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1 Just before the switch is opened at t = 0 in Figure P5.21, the current through the inductor is 1.70 mA in the direction shown. Vs = 12 V L = 0.9 mH R = 6 k2 R2 = 6 k2 R = 3 k2 Determine the time constant of the circuit for t > 0. t= 0 R2 iL Vs R1 L RV33

1 Just before the switch is opened at t = 0 in Figure P5.21, the current through the inductor is 1.70 mA in the direction shown. Vs = 12 V L = 0.9 mH R = 6 k2 R2 = 6 k2 R = 3 k2 Determine the time constant of the circuit for t > 0. t= 0 R2 iL Vs R1 L RV33

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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1 Just before the switch is opened at t = 0 in Figure P5.21, the current through the inductor is 1.70 mA in the direction shown. Vs = 12 V L = 0.9 mH R = 6 k2 R2 = 6 k2 R = 3 k2 Determine the time constant of the circuit for t > 0.
Transcribed Image Text:1 Just before the switch is opened at t = 0 in Figure P5.21, the current through the inductor is 1.70 mA in the direction shown. Vs = 12 V L = 0.9 mH R = 6 k2 R2 = 6 k2 R = 3 k2 Determine the time constant of the circuit for t > 0.
t= 0 R2 iL Vs R1 L RV33
Transcribed Image Text:t= 0 R2 iL Vs R1 L RV33
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