# Unsatisfactory explanation for the EMF measurement of a battery

Experimentally, I have seen how hooking up a battery to a simple circuit just with a high-resistance voltmeter raises the voltage reading (allegedly to a level equal to the EMF of the battery).

However, I find the explanation for why the reading rises, much less to an EMF, very unconvincing. We were told that the internal resistance and the necessary potential drop is ignored, because there is no current in the said circuit, hence why the voltmeter measures an EMF. How can this make sense? There clearly must be some current, albeit very little, flowing, for the high-resistance voltmeter to even have a reading, and that little current will still experience resistive forces from the internal resistance of the cell - so the EMF should not be attainable. Or is there a mechanism by which, when there is very little current, resistors are ignored, hence no work has to be done to traverse them?

Clearly I am wrong, as experimentally I saw the voltage rise in that super simple cell. My point of view suggests that there shouldn't be a difference between the reading in said circuit and the potential difference in a circuit consisting of 3 resistors (the voltmeter in this case measures the drop between these 3 resistors, note there is no other significant source of resistance other than the internal resistance). Really, in my theoretical understanding they should produce an equal reading, but they don't.

So, to be honest, not only do I believe that the reading we saw shouldn't have been the EMF, but not even any different from the reading in a normal circuit, as described (the latter belief clearly conflicts with reality).

I an eager to know what I am thinking wrongly about. Please ask if I can help clarify anything!

Thank you very much :)!

If the internal resistance of the battery is $R_b$ and the resistance of your voltmeter is $R_m$ then the voltage you measure will be:
$$V = \frac{R_m}{R_m + R_b} E$$
The internal resistance of, for example a manganese alkaline battery is less than an ohm, while the resistance of a voltmeter is going to be 10s if not 100s of KOhms, so in practice the measured voltage is going to be less than the true EMF by less than one part in $10^4$. Since you probably can't read the meter that accurately the internal resistance can be ignored.