Please show all steps The figure on the left shows a circuit

The figure on the left shows a circuit that contains three identical resistors with resistance R = 12 ohm, two identical inductors with inductance L = 3.5 mH, and an ideal battery with emf epsilon = 17 V. a) What is the current through the battery just after the switch is closed? b) What is the current through the battery long after the switch has been closed? c) Now consider the circuit on the right with only one of these inductors and one of these resistors connected in series with the battery. How long will the current take to reach a third of its final equilibrium value? Why? a) I = epsilon/R = 17/12 = 1.416 b) I = 17/4 = 4.25 A

Solution

a)Just after the switch is closed,

The current will be:

I = e/R = 17/12 = 1.417

I = 1.417 A

since the drop across the other resistors is zero because of the presence of indcutors in the branch.

b)Long time after the switch is closed

1/Req = 1/12 + 1/12 + 1/12 = 3/12

Req = 4 Ohm

I = e/Req = 17/4 = 4.25 A

Hence, I = 4.25 A

c)we know that,

I = I(max) (1 - e^-tR/L)

1/3 I(max) = I(max) (1 - e^-tR/L)

0.33 = 1 - e^-tR/L

e^-tR/L = 0.67

taking natural log both sides

-tR/L = -0.4005

t = 0.4005 x L/R = 0.4005 x 3.5 x 10^-3/12 = 0.117 ms

Hence, t = 0.117 ms