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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\frac{V}{9 R} + \frac{V}{R} e^{\frac{- t}{CR}}

\frac{V}{9 R} + \frac{V}{27 R} e^{\frac{- t}{3 RC}}

\frac{4 V}{27 R} e^{\frac{- t}{CR}}

\frac{V}{R} + \frac{V}{27 R} e^{\frac{- t}{3 RC}}

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detailed solution

Correct option is B

Current in the branch of capacitor at any instant is $I_{C} = \frac{d q}{d t} = \frac{C V}{3} [- e^{- t / 3 R C}] [- \frac{1}{3 R C}] = \frac{V}{9 R} e^{- t / 3 R C}$

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

Applying KVL to the loop ABEFA,

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

$I (6 R) + \frac{V}{9 R} e^{- t / 3 R C} (R) + \frac{1}{C} \times \frac{C V}{3} (1 - e^{- t / 3 R C}) - V = 0$

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

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

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

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