That's fine. And sure, there are many equalities that you can add into that statement.... but I'm not exactly sure what makes my statement less accurate. I guess you could say that you're 'adding more relevant information' rather than mine being wrong.
Where I got my statement from was the [simplified] equation V = E - Ir. If r (internal resistance) decreases, then V (voltage) increases
No arguing the equation, just that when you look at the closed circuit of a battery light system you have to take more than just the resistance of the cell into account. Obviously we know there is no perfect power source - they all have an internal resistance of some sort - and this resistance is what causes cells to heat up at higher loads - the cell drops votage just like the bulb.
An accurate way for view the closed circuit of the system for a light would be:
Battery -> R battery -> R case and wires -> R switch -> R bulb -> Battery
This is the best picture I could find:
So for any level of current voltage is dropped across every resistive component in the closed loop: the battery, the case, the wires, the switch, and the bulb. We design a system such that the bulb ideally drops all of the voltage, but since that is not real the system is impacted by small voltage drops along the way.
For the RCR123 battery versus the IMR16340 battery both technically have the same nominal Vs of 3.7 volts; however, the IMR16340 has a lower internal resistance. So in the scope of the closed loop system of the flashlight when you add up all of the resistances you will see that the IMR cell will "drop" less voltage than the RCR123.
The Vs - Vinternal = Vin.
So my statement that the battery
drops less voltage is more accurate than your statement that the battery
voltage increases because it doesn't increase. They are both providing nominal 3.7 volts.
