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:
$\tau =\frac{L}{R}=\frac{1}{20}=0.05\mathrm{s}$
Steady-state current:
$i_0=\frac{e}{R}=\frac{2}{20}=0.1\mathrm{A}$
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
$$\begin{gathered} \frac{di}{dt}=-(i_0\times \left(\frac{-1}{\tau }\right)e^{-t/\tau }) \\ =\frac{i_0}{\tau }{e}^{-t/\tau } \end{gathered}$$
(a) At time t = 100 ms,
$\frac{di}{dt}=\frac{0.1}{0.05}\times e^{-0.1/0.05}=0.27\mathrm{A}/\mathrm{s}$

(b) At time t = 200 ms,
$$\begin{gathered} \frac{di}{dt}=\frac{0.1}{0.05}\times e^{-0.2/0.05} \\ =0.0366\mathrm{A}/\mathrm{s} \end{gathered}$$

(c) At time t = 1 s,
$$\begin{gathered} \frac{di}{dt}=\frac{0.1}{0.05}\times e^{-1/0.05} \\ =41\times {10}^{-9}\mathrm{A}/\mathrm{s} \end{gathered}$$