Network Capacitors: Calculate Equiv. Cap & Battery Charge

[purduegirl]

Homework Statement

See capacitor design in attachment.

Calculate the equivalent capacitance of the circuit shown in the diagram above; where C1 = 4.30 uF, C2 = 7.15 uF, C3 = 2.25 uF, and C4 = 3.75 uF.

How much charge is drawn from the battery?

Homework Equations

Parallel C = c1 +c2...
Series C = 1/c1 +1/c2 ...

C=Q/V

The Attempt at a Solution

For the first question, I drew the capacitor in a form that is more familiar than the triangle. I made C2 and C3 parallel to each other, and C4 in between them in the circuit. From there, I added C2 and C3 together, and since they are in parallel, I got the answer to be 9.4 uF. From there all the circuits are in series so I added 1/4.30 uF + 1/94 uF + 1/3.75 uF and got
1/.6056 or 1.65 uF. This however, is not the correct answer.

For the second question, I used C=Q/V, however, having C wrong, my answer was wrong.

 

[Doc Al]

I made C2 and C3 parallel to each other, and C4 in between them in the circuit.

This is incorrect. C2 is not parallel to C3 alone; C2 is parallel to the combination of C3 + C4. First combine C3 and C4, which are in series, with their equivalent capacitance.

 

[purduegirl]

Okay, I see what I was doing wrong. My next question is What is the voltage across C2 immediately after switch S is closed as in the diagram below? (Note, V=3.0 V.)

For this one, I calculated the Q = V*C. I got 8.59 uC for the charge, which is the answer for part three of charge pulled from the battery. I then set up my equation of V2 = Q2/C2
I got 1.20 V for the answer, but that isn't correct.

 

[Doc Al]

For this one, I calculated the Q = V*C. I got 8.59 uC for the charge, which is the answer for part three of charge pulled from the battery. I then set up my equation of V2 = Q2/C2
I got 1.20 V for the answer, but that isn't correct.

That charge is distributed across both C2 and C3/C4. If you use C2 alone, you need the charge on C2 alone. But you can also use the equivalent capacitance of C2 & C3+C4 (which you had to figure out earlier), then you can use the total charge.

 

[purduegirl]

So what you are saying is, that I would take the charge calculated earlier (8.59×10-6 C) and divide it by C of C2 & C3+C4.

I tried that and got 1.003 V and again it wasn't correct. I might not get what you are trying to say.

 

[Gear300]

Design makes it more confusing than it is. C3 and C4 are in series while at the same time, they are parallel to C2. If you take your finger and trace the circuit, you'll notice that when taking the path with C3 you'll run into C4; that does not happen with the path carrying C2.

 

[purduegirl]

I get that part. I don't understand how to the voltage just across C2.

 

[Doc Al]

Realize that, since they are in parallel, the voltage across C2 is the same as that across C3 + C4. So you can just find the voltage across their equivalent capacitance.

 

[purduegirl]

So what I'd have here is the C(3+4) = Q/V
C(3+4) = 1/C = 1/2.25 uF + 1/3.75 uF
1/C = 0.7111
C = 1.41E-6 F
So it would be 1.40 uF = 8.59×10-6 C/V
V = 8.59×10-6 C/ 1.41E-6 F
V = 6.09 V

This still isn't correct. What am I still doing wrong here?

 

[LowlyPion]

So what I'd have here is the C(3+4) = Q/V
C(3+4) = 1/C = 1/2.25 uF + 1/3.75 uF
1/C = 0.7111
C = 1.41E-6 F
So it would be 1.40 uF = 8.59×10-6 C/V
V = 8.59×10-6 C/ 1.41E-6 F
V = 6.09 V

This still isn't correct. What am I still doing wrong here?

You need to find the equivalent capacitance. That is a 3 step process here before you apply any voltages.
First C3 and C4 in series as Doc Al suggested.
That result - that equivalent C - gets combined with what and how - series or parallel?
And so on until you have an equivalent C for the whole mess.

You have the formulas there already in your 1st post.

