Danny,

I think you have a misconception of the thevenin and norton theorems.

Forget (for a moment) the Voltmeter. Look at the circuit consisting of a battery and a load resistor. Now, just as a thought experiment) "open" the battery. YOu can model the battery as an ideal voltage source plus a series resistor (think of this resistor as the ESR). Now replace the original circuit (battery + load resistor) by the ideal voltage source, ESr and load resistor.

You can now either

1) model this circuit as a voltage source with a lower ideal voltage plus a lower ESR by applaying thevenin's theorem

or

2) model this circuit by an ideal current source with an equivalent ideal current and equivalent parallel resistance.

You need to understand that while these circuits do look very dissimilar, they in fact behave the same mathamatically and physically - provided the theorems have been applied correctly.

Now the voltmeter comes back into the game. This is entirely different from the norton/thevenin problem.

A typical voltmeter has ~1MegOhm input resistance. If you put that in series with 100 Ohm, the additional resistance is 0.01% of the voltmeter's resistance, meaning it will not influence the measurement at all (unless the precision of your voltmeter is better than 0.01% - i doubt it). Therefore the current will be negilgible and you measure the battery's open-circuit voltage. As Green Giant already correctly stated: This gives you no information about the state of the battery.

If you connect the resistor parallel to the battery and the voltmeter parallel to both, then you can read the voltage of the battery under load. How much information that gives depends on many factors. If the voltage measured this way breaks down to less than e.g. 50% of the open-circuit voltage, you may take the battery for empty - or the battery may be obverloaded which may happen if the nominal load current of the battery is **much **smaller than the current you draw through the 100 Ohm resistor.

If the voltage dropes just a little (e.g. 5%) either the battery is still in good shape or the load is way too light compared to the battery's nominal load.

It all depends on the characteristics of the batttery. Without that information you can neither say 100 Ohm is to much load, is too little load or is just the right load.

If you've been told to use 100 Ohm by someone at your workplace:

1) they probably gave you this value from experience

2) you should ask the people at work why to use just this value - from your previous posts it looks like you seem to be afraid of asking the people over there. Why? A good question deserves a good answer and even respect for someone acknowleding that he doesn't understand something but is willing to learn. Or ist it that "work" is "school" and your questions belong into the homeworks section?

Harald