# Already-charged Capacitor Voltage time function

1. May 19, 2013

### libelec

1. The problem statement, all variables and given/known data

Find V(a) as a function of time in the following circuit when the switch is opened. Originally, the capacitor is charged to Vc = 1/101 * V1.

3. The attempt at a solution

I know that when the switch is opened, the capacitor is going to get charged again, and V(a) is going to be the same as its voltage. I know the general expression for the voltage of a charging capacitor is:

Vc(t) = V0 + V * (1 - e-t/RC)

Where Vc is the voltage of the capacitor, V0 is its original voltage, V is the voltage of the source, R the resistance and C its capacitance.

So, V(a)(t) is going to be the same as that. The problem is that I know that, in its stationary, Vc has to be equal to V, yet the expression I found doesn't do that. In this problem, I found that V0 = 0,29V, V = 30V, C = 2200μF and R... is it 100Ω? 100100Ω?

In an experiment I made (from which this questions comes), after 160 seconds V(a) was around 30V, but if I replace 160s in t in that expression, I get something closer to 17V. I thought that it has to be because I'm not calculating the R of the equivalent RC circuit right. Is that it?

Thanks

Last edited: May 19, 2013
2. May 19, 2013

### rude man

Circuit makes no sense. Opening the switch does nothing since it's permanently shorted by a wire in parallel with it.

Also, Va is shown as a source on top of the output voltage which also makes no sense.

3. May 19, 2013

### libelec

Yeah, sorry. V(a) isn't supposed to be a a source, it was just the only way I had to draw it.

This is the circuit written on the assignment, I'll edit it:

Pay no attention to the rest of the circuit. Just the bottom half.

4. May 19, 2013

### libelec

Nobody?

5. May 20, 2013

### Staff: Mentor

Concerning your question: V0 and V should be the same if the final voltage is V0. If your capacitor starts with a non-zero charge, modify the exponential instead of the prefactors:
$$V_c(t)=V_0 \left(1-e^{\frac{t-t_0}{RC}}\right)$$
Alternatively, let t=0 refer to some unknown point before the experiment starts, and determine t where the capacitor is charged to its initial voltage.

6. May 20, 2013

### libelec

The problem is that the previous voltage over the capacitor is the one it has in its stationary while the switch L1 is closed. Then when its opened, is when the experiment begins. So that rule for the voltage over the capacitor isn't valid for the previous situation while it is loading at the beginning, only when it is after the switch is opened.

Is this OK?

7. May 20, 2013

### Staff: Mentor

If I understand your problem, you're worried about how the initial potential across the capacitor will affect the expression that describes the potential after the switch opens. Your equivalent circuit would look something like this:

The thing to do is to find the ΔV that will applies. That is, suppose the capacitor potential starts at V1 and and will head towards a final value of V2. Then ΔV = V2 - V1, and the capacitor potential will follow the curve:
$$Vc(t) = V1 + ΔV e^{-t/\tau}$$

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8. May 20, 2013

### libelec

I understand. So it would be something like:

Vc = V1 + (V2 - V1) * e-t/τ

Is this correct? One final question: the total R in this circuit, is it 10100ohm?

9. May 20, 2013

### Staff: Mentor

Yes, see my post above
No, the resistance that the capacitor "sees" will depend upon the position of the switch. When the switch is open, the 100Ω resistor effectively disappears from the circuit. While the switch is closed, there is a voltage divider situation. You should be able to work out the effective resistance and initial potential by finding the Thevenin equivalent model for the source and resistor network.

10. May 20, 2013

### libelec

Thank you very much.