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
\[i = 2 \times e^{- 1} = \frac{2}{e} A\]

(b) rate of change of current at t = 0.3 s
\[\frac{di}{dt} = \frac{- i_0}{RC} \cdot e^{- t/RC}\]
When t = 0.3 s, we have:
\[\frac{di}{dt} = \frac{2}{0 . 30} \cdot e^\left( \frac{- 0 . 3}{0 . 3} \right) = \frac{- 20}{3e} A/s\]

(c) approximate current at t = 0.31 s

\[i = 2 e^\left( \frac{- 0 . 3}{0 . 3} \right) \]

\[ = \frac{5 . 8}{3e} A \left( \text{ approx }. \right)\]