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  • Paragraph-1 When a capacitor is connected to a battery through a resistance <

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When a capacitor \mathrm{C} is connected to a battery through a resistance \mathrm{R}, the plates of the capacitor will acquire equal and opposite charge and potential difference across it become equal to the emf of battery. The process called charging takes sometime and during this time there is an electric current through the resistance. If at any time t, q is the charge on capacitor, then from Kirchoff's law.

\mathrm{R \frac{d q}{d t}+\frac{q}{c}=E }

In the circuit shown, the battery is ideal with emf \mathrm{V}. The capacitor is initially uncharged. The switch \mathrm{S} is closed at \mathrm{t}=0.

Question:-

The current in EB at time is

Option: 1

\mathrm{ \frac{V}{6 R}\left(3 e^{-2 t / 3 R C}-1\right)}


Option: 2

\mathrm{ \frac{V}{2 R}}


Option: 3

\mathrm{\frac{V e^{-2 t / 3 R C}}{R}}


Option: 4

\mathrm{\frac{V}{6 R}\left(e^{-2 t / 3 R C}-3\right)}


Answers (1)

best_answer

Let the current in EB at time t is \mathrm{i_2}.
Using KVL in loop BCDEB

\begin{aligned} & \mathrm{-i_1 R-\frac{1}{C}-i_2 R=0 }\\ \Rightarrow & \mathrm{i_2=-i_1-\frac{q}{C R} }\end{aligned}

Putting the value of \mathrm{i_1} and q from above we get
\mathrm{i_2=\frac{V}{6 R}\left(e^{\frac{-2 t}{3 R C}}-3\right) }

Posted by

shivangi.bhatnagar

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