Question
By evaluating $\int\text{i}^2\text{Rdt},$ show that when a capacitor is charged by connecting it to a battery through a resistor, the energy dissipated as heat equals the energy stored in the capacitor.
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Answer
$\int\text{i}^2\text{Rdt}=\int\text{i}_0^2\text{Re}^{\frac{-2\text{t}}{\text{RC}}}\text{dt}=\text{i}_0^2\text{R}\int\text{e}^{\frac{-2\text{t}}{\text{RC}}}\text{dt}$
$=\text{i}^2_0\text{R}\Big(\frac{-\text{RC}}{2}\Big)\text{e}^{\frac{-2\text{t}}{\text{RC}}}=\frac{1}{2}\text{Ci}_0^2\text{R}^2\text{e}^{\frac{-2\text{t}}{\text{RC}}}=\frac{1}{2}\text{CV}^2$ (Proved).
$=\text{i}^2_0\text{R}\Big(\frac{-\text{RC}}{2}\Big)\text{e}^{\frac{-2\text{t}}{\text{RC}}}=\frac{1}{2}\text{Ci}_0^2\text{R}^2\text{e}^{\frac{-2\text{t}}{\text{RC}}}=\frac{1}{2}\text{CV}^2$ (Proved).
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