One application of an R L circuit is the generation of time-varying high voltage

Question

One application of an RL circuit is the generation of time-varying high voltage from a low-voltage source. As shown in the figure below, R1 = 1030. , R2 = 12.0 , L = 2.10 H and E = 18.00 V.

a) What is the current in the circuit a long time after the switch has been in position A?

Now the switch is thrown quickly from A to B.

b) Compute the initial voltage across R1.

c) Compute the initial voltage across R2.

d) What is the initial voltage across the inductor?

e) How much time elapses before the voltage across the inductor drops to E = 18.00 V?

Explanation / Answer

(a)

Current after a long time = 18/12 = 1.5 A

Current will be 1.5 in the circuit.

(b)

Voltage across R1 = 1030*1.5 = 1545 V

(C)

Voltage across R2 = 12*1.5 = 18 V

(d)

Initial voltage = 1545+18 = 1563 V

(e)

V(t)=1563*e^-t*1042/2.1

18=1563*e^-t*1042/2.1

ln(18/1563)=-t*1042/2.1

-4.464=-t*1042/2.1

t=8.997*10^-3 s or 8.993 ms

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