In an electrical circuit consisting of just a source and some load, when current hits the load, the voltage drops. This means that the potential energy of the charge drops. If the potential energy drops, should the kinetic energy of the charge increase to conserve energy? If this was the case, why is the current before and after the resistance the same?

In the water pump analogy, the source applies the pressure. Voltage is equivalent to pressure. Current is equivalent to water flow. A resistor is equivalent to a restricted pipe. The pressure before the resistor (wide pipe) is larger than the pressure inside the restricted pipe (smaller cross-sectional area, lower pressure, higher velocity).

However, there is a voltage drop. So if voltage is equivalent to pressure, the pipe before the resistance is wide, inside the resistance it is small, but on the other side of the resistance, the pipe should have an area between "wide" and "small". In order for the mass flow rate of the water (current) to remain the same in this narrower section of pipe, velocity must increase.

So in an electrical circuit, after some resistance, in order for current to remain the same, less charge flows, but each charge is moving faster?


1 Answer 1


The drop in potential energy is dissipated as heat in the resistor, mechanical energy of the charges is not conserved.

In the water analogy, the current is equivalent to Moved volume per unit time, not velocity of the molecules.


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