $V_C = I_{\max} X_C$.Since $X_C = \frac{1}{\omega C}$ and $X_L = \omega L$,at resonance $\omega = \frac{1}{\sqrt{LC}}$.Thus,$X_C = X_L = \sqrt{\frac{L}{C}}$.$W_C = \frac{1}{2} C (I_{\max} X_C)^2 = \frac{1}{2} C I_{\max}^2 (\frac{L}{C}) = \frac{1}{2} L I_{\max}^2$.Wait,let's re-evaluate: $W_C = \frac{1}{2} C V_C^2 = \frac{1}{2} C (I_{\max} \cdot \frac{1}{\omega C})^2 = \frac{1}{2} C I_{\max}^2 \cdot \frac{1}{\omega^2 C^2} = \frac{I_{\max}^2}{2 \omega^2 C}$.Substituting $\omega^2 = \frac{1}{LC}$,we get $W_C = \frac{I_{\max}^2}{2 (1/LC) C} = \frac{1}{2} L I_{\max}^2$.Therefore,$W_C = W_L$,so the ratio is $1:1$.

In an $LCR$ series circuit,$C = 2 \mu F$,$L = 1 \ mH$,and $R = 10 \ \Omega$. When the current in the circuit is maximum,what is the ratio of the energy stored in the capacitor to the energy stored in the inductor?

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