# Find equivalent capacitance

1. Mar 12, 2017

### Kronos

1. The problem statement, all variables and given/known data
Find the equivalent capacitance between points A & B

2. Relevant equations

Ceq1 = C + C
1/Ceq2 = 1/2C + 1/2C + ...

3. The attempt at a solution

It is my understanding that each capacitor top and bottom are connected in parallel, so their equivalent is 2C.

Now, they are all connected in series, so their equivalent would be:
1/Ceq2 = 1/2C + 1/2C + 1/2C + 1/2C

Thus,
Ceq2 = C/2

Is my reasoning acceptable?
Thank you.

2. Mar 12, 2017

### haruspex

If the effective capacitance sought were between the left hand side and the right hand side, that would work. But it isn't.

Is there more context here, like an applied varying potential?
If not, think about potentials. What might the potential difference be between the right hand side of the top left capacitor and the right hand side of the bottom left capacitor?

3. Mar 12, 2017

### Kronos

The potential difference would just be zero since the capacitors are at steady state. Thus, there is no current flowing through the wire w/ resistor, and by Ohm's law:
V = I*R
V = (0) * R
V = 0

4. Mar 12, 2017

### haruspex

5. Mar 13, 2017

### Kronos

Yes. That means that the capacitor on the top left and the capacitor on the bottom left are connected in series, correct? Since the same charge will flow through them.

6. Mar 13, 2017