Maximizing Capacitance: Series vs. Parallel Methods for Battery Connections

[bertholf07]

A network of capacitors can be connected to a given battery in two fundamentally different ways-in serious and in parallel inlcuding any number of different combination of each. As a specific example take a seris set of capacitive blocks C1, C2, and C3 containing respectively one, two or three of the same capacitors in series in each cell. The total equivalent capacitance of the three cells in series must be one farad (1F). The capacitors may only be chosen from a slection of two, three, four , six, eight, nine or twelve farads. Parallel combinations within each cell may be used for other numbers. (Hint: this is comparable to solving for ni = 2, 3, 4, 6, 8, 9, 12 the following equation:) 1/n1 + 2/n2 + 3/n3 = 1

 

[chroot]

As a specific example take a seris set of capacitive blocks C1, C2, and C3 containing respectively one, two or three of the same capacitors in series[/color] in each cell.

Did you copy this problem down incorrectly? The word in red doesn't seem to agree with the rest of the problem.

- Warren

 

[bertholf07]

Yes I doubled checked it is right, basically what i get form this it there will be 3 capacitors in a series and those will be the 3 cells. Now in each cell there may be parallel capacitors..


-----|C1|--|C2|--|C3|------
| |
| |
|_________EMF___________|

 

[chroot]

The question does not make any sense, then. The problem goes on to state

Parallel combinations within each cell may be used for other numbers.

and

1/n1 + 2/n2 + 3/n3 = 1

which would indicate that the capacitors inside each cell are wired in parallel.

- Warren

 

[bertholf07]

I copied the question precisely from my worksheet and the total capacitance of the circuit must be 1F

 

[bertholf07]

Correct me if I am wrong but for capacitors in series you do 1/C(eq)=1/C1+1/C2+1/C3 and if they are in parallel it is just C(eq)=C1+C2+C3

 

[chroot]

Okay, well, I have no idea how the cells can be composed of 1, 2, or 3 capacitors in series and simultaneously be composed of capacitors in parallel, but let's just assume the capacitors are all in series.

The first cell is just a single capacitor, so it's total capacitance is just n1.

The second cell is composed of two identical capacitors wired in series, each of capacitance n2. Their total capacitance is:

$$\left( \frac{1}{n_2} + \frac{1}{n_2} \right)^{-1} = \frac{n_2}{2}$$

Similarly, the capacitance of the thid cell, composed of three identical capacitors wired in series, each of capacitance n3, is just n3/3.

The total capacitance of all three cells combined in series is

$$\left( 1/n_1 + 2/n_2 + 3/n_3 \right)^{-1}$$

Does that make sense?

- Warren

 

[bertholf07]

Yes I follow that...So the (1/n1+2/n2+3/n3)^-1 would have to equal the total capacitance of the circuit which would be 1F

 

[bertholf07]

I think I may not have explain myself the best but they in series and simultaneously be composed of capacitors in parallel. What you have are three cells which are you capacitors in series...now you can have those capacitors to be either 2, 3, 4, 6, 8, 9, or 12. Now within those three cells you can have one or all three of those having another or many other capacitors in parallel with just that one individually.

 

[chroot]

Yes, bertholf07. Can you find three values for n1, n2, and n3 to satisfy that equation? Hint: I wrote everything in twelfths to make the pattern easier to spot. In other words, write 2 as "12/6" and 3 as "12/4" and so on.

- Warren

 

[chroot]

I think I may not have explain myself the best but they in series and simultaneously be composed of capacitors in parallel. What you have are three cells which are you capacitors in series...now you can have those capacitors to be either 2, 3, 4, 6, 8, 9, or 12. Now within those three cells you can have one or all three of those having another or many other capacitors in parallel with just that one individually.

This does not appear to be what the problem stated. The problem stated "take a seris set of capacitive blocks C1, C2, and C3 containing respectively one, two or three of the same capacitors in series in each cell."

I should also mention that you don't need any additional capacitors in parallel to make the whole series combination equal to 1F.

- Warren

 

[bertholf07]

No I don't follow can you give me an example

 

[chroot]

You should be able to find a set of values that work with just a bit of trial and error.

- Warren

 

[bertholf07]

ok let me try

 

[bertholf07]

so how about this if n1=2 n2=8 n3=12

 

[chroot]

Looks good, bertholf07. There are a bunch of different valid solutions.

- Warren

 

[bertholf07]

Thanks for your help Warren greatly appreciated