Physics
posted by Sam .
A 1.0 KΩ resistor is connected in series with a 10.0 mH inductor, a 30.0 V battery, and an open switch. At time t=0, the switch is suddenly closed. What are the voltage drops VR and VL 20.00 µs after the switch is closed?

VL = Vap / e^t/T,
T=L/R=10*10^3h/1*10^3Ohms=10*10^6s.
t/T = 20*10^6s / 10*10^6s = 2.
VL = 30 / e^2 4.06 volts.
VR = 30  4.06 = 25.94 volts.
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