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 Q_capacitor = 0. At time t = 0, Switch #2 remains OPEN but Switch 1 is suddenly CLOSED. List the instantaneous values at t = 0 of: V_R1 = Volts. V_R@ = Volts. I_L = Amps I_C = Amps Switch #2 remains OPEN and Switch 1 has now been CLOSED for a long, long time. List the steady-state values of: V_R1 = Volts. V_R2 = Volts. V_L = Volts. I_L = Amps I_C = Amps V_C = Volts 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: V_R1 = Volts. V_R2 = Volts. V_L = Volts. I_L = Amps I_C = Amps V_C = 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.

(1) VR1 = 0
(2) VR2 = 0
(3) VL = Vo

(4 ) IL = 0
(5) IC = 0
(6) VC = 0

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

(1) VR1 = Vo
(2) VR2 = 0
(3) VL = 0

(4) IL = Vo/R1
(5 )IC = 0
(6) VC = 0

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