Here are the initial conditions for the above circuit: Switch #1 is OPEN, Switch

Question

Here are the initial conditions for the above circuit: Switch #1 is OPEN, Switch #2 is OPEN, the capacitor is discharged, so Qcapacitor = 0.

(a) At time t=0, Switch #2 remains OPEN but Switch 1 is suddenly CLOSED. List the instantaneous values at t=0 of:

VR1 = _________ Volts. IL = _________Amps

VR2 = _________Volts. IC = _________Amps

VL = _________Volts. VC = _________Volts.

(b) Switch #2 remains OPEN and Switch 1 has now been CLOSED for a long, long time. List the steady-state values of:

VR1 = _________Volts. IL = _________Amps

VR2 = _________Volts. IC = _________Amps

VL = _________Volts. VC = _________Volts

(c) Suppose that both Switch #1 and Switch #2 have been closed for a long, long time Switch #1 is now suddenly OPENED. Immediately after Switch #1 is opened, list the instantaneous values of:

VR1 = _________Volts. IL = _________Amps

VR2 = _________Volts. IC = _________Amps

VL = _________Volts. VC = _________Volts

Explanation / Answer

(a)
Initially,
At t = 0, Capacitor acts as a short circuit while Inductor acts as an open circuit.

So, no current will flow through the circuit.

VR1 = 0
VR2 = 0
VL = Vo

IL = 0
IC = 0
VC = 0

(b)
After a long time, Max current will flow through the inductor.

VR1 = Vo
VR2 = 0
VL = 0

IL = Vo/R1
IC = 0
VC = 0

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