Solve the following :

Question:
The electric current in a discharging $\mathrm{R}-\mathrm{C}$ circuit is given by $\mathrm{i}=\mathrm{i}_{0} \mathrm{e}^{-\mathrm{t} R \mathrm{RC}}$ 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 $\mathrm{t}=0.31 \mathrm{~s}$.

Solution:

$i=i_{0} e^{-t / R C}$

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) i $\left.\right|_{t=0.3}=2 \times e(-0.3 / 0.3)=2 / e$ Amperes.

c) $\left.\mathrm{i}\right|_{\mathrm{t}=0.31}=5.8 / 3 \mathrm{e}$ Amperes

Solve the following :

Leave a comment

All Study Material