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  • Shown in the figure is a R-L-C circuit. In steady state the ratio of energy stored in the inductor & capacitor is equal to <img alt="" src="https://cdn.entrance360.com/media/uploads/2

Shown in the figure is a R-L-C circuit. In steady state the ratio of energy stored in the inductor & capacitor is equal to

Option: 1

\mathrm{\frac{1}{4}}


Option: 2

\mathrm{\frac{L}{16 \mathrm{C}}}


Option: 3

\mathrm{\frac{L}{4 \mathrm{C}}}


Option: 4

none of these


Answers (1)

best_answer

In steady state\mathrm{V_{A B}=\varepsilon}   (emf of the battery)

{\Rightarrow U_c=1 / 2 c \varepsilon^2=(1 / 2) c I^2 R^2}

If I be the current through the battery then

\mathrm{I=\varepsilon / R \quad\left[\text { as } R_{e q} \text { battery } A \& C=R\right. \text { ] }}

and current through inductor = I/4.

\mathrm{\Rightarrow U_L=(1 / 2) L[I / 4]^2}

\mathrm{\text { Now } \frac{U_L}{U_C}=\frac{1}{2} L \cdot \frac{I^2}{16} / \frac{1}{2} \mathrm{CI}^2 \mathrm{R}^2=\frac{\mathrm{L}}{16 \mathrm{CR}^2}}

Posted by

Suraj Bhandari

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