Physics Capacitance Question?

Question

I'm having trouble with capacitance. Any input would be appreciated. Here is the Question:

You are to connect capacitances C1 and C2, with C1>C2, to a battery, first individually, then in series, and then in parallel.

Rank those arrangements according to the amount of charge stored, greatest first.

Answer 1

When two ressistors are in serie, then you just add them. When they are in parallel, then you add their reciprocal and you get the reciprocall of the equivalent ressistor. When you work with capacitances, things are exactly the other way, you add then when they are in parallel and you add their reciprocal to get the equivalent capacitance when they are in serie.

Hope this helps... I don't know much about Physics, I am struggling against it too, but at least I think I can help you a little bit understand Doug's reply:

When he says: "Ct = 1/((1/C1) + (1/C2)) which is less than either C1 or C2", what he did was:

1/Ct = 1/C1 + 1/C2. He found the reciprocal as the equivalent capacitance adding the reciprocals of C1 and C2
because they were connected in serie.

You get

1/Ct = (C1 + C2)/C1C2

If you want to have Ct, then you have to take the reciprocal of this expression, which is what he wrote or:

Ct = (C1C2)/(C1 + C2) .............. (*)

"which is less than either C1 or C2"

Let's see why this is this way. I am learning at the same time than you, but let's first check an example to see if this is right

If C1 = 3 and C2 = 4, and they are connected in serie, then Ct is 1/3 + 1/4 = 1/Ct . Hence (4 + 3)/12 = 1/Ct and Ct = 12/7

And if fact 12/7 is less than 2. So, Doug is right, which doesn't surprise me since he's very bright. Let's now try to find an explanation to this fact.

Let's come back to the equation (*)

Ct = (C1C2)/(C1 + C2)

Ct = C1 [C2/(C1 + C2)]

And C2/(C1 + C2) = k is a number that is less than 1 since C1 and C2 are positive. And, if I multiply C1 by k, where k<1, then I get kC1 that is less than C1. And the same reasoning can be done for C2.

When he wrote:

"If you connect them in parallel, the total capacitance is
Ct - C1 + C2", I guess that he actually ment Ct = C1 + C2

Ilusion

Answer 2

Q = V * C Remember that.
So if you connect them in series, the total capacitance is
Ct = 1/((1/C1) + (1/C2)) which is less than either C1 or C2
If you connect them in parallel, the total capacitance is
Ct - C1 + C2 which is greater than either C1 or C2
If you connect them one at a time, the total charge on each capacitor will be the same as if they were connected in parallel.

HTH

Doug

Answer 3

greatest to least:

C1 and C2 in parallel, C1 alone, C2 alone, C1 and C2 in series