Question

An L-R circuit has L = 1.0 H and R = 20 Ω. It is connected across an emf of 2.0 V at t = 0. Find di/dt at (a) t = 100 ms, (b) t = 200 ms and (c) t = 1.0 s.

Answer 1

Given:-

Inductance, L = 1.0 H

Resistance in the circuit, R = 20 Ω

Emf of the battery = 2.0 V

Now,

Time constant:-

\[\tau = \frac{L}{R} = \frac{1}{20} = 0 . 05 s\]

Steady-state current:-

\[i_0 = \frac{e}{R} = \frac{2}{20} = 0 . 1 A\]

Current at time t:-

i = i₀(1 − e^−t/τ)

or

i = i₀ − i₀(e^−t/τ)

On differentiating both sides with respect to t, we get

\[\frac{di}{dt} = - ( i_0 \times \left( \frac{- 1}{\tau} \right) e^{- t/\tau} )\]

\[ = \frac{i_0}{\tau} e^{- t/\tau}\]

(a) At time t = 100 ms,

\[\frac{di}{dt} = \frac{0 . 1}{0 . 05} \times e^{- 0 . 1/0 . 05} = 0 . 27 A/s\]

(b) At time t = 200 ms,

\[\frac{di}{dt} = \frac{0 . 1}{0 . 05} \times e^{- 0 . 2/0 . 05} \]

\[ = 0 . 0366 A/s\]

(c) At time t = 1 s,

\[\frac{di}{dt} = \frac{0 . 1}{0 . 05} \times e^{- 1/0 . 05} \]

\[ = 41 \times {10}^{- 9} A/s\]