A \(5.0~\mu\text{F}\) capacitor having a charge of \(20~\mu\text{C}\) is discharged through a wire of resistance \(5.0~\Omega\). Find the heat dissipated in the wire between 25 to 50 \(\mu\text{s}\) after the connections are made.
\(40\left(\frac{1}{e^2} - \frac{1}{e^4}\right)~\mu\text{J}\)
\(40\left(\frac{1}{e} - \frac{1}{e^2}\right)~\mu\text{J}\)
\(40\left(\frac{1}{e^2} - \frac{1}{e^3}\right)~\mu\text{J}\)
Solution:
The remaining energy in the capacitor is \(U(t) = \frac{q_0^2}{2C}e^{-2t/tau}\), where \(tau = RC = 25~\mu\text{s}\). The heat dissipated is \(H = U(t_1) - U(t_2) = \frac{q_0^2}{2C}\left(e^{-2} - e^{-4}\right)\) where \(frac{q_0^2}{2C} = 40~\mu\text{J}\).
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