Questions

# In the circuit shown in figure, switch S is closed at time t = 0. find the current through the cell at any time t.

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a
$\frac{\mathrm{V}}{9\mathrm{R}}+\frac{\mathrm{V}}{\mathrm{R}}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{CR}}}$
b
$\frac{\mathrm{V}}{9\mathrm{R}}+\frac{\mathrm{V}}{27\mathrm{R}}{\mathrm{e}}^{\frac{-\mathrm{t}}{3\mathrm{RC}}}$
c
$\frac{4\mathrm{V}}{27\mathrm{R}}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{CR}}}$
d
$\frac{\mathrm{V}}{\mathrm{R}}+\frac{\mathrm{V}}{27\mathrm{R}}{\mathrm{e}}^{\frac{-\mathrm{t}}{3\mathrm{RC}}}$

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detailed solution

Correct option is B

Current in the branch of capacitor at any instant is ${I}_{C}=\frac{dq}{dt}=\frac{CV}{3}\left[-{e}^{-t/3RC}\right]\left[-\frac{1}{3RC}\right]=\frac{V}{9R}{e}^{-t/3RC}$

Let I be the current through the cell at any instant ‘t’.

Applying KVL to the loop ABEFA,

$I\left(6R\right)+{I}_{c}\left(R\right)+\frac{q}{C}-V=0$

$I\left(6R\right)+\frac{V}{9R}{e}^{-t/3RC}\left(R\right)+\frac{1}{C}×\frac{CV}{3}\left(1-{e}^{-t/3RC}\right)-V=0$

$I\left(6R\right)=V\left[1-\frac{1}{9}{e}^{-t/3RC}-\frac{1}{3}\left(1-{e}^{-t/3RC}\right)\right]$

$I=\frac{V}{6R}\left[\left(1-\frac{1}{3}\right)+{e}^{-t/3RC}\left(\frac{1}{3}-\frac{1}{9}\right)\right]$

$I=\frac{V}{6R}\left[\frac{2}{3}+\frac{2}{9}{e}^{-t/3RC}\right]=\frac{V}{9R}+\frac{V}{27R}{e}^{-\frac{t}{3RC}}$