Question

The electric current in a discharging R−C circuit is given by i = i₀ e^−t/RC where i₀, R and C are constant parameters and t is time. Let i₀ = 2⋅00 A, R = 6⋅00 × 10⁵ Ω and C = 0⋅500 μF. (a) Find the current at t = 0⋅3 s. (b) Find the rate of change of current at at 0⋅3 s. (c) Find approximately the current at t = 0⋅31 s.

Answer 1

Electric current in a discharging R-C circuit is given by the below equation:
i = i₀ ⋅ e^−t/RC ...(i)
Here, i₀ = 2.00 A
R = 6 × 10⁵ Ω
C = 0.0500 × 10⁻⁶ F
= 5 × 10⁻⁷ F

On substituting the values of R, C and i₀ in equation (i), we get:
i = 2.0 e^−t/0.3 ...(ii)

According to the question, we have:

(a) current at t = 0.3 s
$i=2\times e^{-1}=\frac{2}{e}\mathrm{A}$

(b) rate of change of current at t = 0.3 s
$\frac{di}{dt}=\frac{-i_0}{RC}·e^{-t/RC}$
When t = 0.3 s, we have:

$\frac{di}{dt}=\frac{2}{0.30}·e^{\left(\frac{-0.3}{0.3}\right)}=\frac{-20}{3e}\mathrm{A}/\mathrm{s}$

(c) approximate current at t = 0.31 s
$$\begin{gathered} i=2e^{\left(\frac{-0.3}{0.3}\right)} \\ =\frac{5.8}{3e}\mathrm{A}\left(\mathrm{approx}.\right) \end{gathered}$$