- 182

- 8

- Summary
- Resistor R charging a capacitance C. The energy dissipated by R is equal to the energy stored by C.

Assume that a resistor R charges a capacitor C, whose other terminal is connected to the ground.

The charge at time t = 0 is assumed to be null and the supply voltage is equal to V.

We have, as is well known, ##i = \frac{V}{R} e^{-\frac{t}{RC}}##. Integrating ##\frac{i^2}{R}## between t = 0 and infinity, I found that the energy dissipated by R is ##\frac{1}{2} CV^2##. That is, the same as the potential energy stored in the capacitor after it has completed its charging.

Is there an intuitive way to understand this equality ?

The charge at time t = 0 is assumed to be null and the supply voltage is equal to V.

We have, as is well known, ##i = \frac{V}{R} e^{-\frac{t}{RC}}##. Integrating ##\frac{i^2}{R}## between t = 0 and infinity, I found that the energy dissipated by R is ##\frac{1}{2} CV^2##. That is, the same as the potential energy stored in the capacitor after it has completed its charging.

Is there an intuitive way to understand this equality ?