# Power delivered to a Resistor is $V^2/R$. Does size of battery matter?

I had connected a thin Resistor wire to series combination of six coin cells or to a small sized 9V battery. The wire got hot in a normal way BUT when I connected the same resistance wire to a large sized 9V car battery, rate of heat production was very large and the resistance wire got blown off. Now I wonder in fundamental physics course when we come across the equation for power $$P= V^2/R$$, there is nothing which hints towards this observation. So someone please let me know how to explain this observation.

• Careful definition of symbols is important. $V$ is not the battery voltage. Consider reviewing Ohm's law and internal resistance. Commented Dec 7, 2023 at 4:58

$$V$$ in the equation $$P=V^{2}/R$$ is the voltage across the resistor terminals, not the battery voltage. The battery voltage is the voltage measured across the battery terminals with nothing connected to the battery, a.k.a the open circuit voltage or battery emf, not the voltage across the resistor. That's because all real batteries have internal resistance which causes a voltage drop inside the battery when current is draw by the resistor connected to the battery.
In the figure below $$r$$ is the internal battery resistance, $$E$$ is the battery emf, $$R_L$$ is the load resistor (your wire) and $$V_T$$ is the terminal voltage across the load resistor. The power dissipated in the load resistor is $$P=V_{T}^{2}/R_L$$
For small batteries $$r$$ is much greater than $$r$$ in large batteries, like car batteries. That makes the voltage drop across $$r$$ in a small battery much greater than a large battery, resulting in $$V_T$$ being less for the small battery and less power dissipated in the load resistor. That's why your wire got so much hotter when connected to the car battery than the combination of coin cells.