# Is there any intuitive or mathematical explanation to this case regarding Kirchoff's second law?

I was trying to prove Kirchoff's Voltage law, which says that in a series circuit, $\Delta V_{battery} = \Delta V_1 + \Delta V_2 ... + \Delta V_n$, where n is a number of resistors. Before proving this, I tried to think of what will happen if this law wasn't true. And, I got a very important point: Conservation of energy. The sum of individual $\Delta V$s cannot be greater than $\Delta V_{battery}$, because energy lost can't be greater than energy gained. I thought that this was a strong explanation to the law, but I am stuck at one last problem.

Suppose I only have one resistor in the circuit. So, if I apply this law, $\Delta V_{battery} = \Delta V_{resistor}$. And clearly $\Delta V_{resistor}$ cannot be greater than $\Delta V_{batery}$. But, my question:

Can $\Delta V_{resistor}$ be less than $\Delta V_{battery}$? Or, in other words, can energy lost in the resistor be less than the energy gained through the battery? If this is not true, then why is it compulsory for the charge to loose all the energy it has gained in a circuit?

Because conservation of energy is a fundamental law of physics, and no one has ever found an exception to it. If you take into account all the factors, energy is conserved. Simple as that. That being said, in real life situation it may as well be that $\Delta V_{battery} \ne \Delta V_{resistor}$, but that's because we haven't taken into account all the factors (heating of the wires etc. $\Delta V_{battery} = \Delta V_{resistor}$ is an idealisation). So your question ultimately boils down to: why is energy conserved? Well, we don't know! It makes no sense to ask this, it's just what we observe! (Not to say it isn't an interesting philosophical pondering)