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  • Shown in the figure is a parallel R, L, C circuit with key K1 closed and K2 opened. When K1 is opened and K2 is closed simultaneously , The maximum

Shown in the figure is a parallel R, L, C circuit with key K1 closed and K2 opened. When K1 is opened and K2 is closed simultaneously , The maximum charge stored in it is:

Option: 1

\mathrm {\frac{E \sqrt{L C}}{R}}


Option: 2

\mathrm { \frac{\mathrm{E} \sqrt{2 \mathrm{LC}}}{\mathrm{R}}}


Option: 3

\mathrm {\frac{E \sqrt{L C}}{2 R}}


Option: 4

\mathrm { \frac{2 \mathrm{E} \sqrt{\mathrm{LC}}}{\mathrm{R}}}


Answers (1)

best_answer

The current flowing along the inductor L in steady state before closing the switch K2 is given as
\mathrm {i=\frac{E}{R}}
(\because the inductor offers no resistance at steady state).
 The maximum energy stored in the inductor

=\mathrm {U_L=\frac{1}{2} L i^2=\frac{L}{2}\left(\frac{E}{R}\right)^2=\frac{L E^2}{2 R^2}}

Since the maximum energy UC stored in the capacitor after closing the switch K2 is equal to UL according to conservation of total energy,

\begin{aligned} & \mathrm{U}_{\mathrm{C}}=\frac{1}{2} C \mathrm{C}^2=\mathrm{U}_{\mathrm{L}} \\ & \Rightarrow \frac{1}{2} C \mathrm{~V}^2=\frac{\mathrm{LE}^2}{2 \mathrm{R}^2} \\ & \Rightarrow \mathrm{V}=\frac{\mathrm{E}}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \end{aligned}

where V = maximum potential difference across the capacitor
(Maximum charged stored in the capacitor)

=\mathrm {q_{\max }=C V=\frac{E \sqrt{L C}}{R}}

Posted by

seema garhwal

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