Question:
The electric current in a charging $R-C$ circuit is given by $i=i_{0} e^{-U R C}$ where $i_{0}, R$ and $C$ are constant parameters of the circuit and $t$ is time. Find the rate of change of current at (a) $t=0$, (b) $t=R C$, (c) $t=10 R C$.
Solution:
We have, $i=i_{0} e^{-t / R c}$
Rate of change of current $=\frac{d i}{d x}=\frac{d}{d x}\left(i_{0} e^{-\frac{t}{R C}}\right)=-\frac{i_{0}}{R C} \times e^{-\frac{t}{R C}}$
(a) When $\mathrm{t}=0, \mathrm{di} / \mathrm{dt}=-\mathrm{i}_{0} / \mathrm{RC}$
(b) When $\mathrm{t}=\mathrm{RC}, \mathrm{di} / \mathrm{dt}=-\mathrm{i} 0 / \mathrm{RCe}$
(c) When $\mathrm{t}=10 \mathrm{RC}, \mathrm{di} / \mathrm{dt}=-\frac{\mathrm{i}_{0}}{R C e^{10}}$