Engineering Voltage drop across a capacitor in a circuit

[marstery]

Find the voltage drop across the capacitor

In a circuit where a voltage source is connected to a capacitor and resistor in series and then connected to another resistor, shouldn't the current going through the capacitor be 0ma after 5 time constants, resulting in a Vc being 0?

 

[MichaelXY]

But you must consider the initial current at time 0-5.

 

[marstery]

The initial conditions are that the switch has been in position 1 for a long time and I am supposed to find Vc(0) and V0(0-).

Since it has been a long time the capacitor i assumed that it acts as an open circuit so the current through the capacitor is 0ma and the current around the outer window is 12v/4k=3ma so V0(0-) is 3v. Current Ic is zero but was flowing to the right and Vc(0) is 12v since it has been a long time and the capacitor has charged fully.

I don't really know how to find the impedance, but instead of what's above I would take the current going into node A, 12v/[( R1 || cap's impedance )+R2], and set that equal to the current going through the capacitor + current through R1? I can't really picture it.

 

[MichaelXY]

Well that circuit is much different than your original description. Look at what is the voltage source for the cap is with switch in pos 1. Also note the caps neg ref.

 

[CEL]

The initial conditions are that the switch has been in position 1 for a long time and I am supposed to find Vc(0) and V0(0-).

Since it has been a long time the capacitor i assumed that it acts as an open circuit so the current through the capacitor is 0ma and the current around the outer window is 12v/4k=3ma so V0(0-) is 3v. Current Ic is zero but was flowing to the right and Vc(0) is 12v since it has been a long time and the capacitor has charged fully.

I don't really know how to find the impedance, but instead of what's above I would take the current going into node A, 12v/[( R1 || cap's impedance )+R2], and set that equal to the current going through the capacitor + current through R1? I can't really picture it.

As you said, i_C(0^-) = 0 and the current through the resistors is 3mA.
What is the voltage drop across the 3k resistor? The capacitor is in parallel with that resistor, so v_C(0^-) = V_R
When the switch commutes, the capacitor is in parallel with the 1k resistor and starts to discharge through it. What is the steady state voltage(after 5 time constants) across that resistor?

 

[marstery]

Michael, i said series when i really meant parallel. minor details, right? I'm a little confused about the direction that the capactor charges and discharges. does that happen in the same direction, or opposite directions?

 

[marstery]

Oooh, i get the voltage now--so Vc=Vr=9v. When you say that the capacitor discharges through the 1k resistor doesn't that oppose what M..XY was saying about the capacitor being directional?

The steady state voltage of R2 should be 3v. So when the switch is thrown the cap discharges according to Ic = Io e^-t/CR and the voltage can be found by summing the current from the capacitor and the 3ma from the voltage source, yes? If that is the case then when t=0 Ic = Io*1 so the new current through the 1k resistor would be 6ma and therefore V0(0+) would be 6v. is that right?

 

[CEL]

Oooh, i get the voltage now--so Vc=Vr=9v. When you say that the capacitor discharges through the 1k resistor doesn't that oppose what M..XY was saying about the capacitor being directional?

The steady state voltage of R2 should be 3v. So when the switch is thrown the cap discharges according to Ic = Io e^-t/CR and the voltage can be found by summing the current from the capacitor and the 3ma from the voltage source, yes? If that is the case then when t=0 Ic = Io*1 so the new current through the 1k resistor would be 6ma and therefore V0(0+) would be 6v. is that right?

Wrong! Since the resistor is in parallel with the capacitor it's voltage at 0^+ must be -9V(notice that the polarity of the capacitor is reversed). So there is a -9mA current through it (3mA from the source and -12mA from the capacitor.
And Michael never said that the capacitor was directional. He said that you should consider the reference voltage.
And I was wrong when I said that the capacitor would discharge through the resistor. In reality it will be charged in the reverse sense from the source, going from -9V to +3V.