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.
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?
1 answer
1 answer