Question

An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 Ω resistor and a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the circuit is switched on.

Answer 1

Given:
Self-inductance, L = 5.0 H
Resistance, R = 100 Ω
Emf of the battery = 2.0 V

At time, t = 20 ms (after switching on the circuit)
t = 20 ms = 20 × 10⁻³ s = 2 × 10⁻² s
The steady-state current in the circuit is given by
$i_0=\frac{2}{100}$
The time constant is given by
$\tau =\frac{L}{R}=\frac{5}{100}$ s
The current at time t is given by
i = i₀(1 − e^−t/τ)
$$\begin{gathered} i=\frac{2}{100}\left(1-e^{\left(\frac{-2\times {10}^{-2}\times 100}{5}\right)}\right) \\ =\frac{2}{100}(1-e^{-2/5}) \\ =\frac{2}{100}(1-0.670) \end{gathered}$$
= 0.00659 = 0.0066
Now,
V = iR = 0.0066 × 100
= 0.66 V