An inductor is connected to the terminals of a battery that has an emf of 12.0 V and negligible internal resistance. The current is 4.86 mA at 0.725 ms after the connection is completed. After a long time, the current is 6.45 mA.

(a) What is the resistance R of the inductor?
(b) What is the inductance L of the inductor?

(a) 1860 Ω

(b) 0.16H

This is an RL circuit as it contains an inductor with and inherent resistance.

The current, I, flowing through such circuit is given by;

I = [1 - ]          --------------(i)

Where;

E = emf of the circuit = 12.0V

R = Resistance of the inductor

L = inductance of the inductor

t = time taken for the flow of current

(a) After a long time, i.e t → ∞, the current becomes saturated (i.e maximum) and the value is 6.45mA.

At this time, (from equation (i)) becomes 0 and current I, becomes;

I =   -------------------(ii)

Now substitute the value of E and I into equation (ii) as follows;

6.45 x 10⁻³=

R =

R = 1860Ω

Therefore, the resistance of the inductor is 1860 Ω

(b) To get the inductance L of the inductor, we substitute I = 4.86mA at time t = 0.725ms into equation (i) as follows;

4.86 x 10⁻³ =  [1 - ]

4.86 x 10⁻³ x 1860 = 12.0[1 - ]

0.7533 = [1 - ]

1  - 0.7533 =

0.2467 =

Take the natural log of both sides

ln(0.2467) = -1860 x 0.00468 / L

-1.3996 = -8.7048 / L

L = -1.3996 / -8.7048

L = 0.16H

Therefore, the inductance of the inductor is 0.16H

answer; (arch); good luck