Why is this a circuit in parallel if it is topologically different than a parallel circuit

Question

I'm struggling to understand why the left circuit is equivalent to a parallel circuit. I'm trying every deformation in my mind I can think of but I can't seem to get out the second circuit.

Secondly, when we say the circuit is parallel, I don't understand with respect to what. The parallelism of the circuit is such that it requires a pass through the entire circuit in both directions which is sort of useless.

Perhaps mathematically the first circuit works out to be equivalent to a parallel circuit, but then that is startling because it shows the mathematical definition of parallel circuits includes nonintuitive parallel circuits.

Why would I think they are mathematically parallel?

Firstly, we start by defining a parallel capacitor:

A parallel capacitor is any equivalent capacitor, which, once analyzed, emerges with an equivalent capacitance of the form: $$C_{eq} = \sum C_i$$

Now, for our circuit:

$$\epsilon - \frac{q_1}{C_1} = \epsilon - \frac{q_2}{C_2}=0$$

And then

$$\frac{q_1}{C_1} = \frac{q_2}{C_2}$$

From this we can obtain a few things. Firstly, we know that $q_1 + q_2 = q$. Secondly, we can see that $$V = V_1 = V_2$$

From this, we can simply define, with no particular topological orientation in the circuit, an equivalent capacitor:

$$C_{eq} = \frac{q_1 + q_2}{V} = \frac{q_1}{V} + \frac{q_2}{V} = C_1 + C_2$$

This satisfies the definition of parallel capacitor.

However this was not obvious from the topology of the original circuit. It only emerged mathematically. Can someone explain what my intuition for parallel circuits is missing?

Answer 1

The topology of the two graphs is the same. Observe that you are allowed 3D homotopies not just 2D. So you lift up $C_2$ out of the plane and put it back on the right side.

Answer 2

The capacitors $C_1$ and $C_2$ are connected in parallel because their two ends are directly connected.

You can verify this easily. Start from one end of $C_1$ and try tracing the wire to $C_2$. Next, start from the other end of $C_1$ and notice how you can still trace the wire back to $C_2$.

On the other hand, if the two capacitors are connected only by one end, they are connected in series.

Answer 3

Considering two-port circuit elements such as capacitors, batteries, resistors, and inductors, two elements are in parallel if each end of the first element is connected to each end (but only one end) of the other. The actual schematic drawing, with lengths of lines and placement of elements on the left, right, top , or bottom is irrelevant.

For the drawing on the left, the battery is in parallel with $C_2$ because each end of $C_2$ is connected to an end of the battery. Now, you can reverse their positions with no change in behavior. The battery is also in parallel with $C_1$ because each end of $C_1$ is connected to an end of the battery. Consequently, $C_1$ must be in parallel with $C_2$, which an examination of their connection points with each other confirms.

Now we are free to draw all 3 in parallel, in any orientation on the page we choose. The diagram on the right shows $C_1$ in parallel with $C_2$ and the battery.