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

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

Option: 1

\mathrm {\frac{L }{16R^2C}}


Option: 2

\mathrm {\frac{L }{16R^2C}}


Option: 3

1


Option: 4

\mathrm {\frac{ 16R^2C}{L}}


Answers (1)

best_answer

Consider the current distribution in various
branches in the steady state as shown.
Now, for capacitor, \mathrm{U_c}=\frac{1}{2} \mathrm{C} \varepsilon^2
& for inductor UL = \mathrm{U_L}=\frac{1}{2}\mathrm{L}\left(\frac{\varepsilon}{\mathrm{4 R}}\right)^2

Hence, \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{\mathrm{L}}{16 \mathrm{R}^2 \mathrm{C}}
 

Posted by

Sayak

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