# How does this circuit behave?

A,B are connected to a battery with $$V$$ voltage. When $$S$$ is open we can just calcualte the total capacitance simply. What happens when it's close? I don't see which capacitors are connected series and parallel. I attemped this:

I know that everything in the green must get to equal potential, and charge will move according to the capacitance of each. But how to calcualte that?

Edit - is this another way to describe the circuit?

• Don't you think that $C_1$ and $C_2$ form a parallel combination when the switch is closed? You should be able to take it from there. Commented Feb 28 at 17:47
• I'm not actually sure. I've never understood what is the definition of parallel and series connections. I just know series has the same current and parallel has the same voltage, and I "see" which is the case
– amit
Commented Feb 28 at 17:55
• 'Series' and 'parallel' are ways of connecting two-terminal components. When one terminal of each component is connected to one terminal of every other component, and the other terminal of the component is connected to the other terminal of every other component, then the components are connected in parallel. That's the case for $C_1$ and $C_2$, is it not? Can you spot another parallel combination? And how are these combinations themselves connected to each other? Commented Feb 28 at 18:11
• See my edit on that. I guess thats what you meant but is there a general rule for that? How can I know I can change the wire to this?
– amit
Commented Feb 28 at 18:19
• "How can I know I can change the wire to this?" The plates in your green box are at the same potential because they are all connected together by wires with no potential differences across them, and the same plates in your hand-drawn diagram are at the same potential as each other because they are all connected together, even though the detailed way in which this is done is different. Commented Feb 28 at 18:28

In the original diagram, the voltage across capacitance $$C_1$$ is $$\Delta V_1 = V_A - V_{\text{junction}}$$, where $$V_{\text{junction}}$$ is the potential between at the junction just to the right of $$C_1$$. Because there is no potential drop across a resistance-less wire, the potential at the junction just to the right of $$C_2$$ is also $$V_{\text{junction}}$$, so the voltage across $$C_2$$ is $$\Delta V_2 = V_A - V_{\text{junction}}$$.

Similarly, the voltage across $$C_3$$ is $$\Delta V_3 = V_{\text{junction}} - V_B$$ and across $$C_4$$ is $$\Delta V_4 = V_{\text{junction}} - V_B$$.

If the voltage across two or more capacitances is the same, then they are connected in parallel, so here clearly the only combinations are $$C_1/C_2$$ and $$C_3/C_4$$. Their equivalents (say $$C_{eq, 12}$$ and $$C_{eq, 34})$$ have different voltages across them ($$V_A - V_{\text{junction}}$$ and $$V_{\text{junction}} - V_B$$ respectively), so they are NOT connected in parallel.

However, the total voltage across these two (equivalent) components is $$V_A - V_B = (V_A - V_{\text{junction}}) + (V_{\text{junction}} - V_B)$$, and when two capacitances are connected such that $$\Delta V = \Delta V_1 + \Delta V_2$$ (and that they have the same charge $$Q_1$$ = $$Q_2$$), they are connected in series.

$$C_{eq, 12} \ (C_1 || C_2)$$ and $$C_{eq, 34} \ (C_3 || C_4)$$ are in series.

The variation you drew maintains all these descriptions of the circuit, so it is indeed equivalent.

Hope this helps.

When S is open C1 and C3 are in series and C2 and C4 are in series. Then the equivalent capacitance of the combination of C1 and C3 is in parallel with the equivalent capacitance of C2 and C4.

When S is closed, C1 and C2 are in parallel and C3 and C4 are in parallel, as shown in your last diagram. Then the equivalent capacitance of the combination of C1 and C2 is in series with the equivalent capacitance of the combination of C3 and C4.

Hope this helps.