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The contact switch in this circuit has been open for a very long time. The switch is connected to the circuit at time t=0.
What is the voltage across the inductor (VL) immediately after the switch is closed?
The answer to this question is 3 V, but why isn't it VL = 5 V?
I thought the inductor would have the voltage source as it's own voltage immediately after it is connected to the voltage source?
No because at t = 0, there is no current flowing through the inductor but there is a current flowing through both resistors.
Hence at t = 0 you have a voltage divider VL = 3/(3+2) * 5V = 3V
Extra credit: if the switch is closed for a long time, the voltage accross the inductor is 0. Hence no current is flowing through the 3kOhm resistor, and the current flowing through the inductor is 5/2k = 2.5mA. The switch is opened: the current flowing through the inductor is still 2.5mA. Because the switch is open, this is also the current flowing through the 3kOhm resistor. Hence the voltage accross the inductor is -(3k*2.5mA) = -7.5V.
