Two resistors R1 = 15.0 , and R2 = 10.5 are connected with a 340 mH inductor, a

Question

Two resistors R1 = 15.0 , and R2 = 10.5 are connected with a 340 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.

(a) Determine the time constant for this circuit.

s

(b) Calculate the current in the two resistors and the inductor a long time after the switch is closed.

IR1 = A

IR2 = A

IL = A

(c) What is the voltage across the two resistors and the inductor a long time after the switch is closed?

VR1 = V

VR2 = V

VL = V

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.)

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?

VR1 = V

VR2 = V

VL = V

Explanation / Answer

a) Time constant

T = L/R1 = 0.340 / 15 = 0.023 sec

b)

after long time switch is closed

current in the circuit = i = V/R1 = 12 / 15 = 0.8 A

iR1 = 0.8 A

iL = 0.8 A

iR2 = 0 A                 since R2 is not connected

c)

after long time switch is closed

Voltage in R1 = iR1 R1 = 0.8 x 15 = 12 volts

Voltage in L = VL = 0             since inductor behaves as a short circuit after long time

Voltage in R2 = 0                  since R2 is not connected

d)

after AC is closed

T = time constant = L/(R1 + R2) = 0.340 / (15 + 10.5) = 0.013 sec

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