Question

A coil of resistance 40 Ω is connected across a 4.0 V battery. 0.10 s after the battery is connected, the current in the coil is 63 mA. Find the inductance of the coil.

Answer 1

Given:
Resistance, R = 40 Ω
Emf of the battery, E = 4 V

Now,
The steady-state current in the LR circuit is given by
$i_0=\frac{4}{40}=0.1\mathrm{A}$
At time, t = 0.1 s, the value of current i is 63 mA = 0.063 A
The current at time t is given by
i = i₀(1 − e^−t/τ)
⇒ 0.063 = 0.1(1 − e^−tR/L)
⇒ 63 =100(1 − e^−4/L)
⇒ 63 = 100(1 − e^−4/L)
⇒ 1 − 0.63 = e^−4/L
⇒ e^−4/L = 0.37
⇒ $-\frac{4}{L}$ = ln (0.37)
⇒ L = $\frac{-4}{-0.994}$

= 4.024 H
= 4 H