How Is Charge Calculated in a Two Loop RC Circuit After the Switch Opens?

[Gee Wiz]

Homework Statement

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 )

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?

Homework Equations

Q(t)=Qmax(1-e^(-t/rc))

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 I am messing up getting rc. My final answer comes out as negative each time. (clearly not correct)

 

[rude man]

1.

The Attempt at a Solution

I found earlier that the Q at infinity is 180.496uC. QUOTE]

Definitely, your Q at infinity is not 180.496 μC. It's a lot closer to zero ...

 

[Gee Wiz]

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

 

[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

 

[gneill]

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?

 

[Gee Wiz]

decrease, because it will be discharging.

 

[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?

 

[gneill]

decrease, because it will be discharging.

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

 

[Gee Wiz]

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

 

[gneill]

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?

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.

 

[Gee Wiz]

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

 

[gneill]

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

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

 

[gneill]

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

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

 

[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.

 

[gneill]

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.

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

 

[Gee Wiz]

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

 

[gneill]

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

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

 

[Gee Wiz]

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

 

[gneill]

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

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

 

[Gee Wiz]

It has exponential shape (um..yes I am 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

 

[gneill]

It has exponential shape (um..yes I am 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

Yes, very good. Now, at t=0 (when the switch opens) is q = 0 and does it then increase to a maximum value?

 

[Gee Wiz]

When you say when the switch opens, was it closed before?

 

[gneill]

When you say when the switch opens, was it closed before?

I am following the logic of the problem statement. The switch was closed for a long time, then opened at time t=0.

 

[Gee Wiz]

okay, well it then q would be at its max value and from there would be exponentially decreasing towards 0

 

[gneill]

okay, well it then q would be at its max value and from there would be exponentially decreasing towards 0

Yes, that's right. So what equation would describe that?

 

[Gee Wiz]

isn't it this one q(t)=q(max)(1-e^(-t/(r*c)))?

 

[gneill]

isn't it this one q(t)=q(max)(1-e^(-t/(r*c)))?

No. Since $e^0$ is 1, the curve starts from 0 and rises. This is because (1 - 1) = 0.
You want the curve to start with the maximum value and then decay down to zero. So get rid of the "1 - " bit and use just the $e^{-x}$.

 

[Gee Wiz]

Ahhh, okay. So the formula i gave is when q starts at zero. But, this problem q is starting out maxed out. so, from what you said it would just be q(t)=q(max)(e^(-t/(r*c)))?

 

[gneill]

Ahhh, okay. So the formula i gave is when q starts at zero. But, this problem q is starting out maxed out. so, from what you said it would just be q(t)=q(max)(e^(-t/(r*c)))?

Yes. I'm surprised that you didn't know this from your text or notes; surely there must be examples there?

 

[Gee Wiz]

We don't have a textbook for my class. The notes only have the original equation i posted. But, i think i should have been able to deduce this one since the current formula given is I(t)=I(0)(e^(-t/(r*c)))

 

[Gee Wiz]

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

Okay, so for this part (figuring out R for the time constant) do i just include the resistors that are in the new circuit with the capacitor. Let me try to explain, when the switch is open i can basically ignore that wire and everything that feeds off only that wire. So, in this case the battery, R4, and R1. Is that correct logic?

 

[gneill]

Okay, so for this part (figuring out R for the time constant) do i just include the resistors that are in the new circuit with the capacitor. Let me try to explain, when the switch is open i can basically ignore that wire and everything that feeds off only that wire. So, in this case the battery, R4, and R1. Is that correct logic?

Yes. Current can only flow when there is a closed loop.

 

[Gee Wiz]

Awesome. Thank you so much. The whole talk through was very helpful, and it was better than just getting the right answer.

 

[gneill]

Awesome. Thank you so much. The whole talk through was very helpful, and it was better than just getting the right answer.

I'm very happy to have helped

 

[rude man]

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

Did we ever come up with the right Qmax across C? Last I saw was Qmax = CV but V is the battery so that equation is incorrect.