# Why does the potential drop across a battery and resistor equal the emf of the battery?

In this diagram you can see the potential difference across the battery and resistor is the same as the pd created by the battery (the battery and resistor are representing a battery with internal resistance):

I know the route through the voltmeter is not actually a route, so I'm guessing here is no current? But why* is there no difference in the voltmeter reading, whether you put the voltmeter across the battery and resistor or just the battery?

The following diagram makes sense to me as some pd should be dropped over the internal resistance (the 680ohm resistor):

*(physically, when I put the numbers into the equations it make sense mathematically, but I don't understand the physics side)

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The resistance through the voltmeter is so much greater than that of the resistor that the resistor's resistance has negligible impact on the circuit. A precise-enough voltmeter would see a very small difference, on the order of micro or nanovolts.

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So why is there a difference between the first and second diagrams, in terms of the precision of the voltmeter? –  Jonathan. May 14 '12 at 22:36
When you attach external resistance to battery, the voltage of battery shall drop and this drop would be rather small for, say, few hundred ohms. –  Pygmalion May 14 '12 at 22:42
But why does it matter whether the resistor is negligible to the voltmeter, surely it should matter whether the resistor is negligible to the emf? –  Jonathan. May 14 '12 at 23:02
Except that ohms and volts are two different units, related to each other only through other units. You need to compare ohms to ohms. –  Ignacio Vazquez-Abrams May 14 '12 at 23:06

You've far overestimated internal resistance of the battery. According to Wikipedia typical internal resistance is of order $0.1 \; \Omega ... 1 \; \Omega$

EDIT: Supposing the current through voltmeter is zero, then current through external resistance is

$$I = \frac{U_0}{R_\text{i}+R_\text{e}}$$

therefore, the voltage on the battery is

$$U = U_0 - R_\text{i} I = U_0 \frac{R_\text{e}}{R_\text{i}+R_\text{e}}.$$

You see, if $R_\text{i} \ll R_\text{e}$, $U \approx U_0$.

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I know this, it's just the default resistance of a resistor on the software used to draw/simulate the circuit. –  Jonathan. May 14 '12 at 22:34
What kind of software assumes internal resistance of $680 \Omega$?? Actually, internal resistance of battery does increase as battery is slowly emptying, but IMHO the software should give you the internal resistance of full battery. –  Pygmalion May 14 '12 at 22:39
It doesn't have a special internal resistance resistor. I just dragged out a battery and resistor. Then just decided to use it as internal resistance. I would change it but my internet is slow and uploading images is painful. –  Jonathan. May 14 '12 at 23:00

After going again through your questions I figured I misunderstood your question.

The idea behind voltmeters is that they have a very large internal resistivity. If you have some "normal" resistance $R$ and voltmeter $R_V \gg R$ in series, the voltage drop on voltmeter shall be $$U = U_0 \frac{R_\text{V}}{R_\text{V}+R} \approx U_0.$$
I think that your program assumes that internal resistivity of voltmeter is simply $R_V = \infty$, therefore current through voltmeter is simply $I_V = 0$. This is exactly the point. We don't want for the measurement appliance to influence our system, e.g. we want that current through voltmeter is practically zero, and that voltage drop on ammeter is practically zero.