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

Then the current in be at time $\mathrm{t}$ . What is the limiting value at $\mathrm{t \rightarrow \infty}$ is

Option: 1
$\mathrm{\frac{V}{3R}}$

Option: 2
$\mathrm{\frac{V}{2R}}$

Option: 3
$\mathrm{\frac{V}{3R}}$

Option: 4
$\mathrm{\frac{V}{R}}$

Answers (1)

best_answer

The situation at time t will be as shown in the diagram
Now, applying Kirchoff's loop law on loop abef

$\begin{aligned} & -\left(\mathrm{i}-\mathrm{i}_1\right) \mathrm{R}-\mathrm{i} \mathrm{R}+\mathrm{V}=0 \\ & \Rightarrow 2 \mathrm{i} R-\mathrm{i}_1 \mathrm{R}=\mathrm{V \quad \dots(1)} \end{aligned}$

For loop bcde

$\begin{aligned} & -\mathrm{i}_1 \mathrm{R}-\frac{q}{c}+\left(i-i_1\right) R=0 \\ & \mathrm{i} \mathrm{R}-2 \mathrm{i}_1 \mathrm{R}=\mathrm{q} / \mathrm{C \quad \dots(2)} \end{aligned}$

Eliminating i between (1) and (2)

$\begin{aligned} & \mathrm{2\left(q / c+2 i_1 R\right)-i_1 R=V} \\ & \mathrm{3 i_1 R+2 q / C=V }\end{aligned}$

But, $\mathrm{i_1=\frac{d q}{d t}}$ as the current $\mathrm{i_1}$ is charging the capacitor.

$\mathrm{\therefore \frac{d q}{d t}+\frac{2}{3 R C} \quad q=\frac{V}{3 R} }$

Integrating, we get

$\mathrm{(a)~ q=\frac{v C}{2}\left(i-e^{-2 t / 3 R C}\right) }$

$\mathrm{(b) ~i_1=\frac{d q}{d t}=\frac{V}{3 R} e^{-2 t / 3 R C} }$

Also current in be = $\mathrm{i-i_1}$

As $\mathrm{\mathrm{t} \rightarrow \infty}$ , this value becomes $\mathrm{\frac{V}{2R}}$ .

Posted by

manish painkra

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