 

[alphysicist]

Hi purduegirl,

So what I'd have here is the C(3+4) = Q/V
C(3+4) = 1/C = 1/2.25 uF + 1/3.75 uF
1/C = 0.7111
C = 1.41E-6 F
So it would be 1.40 uF = 8.59×10-6 C/V
V = 8.59×10-6 C/ 1.41E-6 F
V = 6.09 V

This still isn't correct. What am I still doing wrong here?

When you use V=Q/C, you need all of the quantities to refer to the same thing.

In the above you are using the equivalent capacitance of of capacitors (C3 & C4). However, the charge you are using is not the charge that is on capacitors (C3 & C4).

When you calculated the charge 8.59×10-6 C, it was the charge on the equivalent capacitance from combining all four capacitors.

Since that total capacitance came from a series combination of C1 and the total of (C2,C3,C4), that same total charge 8.59×10-6 C is the charge on C1. What is the total charge on the combined equivalent capacitance of (C2,C3,C4)? Do you see what to do now?

 

[purduegirl]

No, not really. I have tried the suggestions of getting the capitance C3 and C4 and then dividing it by the total charge from an earlier post, which didn't work. I'm still really confused.

 

[alphysicist]

No, not really. I have tried the suggestions of getting the capitance C3 and C4 and then dividing it by the total charge from an earlier post, which didn't work. I'm still really confused.

You have already found the total charge over the total combined capacitance to be [itex]8.59\times 10^{-6}[/itex] C.

A couple of questions (no calculator required): Now that you have that total charge, what is the charge on C1? How is the voltage of C1 and the voltage of C2 related?

 

[purduegirl]

The voltage across C1 should be 3V because the source of the charge is coming from the battery. Because C2 is in series with C1, it too should have the same charge on the battery, right?

 

[alphysicist]

The voltage across C1 should be 3V because the source of the charge is coming from the battery.

The voltage across the entire set of capacitors (from one end C1 to the far end of C2) should be 3V, because those are connected to the 3V battery. So how is the voltage of C1 and C2 related?

Let's suppose a mountain is 3km high, and you climb partway (but no halfway) up the mountain one day, and reach the top on the second day. Now think about the vertical distance you climbed each day. Since you reached the top, what has to be true about those two vertical displacements?

Because C2 is in series with C1, it too should have the same charge on the battery, right?

C2 and C1 are not in series with each other; C1 is in series with the combined capacitance of (C2 & C3 & C4).

So C1234 (the total equivalent capacitance) had a charge of [itex]8.59\times 10^{-6}[/itex] C, and C1234 came from combining C1 and C234 in series. Based just on that, what is the charge on C1?




I did have a question; it seemed to me like everyone was answering the question "what is the voltage on C2 when the capacitors are fully charged?", so that is what I was thinking about. But I looked back at your post #3, and you asked what is the voltage across C2 immediately after the switch is closed. Is that what you were trying to find? Is the same diagram as in your original post supposed to be used?

If that is the question, then the words immediately after would be very important, because as long as there is any resistance at all (even in the wires or batteries), that would mean the charge has not had a chance to build up on the capacitors. If that's the case, then the problem would be very different and much more straightforward.

 

[purduegirl]

Yes, that is the question, and the diagram is still supposed to be used.

 

[alphysicist]

Yes, that is the question, and the diagram is still supposed to be used.

Well there is always some resistance in the circuit, so if they ask for the voltage immediately after the switch is thrown, there would be no charge built up yet on the capacitor. What would the voltage across the capacitor be if there is no charge on it?

 

[LowlyPion]

So what I'd have here is the C(3+4) = Q/V
C(3+4) = 1/C = 1/2.25 uF + 1/3.75 uF
1/C = 0.7111
C = 1.41E-6 F
So it would be 1.40 uF = 8.59×10-6 C/V
V = 8.59×10-6 C/ 1.41E-6 F
V = 6.09 V

This still isn't correct. What am I still doing wrong here?

This part is not correct. The 1.4 uf is the equivalent capacitance of just C34. And not all of the charge will be there across just those capacitors.
The 8.59 x 10^-6 c is however, the charge across the equivalent C1234 and that number looks correct to me, but it is based on the equivalent capacitance of 2.86 uf that I calculate for C1234 and the 3 volt of the battery.

Your original statement asked:
"How much charge is drawn from the battery?"

