# How is Kirchhoff's voltage law understood in the water flow analogy?

I met the Kirchhoff circuit laws in the past, but now I'm trying to associate them with a practical representation to be sure to understand them.

Let's start with the Kirchhoff current law: If I say that the electrons are like water going though a pipe, it can clearly be understood that the water flow will be divided when meeting a junction.

The problem is how to understand (practically and physically) the voltage law - what analogy can I use to understand it?

Suppose your pipes form a loop i.e. water can flow through the pipes and get back to where it started. As the water flows round the loop there will be some places where pressure rises (e.g. a pump = battery) and some places where pressure falls (e.g. a restriction = resistor). However if the water goes all the way round the loop back to its starting point the pressure must be the same as when it started. So along the loop the pressure rises and the pressure falls must balance out, so that the total pressure change round the loop is zero.

And this is the same as Kirchoff's Voltage Law. In the hydraulic analogy a pressure change is equivalent to a voltage change. So if you go round a loop measuring all the voltage increases and all the voltage decreases then when you return to the starting point the total voltage change must be zero.

As you said, current is like water flow, similarly voltage is like water level and voltage difference like difference of water level. We know that water flow from higher level to lower level like current flow from higher voltage to lower voltage. Voltage difference means there is a difference of charge, i.e., a difference in the number of electrons. Now, if we consider a circular current loop as two water vessels connected by a narrow tube and in equilibrium we see that the water level difference becomes zero. Similarly, voltage drops around a closed path must be zero in equilibrium, i.e., net flow of electrons is zero.

• Very good analogy, the best I understood here, thank you! Commented Oct 27, 2014 at 22:35
• you are most welcome Chirac. Commented Nov 7, 2014 at 6:37

The Kirchoff Voltage law states that the sum of emfs in a circuit is equal to the total potential drop in the circuit.

So for a simple example, where you a 6V cell, for example, and 2 resistors in series. The 6V cell can be seen as a place where the water is given potential energy - if we imagine a ramp, it would be the water being pumped up to the top of the ramp.

The water then flows along a straight path (wires assuming 0 resistance) until it meets a resistor. A resistor can be seen as another ramp, however since there are 2 resistors, (assume they are the same resistance, however this doesn't really matter) all the potential cannot be dropped across one resistor so this ramp is smaller than the ramp up from the 6V battery. As water falls down this ramp, it loses potential energy - this is 'the same' as the drop in potential difference across a resistor.

You may ask - but the water speeds up when it falls down the ramp? well for this we should assume that there is no gain in kinetic energy and all gained kinetic energy is dissipated falling down the ramp - this can be analogous to current remaining constant in a series circuit.