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

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(A) The current in a series $LCR$ circuit is maximum at the condition of resonance,where the inductive reactance equals the capacitive reactance $(X_L

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$1 : 1$

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(A) The current in a series $LCR$ circuit is maximum at the condition of resonance,where the inductive reactance equals the capacitive reactance $(X_L = X_C)$.At resonance,the voltage across the inductor $(V_L)$ and the capacitor $(V_C)$ are equal in magnitude but opposite in phase,such that $V_L = V_C = Q \cdot V_{source}$,where $Q$ is the quality factor.The energy stored in the inductor is $U_L = \frac{1}{2} L I_{max}^2$.The energy stored in the capacitor is $U_C = \frac{1}{2} C V_C^2$.At resonance,$I_{max} = \frac{V}{R}$. The voltage across the capacitor is $V_C = I_{max} X_C = \frac{V}{R} \cdot \frac{1}{\omega C}$.Since $\omega = \frac{1}{\sqrt{LC}}$,we have $V_C = \frac{V}{R} \sqrt{\frac{L}{C}}$.Substituting these into the energy expressions:$U_L = \frac{1}{2} L (\frac{V}{R})^2$$U_C = \frac{1}{2} C (\frac{V}{R} \sqrt{\frac{L}{C}})^2 = \frac{1}{2} C \frac{V^2}{R^2} \frac{L}{C} = \frac{1}{2} L \frac{V^2}{R^2}$Thus,$U_C = U_L$,which means the ratio $U_C : U_L = 1 : 1$.

In a series $LCR$ 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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