Question

The electric current in a discharging R-C circuit is given by $i=i_o{e}^{\frac{-t}{RC}}$ where $i_o$ , R and C are constant parameters and$\text{t}$ is time. Let $i_o$ = 2.00 A, R=$6\times {10}^5$ Ω and C = 0.500 $\mu$ F.

(a) Find the current at t = 0.3 s.

(b) Find the rate of change of current at t = 0.3 s

(c) Find approximately the current at t = 0.31 s.

Answer 1

Step 1: Given Data:

Equation of current in R-C circuit is, $i=i_o{e}^{\frac{-t}{RC}}$

Where, $i_o$ = 2.00 A,

R=$6\times {10}^5$ Ω and

C = 0.500$\mu$F

Step 2: a) Calculation of current at t=0.3s:

$i=i_o{e}^{\frac{-t}{RC}}⟹i=2\times e^{\frac{-0.3}{6\times {10}^5\times 5\times {10}^{-6}}}$

$i=2\times e^{-1}⟹i=\frac{2}{e}A$

Step 3: b) Calculation of the rate of change of current at t = 0.3 s:

$i=i_o{e}^{\frac{-t}{RC}}⟹\frac{di}{dt}=\frac{d\left(i_o{e}^{\frac{-t}{RC}}\right)}{dt}$

$\frac{di}{dt}=\frac{-i_o}{RC}{e}^{\frac{-t}{RC}}$

At t= 0.3 sec

$\frac{di}{dt}=\frac{-\left(2\right)}{6\times {10}^5\times 0.5\times {10}^{-6}}{e}^{\frac{-t}{6\times {10}^5\times 0.5\times {10}^{-6}}}$

$\frac{di}{dt}=\frac{-20}{3}{e}^{-1}⟹\frac{di}{dt}=\frac{-20}{3e}ampere$

Step 4: c) Calculation of the current at t = 0.31 s:

$i=i_o{e}^{\frac{-t}{RC}}⟹i=2\times e^{\frac{-0.31}{6\times {10}^5\times 5\times {10}^{-6}}}$

$i=2\times e^{-\frac{0.31}{0.3}}$

$i=2\times e^{-1.03}⟹i=\frac{5.8}{3e}ampere$

Thus,

(a) The current at t = 0.3 s= $\frac{2}{e}A$

(b) The rate of change of current at t = 0.3 s= $\frac{-20}{3e}A$

(c) The current at t = 0.31 s= $\frac{5.8}{3e}A$