Question

By evaluating ∫ i2Rdt, 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.

Answer 1

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Answer 2

Explanation:

The capacitor’s charge growth at time t,

Q = Q01-e-tRC

I = dQdt = Q0RCe-tRC

During time t1 to t2, the dissipated heat is

U = Q022Ce - 2t1RC – e -2t2RC

Therefore, Q0 = CV0, U = 12CV02e - 2t1RC – e - 2t2RC

Across a capacitor, the potential difference at any time t,

V = V01-e-tRC

In the capacitor, the energy stored at any time t,

E = 12CV021-e-2tRC2

Therefore, in the capacitor, the energy stored from t1 to t2,

⇒E = 12CV02e-2t1RC-e-2t2RC