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

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

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$1:1$

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(A) In a series $LCR$ circuit,the current is maximum at resonance.At resonance,the inductive reactance equals the capacitive reactance,i.e.,$X_L = X_C$.The energy stored in the inductor is $U_L = \frac{1}{2} L I^2$.The energy stored in the capacitor is $U_C = \frac{1}{2} C V_C^2$,where $V_C = I X_C$.At resonance,the current $I$ is the same through both components. Since $X_L = X_C$,the voltage across the inductor is $V_L = I X_L$ and the voltage across the capacitor is $V_C = I X_C$.Substituting $X_C = X_L = \omega L$ and $X_C = \frac{1}{\omega C}$,we have $U_C = \frac{1}{2} C (I X_C)^2 = \frac{1}{2} C I^2 (\frac{1}{\omega C})^2 = \frac{1}{2} \frac{I^2}{\omega^2 C}$.Since at resonance $\omega^2 = \frac{1}{LC}$,we get $U_C = \frac{1}{2} \frac{I^2}{(1/LC) C} = \frac{1}{2} L I^2$.Thus,$U_L = U_C$,and the ratio of the energies stored is $1:1$.

In a series $LCR$ circuit,$C = 2 \mu F$,$L = 1 \text{ mH}$,and $R = 10 \Omega$. The ratio of the energies stored in the inductor and the capacitor,when the maximum current flows in the circuit,is:

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