Two resistors R 1 = 15.0 , and R 2 = 11.5 are connected with a 350 mH inductor,

Question

Two resistors R1 = 15.0 , and R2 = 11.5 are connected with a 350 mH inductor, a 12.0 V battery and a two-way switch as shown in the diagram below. At t = 0, the switch ab is closed.

Now the switch ab is opened and ac is closed.

(d) Determine the time constant for this circuit after ac is closed. (Enter your answer to at least four decimal places.)

1.32e-2 s

(e) What is the current in the inductor at t = 0.006 s after ac is closed?
A

(f) What is the voltage across the two resistors and the inductor at t = 0.006 s after ac is closed?

Explanation / Answer

d)

time constant, T = L/Rnet

Rnet = R1 + R2 = 15 + 11.5 = 26.5 ohm

So, T = 0.35/26.5 = 0.0132 s

e)

current in the inductor is given by:

I = Io*e^(-t/T)

where Io = V/R1 = 12/15 = 0.8 A

So, I = 0.8*e^(-t/0.0132)

So, I = 0.8*e^(-0.006/0,0132) = 0.51 A

f)

So, VR1 = 0.51*15 = 7.7 V

VR2 = 0.51*11.5 = 5.9 V

VL = 7.7 + 5.9 = 13.6 V

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