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

\[\tau = \frac{L}{R} = \frac{5}{100}\]

The current at time t is given by

i = i₀(1 − e^−t/τ)

\[i = \frac{2}{100}\left( 1 - e^\left( \frac{- 2 \times {10}^{- 2} \times 100}{5} \right) \right)\]

\[ = \frac{2}{100}(1 - e^{- 2/5} )\]

\[ = \frac{2}{100}(1 - 0 . 670)\]

= 0.00659 = 0.0066

Now,

V = iR = 0.0066 × 100

= 0.66 V