A5.6 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

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