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Question
A capacitance C charged to a potential difference V is discharged by connecting its plates through a resistance R. Find the heat dissipated in one time constant after the connections are made. Do this by calculating ∫ i2R dt and also by finding the decrease in the energy stored in the capacitor.
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Solution 1
Let Q0 be the initial charge on the capacitor. Then,
Q0 = CV
The charge on the capacitor at time t after the connections are made,
\[Q = Q_0 e^{- \frac{t}{RC}}\]
\[i = \frac{dQ}{dt} = - \left( \frac{Q_0}{RC} \right) e^{- \frac{t}{RC}}\]
Heat dissipated during time t1 to t2,
\[U = \int_{t_1}^{t_2} i^2 Rdt\]
\[ = \frac{{Q_0}^2}{2C}\left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)\]
Time constant = RC
Putting t1 = 0 and t2 = RC, we get:-
\[U = \frac{{Q_0}^2}{2C}\left( e^{- 0} - e^{- 2} \right)\]
\[ \because Q_0 = CV, \]
\[ U = \frac{CV^2}{2}\left( 1 - \frac{1}{e^2} \right)\]
Solution 2
Heat dissipated at any time = Energy stored at time 0 − Energy stored at time t
\[\Rightarrow U = \frac{1}{2}C V^2 - \frac{1}{2}C V^2 e^{- \frac{2t}{RC}} \]
\[ \because t = RC, \]
\[ U = \frac{1}{2}C V^2 - \frac{1}{2}C V^2 e^{- 2} \]
\[ \Rightarrow U = \frac{1}{2}C V^2 \left( 1 - \frac{1}{e^2} \right)\]
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