# Charging capacitor by another capacitor

by bgq
Tags: capacitor, charging
 Sci Advisor HW Helper Thanks P: 26,157 hi bgq! the energy loss equals I2R, so it's certainly the loss through the resistor from the pf library on capacitor … Inverse exponential rate of charging: A capacitor does not charge or discharge instantly. When a steady voltage $V_1$ is first applied, through a circuit of resistance $R$, to a capacitor across which there is already a voltage $V_0$, both the charging current $I$ in the circuit and the voltage difference $V_1\,-\,V$ change exponentially, with a parameter $-1/CR$: $$I(t) = \frac{V_1\,-\,V_0}{R}\,e^{-\frac{1}{CR}\,t}$$ $$V_1\ -\ V(t) = (V_1\,-\,V_0)\,e^{-\frac{1}{CR}\,t}$$ So the current becomes effectively zero, and the voltage across the capacitor becomes effectively $V_1$, after a time proportional to $CR$. Energy loss: Energy lost (to heat in the resistor): $$\int\,I^2(t)\,R\,dt\ =\ \frac{1}{2}\,C (V_1\,-\,V_0)^2$$