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  • The voltage across the capacitor C varies which time t (figure) after the shorting of the switch S at the moment t = 0 <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/07/0

The voltage across the capacitor C varies which time t (figure) after the shorting of the switch S at the moment t = 0

Option: 1

\frac{E}{3}\left(1-e^{\frac{-2 t}{R C}}\right)


Option: 2

\frac{E}{2}\left(1-e^{\frac{-2 t}{R C}}\right)


Option: 3

\frac{E}{2}\left(1-e^{-\frac{t}{R C}}\right)


Option: 4

None of the above


Answers (1)

best_answer

At any time t, the current distribution is shown in figure

Applying Kirchoff's law to mesh 1 2 3 4 5 6 1, we have \mathrm{I}=\mathrm{i}_1+\mathrm{i}_2 \text { and } \mathrm{i}_2=\frac{\mathrm{dq}}{\mathrm{dt}}

\begin{aligned} & \frac{q}{c}+R i=E \\ & \frac{q}{c}+R\left(i_1+\frac{d q}{d t}\right)=E \\ & \frac{q}{c}+R i_1+R \frac{d q}{d t}=E \end{aligned}

Applying Kirchoff's law to mesh 2 5 4 3 2 we have

\mathrm{I}_1 \mathrm{R}=\frac{\mathrm{q}}{\mathrm{c}}

From eqs (1) and (2) we have

\begin{aligned} & \frac{q}{c}+\frac{q}{c}+R \frac{d q}{d t}=E \\ & \text { or } R \frac{d q}{d t}=E-\frac{2 q}{c} \\ & \frac{d q}{E-\frac{2 q}{c}}=\frac{d t}{R} \end{aligned}

Integrating eq(3) we get

\begin{aligned} & \int_0^q \frac{d q}{\left(E-\frac{2 q}{c}\right)}=\frac{1}{R} \int_0^t d t \\ & -\frac{c}{2} \log e\left[\frac{E-\left(\frac{2 q}{c}\right)}{E}\right]=\frac{t}{R} \\ & \text { loge } \frac{E-\left(\frac{2 q}{c}\right)}{E}=-\frac{2 t}{R C} \end{aligned}

\begin{aligned} & \frac{E-\left(\frac{2 q}{c}\right)}{E}=e^{\frac{-2 t}{R C}} \Rightarrow 1-\frac{2 q}{C E}=e^{\frac{-2 t}{R C}} \\ & \frac{2 q}{C E}=1-e^{-\frac{2 t}{R C}} \\ & V=\frac{q}{c}=\frac{E}{2}\left(1-e^{\frac{-2 t}{R C}}\right) \\ & V=\frac{E}{2}\left(1-e^{\frac{-2 t}{R C}}\right) \end{aligned}

 

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seema garhwal

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