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प्रश्न
A capacitor of capacitance C is given a charge Q. At t = 0, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the second capacitor as a function of time.
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उत्तर
Given:-
Initial charge on first capacitor = Q
Let q be the charge on the second capacitor after time t.
According to the principle of conservation of charge, charge on the first capacitor after time t = Q - q.
Let V1 be the potential difference across the first capacitor and V2 be the potential difference across the second capacitor after time t. Then,
\[V_1 = \frac{Q - q}{C} \]
\[ V_2 = \frac{q}{C}\]
\[ \Rightarrow V_1 - V_2 = \frac{Q - q}{C} - \frac{q}{C}\]
\[ = \frac{Q - 2q}{C}\]
The current through the circuit after time t,
\[i = \frac{V_1 - V_2}{R} = \frac{dq}{dt}\]
\[ \Rightarrow \frac{Q - 2q}{CR} = \frac{dq}{dt}\]
\[ \Rightarrow \frac{dq}{Q - 2q} = \frac{1}{RC}dt\]
\[ \Rightarrow \frac{dq}{Q - 2q} = \frac{1}{RC}dt\]
Integrating both sides within the limits time =0 to t and charge on the second capacitor varying from q=0 to q, we get:-
\[\frac{1}{2} \left[ \ln \left( Q - 2q \right) - \ln Q \right] = \frac{- 1t}{RC}\]
\[\ln \frac{Q - 2q}{Q} = \frac{- 2t}{RC}\]
\[Q - 2q = Q e^{- \frac{2t}{RC}} \]
\[2q = Q\left( 1 - e^{- \frac{2t}{RC}} \right)\]
\[q = \frac{Q}{2}\left\{ 1 - e^{- \frac{2t}{RC}} \right\}\]
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