# Two Loop RC Circuit 2

1. Feb 22, 2013

### Gee Wiz

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

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 22 Ω, R3 = 91 Ω and R4 = 129 Ω. The capacitance is C = 40 μF and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.( picture: https://www.smartphysics.com/Content/Media/Images/EM/11/h11_RC_time.png [Broken])

After the switch has been closed for a very long time, it is then opened. What is Q(topen), the charge on the capacitor at a time topen = 790 μs after the switch was opened?

2. Relevant equations
Q(t)=Qmax(1-e^(-t/rc))

3. The attempt at a solution
I found earlier that the Q at infinity is 180.496uC. I assume that would be my Qmax. My t would be 790us. I think im messing up getting rc. My final answer comes out as negative each time. (clearly not correct)

Last edited by a moderator: May 6, 2017
2. Feb 22, 2013

### rude man

3. Feb 22, 2013

### Gee Wiz

No, but it really is. The online homework marked that as correct

4. Feb 22, 2013

### Gee Wiz

the current approaches zero as time goes to infinity, and q approaches it max value as time goes to infinity. Q is zero in the beginning

5. Feb 22, 2013

### Staff: Mentor

Gee Wiz, do you think that the charge on the capacitor will increase or decrease over time after the switch opens? What does your Relevant Equation predict?

6. Feb 22, 2013

### Gee Wiz

decrease, because it will be discharging.

7. Feb 22, 2013

### Gee Wiz

i think i'm struggling to find the time constant r*c. I know that c is just the capacitance, so 40e-6F in this case. But is r the total resistance of the circuit, or part of it?

8. Feb 22, 2013

### Staff: Mentor

Then consider the curve that your Relevant Equation describes at time increases.

9. Feb 22, 2013

### Gee Wiz

Well, it is exponential. Is that what you're getting at? or am i off the mark?

10. Feb 22, 2013

### Staff: Mentor

r is the total resistance that the capacitor "sees"; it is the equivalent resistance that the network presents to the capacitor at the terminals where it is connected. So in this case it is the sum of the series connected resistors.

11. Feb 22, 2013

### Gee Wiz

So, in this case it would be R1 and R2?

12. Feb 22, 2013

### Staff: Mentor

Exponential is correct. But should it be exponentially deceasing, or increasing to a maximum plateau? What curve does your Relevant equation describe?

13. Feb 22, 2013

### Staff: Mentor

Is there a complete path (a circuit) though R1 and R2 to the capacitor when the switch is open?

14. Feb 22, 2013

### Gee Wiz

Well i believe it would be exponentially decreasing to zero. Because once a capacitor is fully charged (for the circuit that it is in) the current of the through the capacitor is zero. So, any resistor in series with the capacitor would have a net voltage drop of zero. The capacitor has a voltage drop which is calculated v=Q/C. Before the capacitor if fully charged, or rather has no charge t=0, it acts almost like a wire. There is not voltage drop across it, and the current is maxed out.

15. Feb 22, 2013

### Staff: Mentor

So what, then, should be the equation describing the charge (or voltage) on the capacitor after the switch is opened?

16. Feb 22, 2013

### Gee Wiz

Well immediately after it would just be C=Q(max)/V...?

17. Feb 22, 2013

### Staff: Mentor

Yes, that's the initial condition. But what equation will the charge (or voltage) follow after that?

18. Feb 22, 2013

### Gee Wiz

q(t)=q(max)(1-e^(-t/(r*c)))

19. Feb 22, 2013

### Staff: Mentor

Why don't you plot that equation. What shape does it have? Suppose t = 0, what is q(0)?

20. Feb 22, 2013

### Gee Wiz

It has exponential shape (um..yes im aware that is poor description =/ ). It is increasing rapidly at first and then levels off as it approaches its max value. At t=0 (assuming the switch was open before) q=0