Potential energy in an electric circuit

Question

I am studying electrical circuits and how they work. I know that electrons go from the negative pole to the positive pole (where potential difference increases), this means they go from a lower electric potential to upper electric potential.

I know in the electric circuit where there is no resistance the potential difference is equal to $0V$, this means that where there is a resistance there is a potential difference.

I don't understand how resistance's potential difference changes if I add or remove other resistances in the electrical circuit, what happens inside the circuit? How does the potential energy of resistance terminals change?

When the charges go through the resistor, how could they know their flow intensity? How could they know if there is a second or third resistor?

Answer 1

I don't understand how resistance's potential difference changes if I add or remove other resistances in the electrical circuit, what happens inside the circuit? How does the potential energy of resistance terminals change?

@Steeven has given you a nice water pressure and plumbing system analogy. The following is using voltage $V$ (roughly analogous to water pressure), electrical resistance $R$ (roughly analogous to pipe resistance to water flow) and current $I$ (roughly analogous to water flow) and Ohms law that relates the three in the case of electrical circuits.

For a given resistor, the potential difference $V$ across the resistor $R_1$ is $V=IR_1$ per Ohm's law where I is the current through the resistor. If that's the only resistor in the circuit, then the voltage $V$ is also the only voltage source in the circuit. If you add more resistors in series with resistor $R_1$, e.g., $R_2$ and $R_3$, then the current in the circuit will be

$$I=\frac{V}{(R_{1}+R_{2}+R_{3})}$$

And the potential difference across $R_1$ alone will be

$$V=IR_{1}=\frac{VR_1}{(R_{1}+R_{2}+R_{3})}$$

Making the potential difference across $R_1$ less than when it was the only resistor in the circuit.

You can think of the three resistors as roughly analogous to three sections of pipe connected end to end to the output of a fixed pressure water pump. There will be a water pressure drop across each section of pipe, and the sum of the water pressure drops will equal the total applied water pump pressure to the piping system.

Hope this helps.

Answer 2

Think of electric potential as water pressure. Such as in hose or plumbing system.

Think of the resistor as a filter or a constriction in the hose or pipe which "resists" the flow of water.

There is a certain pressure on one side of the filter because a lot of water particles are pushing forward. This pressure squeezes the water through the filter. As soon as a water particle is through, it can continue flowing with no large pressure from behind. On this side of the filter, the pressure is lower.

If you now add another filter on this side, then the water will not have any tendency to pass through it. There is such as low pressure here after the first filter, that nothing is pushing it through the next filter. The second filter is like a wall, and so the water will stop and stay here. But soon, more water molecules arrive from the first filter - so, soon they accumulating water amount here behind the first filter starts building up a pressure on this side, since there is soon not enough space for all the water.

Then the pressure grows. The pressure difference across the first filter is now smaller (the pressure on one side is the same, but that on the other is larger). There is now also a pressure to force water through the second filter. And so, water is squeezed through this second filter as well. On the other side, there is again no pressure.

• With one filter, the entire pressure-drop happened over that filter.
• With two filters, the total pressure-drop is shared between them. There is a smaller pressure-drop across each filter, that sum up to the original pressure-drop.

With a smaller pressure-drop, the water flow is also smaller. And this is how the flow and the pressure around a filter is influenced by other filters being nearby.

In your case, this water-system analogy is very fitting with charges (water particles) flowing in a current (water flow, litres/second e.g.) due to a potential (pressure) difference across resistors (filters) along the wires (hose, pipe).