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Basic Differentiation Rule

In the figure...

Question

In the figure shown below, find the current as function of time $t$ on capacitor $C_{1}$ . Initially $C_{1}$ is uncharged and $C_{2}$ is charged to a potential of $2 E$ .

i = \frac{3 E}{R} e^{⎛ ⎝ \frac{- 2 t}{R C} ⎞ ⎠}

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i = \frac{E}{R} e^{(\frac{- t}{R C})}

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i = \frac{E}{R} e^{⎛ ⎝ \frac{- 2 t}{R C} ⎞ ⎠}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

i = \frac{3 E}{R} e^{(\frac{- t}{R C})}

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Solution

The correct option is A $i = \frac{3 E}{R} e^{⎛ ⎝ \frac{- 2 t}{R C} ⎞ ⎠}$

Charge distribution on each plate is shown in the figure below:

Using, $KVL$ to the given loop we get,

$- \frac{q}{C} + \frac{2 C E - q}{C} - i R + E = 0$

$\Rightarrow 3 E = \frac{2 q}{C} + i R$

$\Rightarrow 3 E C = 2 q + \frac{d q}{d t} R C$

$\Rightarrow \frac{d q}{3 E C - 2 q} = \frac{d t}{R C} . . . (1)$

Integrating eq $(1)$ with proper limits we get,

$\int_{0}^{q} \frac{d q}{3 E C - 2 q} = \int_{0}^{t} \frac{d t}{R C}$

$\Rightarrow [- \frac{ln (3 E C - 2 q)}{2}]_{0}^{q} = \frac{t}{R C}$

$\Rightarrow [ln (3 E C - 2 q)]_{0}^{q} = - \frac{2 t}{R C}$

$\Rightarrow ln (\frac{3 E C - 2 q}{3 E C}) = - \frac{2 t}{R C}$

$\Rightarrow \frac{3 E C - 2 q}{3 E C} = e^{- ⎛ ⎝ \frac{2 t}{R C} ⎞ ⎠}$

$\Rightarrow 3 E C - 2 q = 3 E C e^{- ⎛ ⎝ \frac{2 t}{R C} ⎞ ⎠}$

$\Rightarrow q = \frac{3 E C}{2} ⎡ ⎢ ⎢ ⎣ 1 - e^{- ⎛ ⎝ \frac{2 t}{R C} ⎞ ⎠} ⎤ ⎥ ⎥ ⎦$

So, the current at time $t$ ,

$i = \frac{d q}{d t} = \frac{3 E}{R} e^{- ⎛ ⎝ \frac{2 t}{R C} ⎞ ⎠}$

Hence, option (a) is the correct answer.

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Basic Differentiation Rule

Standard XII Physics

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In the figure shown below, find the current as function of time t on capacitor C1. Initially C1 is uncharged and C2 is charged to a potential of 2E.

In the figure shown below, find the current as function of time $t$ on capacitor $C_{1}$ . Initially $C_{1}$ is uncharged and $C_{2}$ is charged to a potential of $2 E$ .