# Double batteries capactior circuit

Can anyone help me solve this problem?

• seems like a bad question to me, unless the initial charge on the capacitors is specified the rest state is undefined. $V_{NM}+V_{AB} = 5V$ is all that can be derived. – Jasen Jun 8 '17 at 21:10

Ok. First things first:

We have four elements in series so we can arrange them however we like and it's all the same.

We have: -3v+ , C1, +8V- , C2

But it's the same if we have:

+8v-, -3v+, C1, C2

That's equivalent to having one 5v battery (8v - 3v) charging two capacitors in series.

So we have a 5v battery charging a 2uf and a 3uf capacitor.

We will assume capacitors start with no charge. We are looking for the voltage between the two capacitors, let's call it Vo.

We know that both capacitors have the same charge, the same number of q, or electrons attached to its plate since they are in series.

We also know that the voltage of both capacitors must add up to 5v.

We know that for a capacitor q = C*V. this is equal to V = q/C

Let's put all together:

Vc1 = q1/C1

Vc2 = q2/C2

Let's divide this equations:

Vc1/Vc2 = q1/q2 * C2/C1

but we said q1 = q2, so we can cancel them.

Vc1 / Vc2 = C2/C1

That's equation number 1. Equation number 2 is:

Vc1 + Vc2 = 5v.

Now just replace:

Vc1 +C1/C2*Vc1 = 5v

Vc1*(1+2/3) = 5v

Vc1 = 5v/(1+2/3) = 5v/(5/3) = 3v

and Vc1 + Vc2 = 5v --> Vc2 = 2v

Cheers!