Did it ask you to separate the charge distribution across all plates of the network? Or what the internal node voltages may be? Because I guess I'm not seeing that in the OP.

 

[alphysicist]

Hi LowlyPion,

This part is not correct. The 1.4 uf is the equivalent capacitance of just C34. And not all of the charge will be there across just those capacitors.
The 8.59 x 10^-6 c is however, the charge across the equivalent C1234 and that number looks correct to me, but it is based on the equivalent capacitance of 2.86 uf that I calculate for C1234 and the 3 volt of the battery.

Your original statement asked:
"How much charge is drawn from the battery?"

Did it ask you to separate the charge distribution across all plates of the network? Or what the internal node voltages may be? Because I guess I'm not seeing that in the OP.

I pointed that error out in one of my earlier posts, but after I read the question (it was a followup question in post #3), I don't think any calculation is required at all. What do you think?

 

[LowlyPion]

I don't think any calculation is required at all. What do you think?

Well, no, since you asked. I think it is already solved, if that is the question in the original statement. It seemed things were getting way too complicated.

An interesting follow up exercise is to figure the node voltages and the charge distribution, but I sense solving the problem is all that was asked. The problem as stated seems aimed at teaching the concept of calculating layers of equivalent capacitance with mixed parallel and series components.

Edit: I see now in post #3 there is a follow-up, sorry I missed it. But the term "immediately after" I read to be before the capacitor is fully charged. The charge will build up rapidly, but so will the voltage across C2 corresponding to the build up of charge. But lest we forget, the total charge at C2 is shared at steady state by the C34 leg as well on one side and by the C1 in series above. At least that is my sense of the schematic. Maybe a calculation of the inner node voltage at steady state is what is being asked in that part?

 

[LowlyPion]

OK. Here it is again. I'm pretty sure they are asking for the node voltage at steady state, according to the follow up in Post #3.

That is fairly easy to calculate.
Note that the Total voltage is going to be 3V from top to bottom of the network.

This can be expressed as:
[tex]V_t_o_t = V_c_1 + V_(_c_2_+_c_3_4_)[/tex]

The easiest way to calculate the node voltage is to calculate the voltage drop across C1.

To do this you note that
[tex]V_c_1 = 8.58 * 10^-^6 / C_c_1[/tex] where [tex]C_c_1[/tex] is given as 4.3 uf

That yields 2V across C1.

Checking against the bottom mess namely the equivalent capacitance of C234 that has a value of 8.55 uf, that comes out to be 1V. Happily 2 + 1 = 3 and the voltages check out. Fini.

 

[alphysicist]

LowlyPion,

OK. Here it is again. I'm pretty sure they are asking for the node voltage at steady state, according to the follow up in Post #3.

That is fairly easy to calculate.
Note that the Total voltage is going to be 3V from top to bottom of the network.

This can be expressed as:
[tex]V_t_o_t = V_c_1 + V_(_c_2_+_c_3_4_)[/tex]

The easiest way to calculate that is to calculate the voltage drop across C1.

To do this you note that
[tex]V_c_1 = 8.58 * 10^-^6 / C_c_1[/tex] where [tex]C_c_1[/tex] is given as 4.3 uf

That yields 2V across C1.

Checking against the bottom mess namely the equivalent capacitance of C234 that has a value of 8.55 uf, that comes out to be 1V. Happily 2 + 1 = 3 and the voltages check out. Fini.

That's what I was suggesting to purduegirl at first; however, when I read the question, it asks for the voltage across C2 immediately after the switch is closed. To me, I believe that means before charge has had a chance to build up on the capacitor at all. (There is always some amount of resistance in the circuit.)

So if no charge, then using Q = CV means you need no calculation at all to find the voltage.

(At least, that's how I'm reading the question.)

 

[LowlyPion]

I agree. I found that confusing too, if you see my earlier remarks it is easy to see. But I think the object lesson being taught is calculating equivalent capacitance and calculating node voltages. That was why I went ahead and wrote it out.

As you begin to interpret "immediately" in near relativistic terms, I think that begins to lie outside the scope of the intended material. This must be what is being asked, since it is a pretty elementary problem.