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) as switch 1 is suddenly closed

the inductor becomes open circuit

so current through the circuit = 0

VR1 = 0 IL =0 VR2 =0 IC =0 VL=0 VC =0

(b) Inductor is short circuit

so IL = V0/R1 VR1 = V0 VL =0 [as inductor is short circuit]

NO CURRENT FLOWS THROUGH R2 and C as Switch 2 is open

so VR2 =0 VC =0 IC =0

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