# What is the net capacitance?

1. Jun 30, 2013

### sunflowerzz

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

A 9.0 uF and 4.0 uF capacitors are connected in parallel, and this combination is connected in series with a 12.0 uF capacitor.
a) what is the net capacitance?
b) if 32 V is applied across the whole network, calculate the voltage across each capacitor.

2. Relevant equations

In parallel, capacitance add together C = C1 + C2
In series, 1/C = (1/C1) + (1/C2)

Q = CV

3. The attempt at a solution

I was able to find part a) to be 6.24 uF.

For part b), this is what I have:

Q = CV = 6.24 * 10^-6 F * 32 V = 2.0 * 10^-4 C

For the 12 uF:
V = (2.0 * 10^-4 C) / (12 * 10^-6 F) = 16.64 V

For the 13 uF (9.0 and 4.0 in parallel):
V = (2.0 * 10^-4 C) / (13 * 10^-6 F) = 15.36 V

My question is will the 9.0 and 4.0 uF have the same voltage as the 13 uF or will I have to do the same for each 9.0 and 4.0 uF capacitor?

Thanks

2. Jun 30, 2013

### Staff: Mentor

Components in parallel always have the same potential difference.

3. Jun 30, 2013

### sunflowerzz

So I assumed right? The 9.0 and 4.0 uF, which combines to 13 uF, will have the same voltage as if it were 13 uF? Is that the same for in series as well?

4. Jun 30, 2013

### Staff: Mentor

Series connected components are not constrained to have the same potential difference. Besides their own (isolated) connection which marks them as series connected, they connect to two different nodes. That's three different nodes that may each have different potentials.

5. Jun 30, 2013

### sunflowerzz

Ok thanks.

Just to clarify,

V (12 uF) = (2.0 * 10^-4 C) / (12 * 10^-6) = 16.64 V

V (13 uF) = (2.0 * 10^-4 C) / (13 * 10^-6) = 15.36 V

V (9.0 uF) and V (4.0 uF) = 15.36 V because components connected in parallel will have the same voltage across the capacitors.

6. Jul 1, 2013

### Staff: Mentor

Yup. Looks good.