Question

A certain circuit consists of an inductor of
50 mH in series with a resistor of 130 Ω. At
one moment, the current in the circuit is 12 A,
and decreasing.
How long will it take for the current to fall
to 4.8 A?
Answer in units of s.

Answer 1

I assume a short circuit so total voltage is 0

V = L di/dt + iR = 0

let i = I e^kt
then when t =0, i = I
di/dt = I k e^kt = k i

V = L I k e^kt + I e^kt R = 0
L k + R = 0
k = -R/L but I bet you knew that

so
k = -130/50*10^-3 = -2.6*10^3
I = 12
so
i = 12 e^-2.6*10^3 t
4.8/12 = .4
ln .4 = -.916 = -2.6*10^3 t
solve for t (not long :)