Solve the following :

Question:

The electric current in a charging $R-C$ circuit is given by $i=i_{0} e^{-t / R C}$ where $\mathrm{i}_{0}, R$ 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) $\mathrm{t}=10 \mathrm{RC}$.

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 $t=0, \mathrm{di} / \mathrm{dt}=-\mathrm{i}_{0} / \mathrm{RC}$

b) When $\mathrm{t}=\mathrm{RC}, \mathrm{di} / \mathrm{dt}=-\mathrm{i}_{0} / \mathrm{RCe}$

c) When t $=10 \mathrm{RC}, \mathrm{di} / \mathrm{dt}=-\frac{\hat{i}_{O}}{R C e^{10}}$

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