A coil of resistance `40 Omega` 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.

Explanation

Here ` R = 40Omega, E = 4V ,`
` t = 0.1, `
` i= 63mA `
` i=i_0(1-e^(-t/tau))`
` rArr 63 xx (10^-3) = 4/40 (1- e^(-tR/L)) `
` rArr 63 xx (10^-3) = (10^-1) (1-e^(-tR/L)) `
` rArr 63 xx (10^-2) = (1- e^(-4/L)) `
` rArr 1-0.63 = e^(4/L) `
` e^(-4/L) = 0.37 `
` rArr -4/L = In (0.37) = - 0.994 `
` L = -4/-0.994 `
` = 4.024 H = 4H. `