In a circuit $C = 2 μ F, L = 1 m H$ and $R = 10 Ω$ . When the current in the circuit is maximum, at the time ratio of the energies stored in the capacitor and the inductor will be :-

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Solution

At maximum current condition
if is at Resonance condition.
$x_{c} = x_{L}$
$\begin{matrix} \Rightarrow \frac{U_{C}}{U_{L}} = \frac{\frac{1}{2} C V^{2}}{\frac{1}{2} L i^{2}} = \frac{C i^{2} R^{2}}{L i^{2}} \Rightarrow \frac{U_{C}}{U_{L}} = \frac{2 U}{1 m} \times 10^{2} = \frac{2 \times 10^{- 6}}{10^{- 3}} \times 100 H e n c e, w e g e t \frac{U_{C}}{U_{L}} = \frac{2}{10} = \frac{1}{5} \end{matrix}$

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In a circuit C=2μF,L=1mH and R=10Ω. When the current in the circuit is maximum, at the time ratio of the energies stored in the capacitor and the inductor will be :-

At maximum current condition if is at Resonance condition.xc=xL⇒UCUL=12CV212Li2=Ci2R2Li2⇒UCUL=2U1m×102=2×10−610−3×100Hence,wegetUCUL=210=15

In a circuit $C = 2 μ F, L = 1 m H$ and $R = 10 Ω$ . When the current in the circuit is maximum, at the time ratio of the energies stored in the capacitor and the inductor will be :-