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

Inductance of Coil is 4 H

Explanation:

Given,

R = 40$\Omega$, E = 4V, t = 0.1, i = 63 mA

i = $i_0 - (1 - e^{tR/2})$

$63 \times 10^-^3 = \frac {40}{4} (1 - e^{-4/L})$

$63 \times 10^-^3 = 10^-^1(1 - e^{-4/L})$

$63 \times 10^-^2 = (1 - e^{-4/L})$

$1 - 0.63 = e^{-4/L}$

$e^{-4/L} = 0.37$

$-\frac {4}{L} =ln(0.37) = -0.994$

$L = \frac {-4}{-0.994}$

4.024 H = 4H