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

\[i_0 = \frac{4}{40} = 0 . 1 A\]

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

\[\Rightarrow- \frac{4}{L}=\ln(0.37)\]

\[\Rightarrow L=\frac{- 4}{- 0 . 994}\]

= 4.024 H

= 4 H