- #1

Jimmy87

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## Homework Statement

A capacitor (A) of capacitance 4.0uF is charged to a potential difference of 20V. An uncharged capacior (B) is connected in parallel to capacitor A. What is:

(a) the initial energy stored on capacitor A

(b) the p.d. across A after B has been connected

(c) the final energy stored in A and B

## Homework Equations

## The Attempt at a Solution

Answers in back of my book are (a) 0.8mJ, (b) 14V (c) 0.53 mJ

I am ok with part (a) - I did E = 1/2 CV^2 = 0.8 mJ (milli Joules) as per the answer

(b) I really don't get this whole circuit problem. For one thing the answer to (b) must be wrong because if you use 14V you don't get answer (c) right. I got 13.3V for (b) which leads to the correct answer they give for (c). The way we were shown these for parallel capacitors where you have one initially charged then add another is to work out the total voltage by dividing the total charge by the total capacitance. The total charge to start with is 8uC (micro coulombs) using Q = CV for the charged capacitor. This gives 13.3V. Since they are in parallel they both get 13.3V. But this is what I really don't get. It implies the 4uF capacitor is charged initally (i.e. there is no power source when B is added. therefore there are only 2 capacitors in the circuit and no power source so how can they be in parallel - surely it is just one loop (i.e. series)? Also the answer of both A and B getting 13.3V adds up to more than the 20V we had to start with. If we do use 13.3V for both capacitors you end up with the correct energy for part (c). But this energy is less than the energy we started with in part (a). Its so confusing, please help!

Thanks