A 5.6 V battery is connected in series with a 41 mH inductor, a 100 Ω resistor, and an open switch. At what time after the switch is closed (t=0) will the current in the circuit be equal to 12 mA?

The current in a direct current resistor inductor circuit is given by:

i (t) = (-ℰ/R) e^{-Rt/L} + ℰ/R

Where i (t) is the current, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and L is the inductor's inductance.

Given values:

ℰ = 5.6V

R = 100Ω

L = 4.1*10⁻²H

Plug in the values to get i (t):

i (t) = - 0.056e^{-2440t} + 0.056

We want to calculate when the current is 0.012A, i. e. find a time t when i (t) = 0.012A. So let us set i (t) equal to 0.012 and solve for t:

-0.056e^{-2440t} + 0.056 = 0.012

0.056e^{-2440t} = 0.044

e^{-2440t} = 0.786

-2440t = ㏑ (0.786)

t = - ㏑ (0.786) / 2440

t = 9.87*10⁻⁵s