# Archived Charge and Energy for Capacitors in Different Configurations

1. Mar 17, 2009

### rowkem

1. The problem statement, all variables and given/known data

How much energy must a 12V battery expend to fully charge a 0.15 uF and a 0.20 uF capacitor when they are (a) in parallel and (b) in series? How much charge flowed through the battery in each case?

2. Relevant equations

i: C(eq)=((1/C1)+(1/C2)+...+(1/Cn))^-1 and C(eq)=C1+C2+...+Cn
ii: U=0.5CV^2
iii: Q=CV

3. The attempt at a solution

I have worked it out but, I prefer not to post my answers. However, I did the following for (a) and (b):

1) Worked out the equivalent capacitance of each circuit using the appropriate equation (i)
2) Used the result(s) of part 1 and plugged that into equation (ii) using the voltage of the battery to get U for each circuit (energy expended by the battery solved)
3) Used the result(s) from part 1 and the given battery voltage in equation (iii) to obtain the charge for each circuit. (charge through battery solved)

So - that's essentially what I did. Was I correct in my methods? Thanks.

2. Feb 7, 2016

### Staff: Mentor

The OP's methods are correct.

Given: $C_1 = 0.15~μF~~~;~~~C_2 = 0.20~μF~~~;~~~U = 12~V$

$C_p = C_1 + C_2 = 0.35~μF$
$C_s = \frac{C_1 C_2}{C_1 + C_2} = 85.7~nF~~(8.57~×~10^-8~F)$

Then the energy expended by the battery in each case is:

$E_p = ½ C_p U^2 = 2.52~×~10^{-5}~J$
$E_s = ½ C_s U^2 = 6.17~×~10^{-6}~J$

and the charge moved in each case:

$Q_p = C_p U = 4.20~μC$
$Q_s = C_s U = 1.03~μC$