The electric current in a discharging $\mathrm{R}-\mathrm{C}$ circuit is given by $\mathrm{i}=\mathrm{i}_{0} \mathrm{e}^{-/ / R C}$ where $\mathrm{i}_{0}, \mathrm{R}$ and $\mathrm{C}$ are constant parameters of the circuit and $\mathrm{t}$ is time. Let $\mathrm{i}_{0}=2.00 \mathrm{~A}, \mathrm{R}=6.00 \times 10^{5} \Omega$ and $\mathrm{C}=0.500 \mu \mathrm{F}$. (a) Find the current at $t=0.3 \mathrm{~s}$. (b) Find the rate of change of current at $t=0.3 \mathrm{~s}$. (c) Find approximately the current at $\mathrm{t}=0.31 \mathrm{~s}$.
$i=i_{0} e^{-t / R C}$
Here, $\mathrm{i}_{0}=2 \mathrm{~A}, \mathrm{R}=6 \times 10^{5} \Omega, \mathrm{C}=5 \times 10^{-6} \mathrm{~F}=5 \times 10^{-7} \mathrm{~F}$
(a) $\mathrm{i} \mid \mathrm{t}=0.3=2 \times \mathrm{e}(-0.3 / 0.3)=2 / \mathrm{e}$ Amperes.
(b) $\mathrm{di} /\left.\mathrm{dt}\right|_{\mathrm{t}=0.3}=-2 \mathrm{e}(-0.3 / 0.3) / 0.3=-20 / 3 \mathrm{e}$
(c) $\left.\mathrm{i}\right|_{\mathrm{t}=0.31}=5.8 / 3 \mathrm{e}$ Amperes