# Measuring the Electromotive Force and Internal Resistance of a Battery?

Here is a problem that emerged some time ago:

We needed to measure the emf(electromotive force) and internal resistance of a car battery. The only instruments were voltmeter and ammeter.

Someone suggested:

1) the first time connect voltmeter and ammeter in series to the battery and write up the results of the measurements.

2)the second time connect voltmeter and ammeter in parallel to the battery and write up the results of the measurements.

Based on these measurements only, calculate emf $E$ and internal resistance $r$ of the battery. To be specific,

1)for the first case $U_1=10V$ and $I_1=0.1A$

2)for the second case $U_2=1V$ and $I_2=1A$

How do you solve this problem?

Arguably, this method is advanced in the sense that the result do not depend on internal resistances(unknown) of voltmeter and ammeter. This last claim does not seem obvious.
Maybe there is an easy way prove(disprove) this claim!

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Internal resistance of a car battery is less than 0.1 ohm. (Think of currents whem starting, about 100 amp!) I doubt You have equipment to determin such a resistance. The emf is voltage for the open battery, simpe measurment with a good multimeter is precise enough (because int. resistance is very low). –  Georg Oct 31 '11 at 11:10
@Georg An ammeter shunt is added. But it is not expected to impact on the results. –  Martin Gales Nov 1 '11 at 6:52
Is this a homework problem or a real-life problem? –  endolith Dec 2 '11 at 21:15

A typical car battery has a voltage around 12V and a very small (less than 0.1 Ohm) internal resistance. If you would connect the ammeter in parallel to the battery it will hopefully trip the internal fuse or just blow up. There are very few specialized ammeters than can measure currents above 100A. So you have to approach this problem differently.

1) the first time connect voltmeter and ammeter in series to the battery and write up the results of the measurements. -> Your voltmeter will show 12V, the ammeter a value close to zero (below it's resolution)

2) the second time connect voltmeter and ammeter in parallel to the battery and write up the results of the measurements. -> The ammeter will be overloaded instantly, the voltmeter will show a short voltage drop until the fuse of the ammeter is triggered or it is destroyed.

From neither of these measurements will you get both quantities. Try this approach:

• Your measure the emf with the voltmeter connected to the battery.
• You connect a known resistance to the battery that a current flows through that resistor (should be at least 1A at 12V to get any kind of precision) and your ammeter. You write down the voltage from the voltmeter and the current from the ammeter.

Now we can calculate the internal resistance of the battery easily: $$U_0 = (R_{external} + R_{internal})\cdot I$$ The voltage $U_0$ is the voltage you measured just with the voltmeter, $I$ is the current from the ammeter. For higher precision you could also include the resistance of the ammeter (<1Ohm) and the voltmeter (>1MOhm) but in practice you can't do much better without a four-wire measurement.

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ammeter is always connected in series while voltmeters are always connected in parallel... –  Vineet Menon Nov 1 '11 at 4:57
@Alexander: An ammeter shunt is added. But it is not expected to impact on the results. –  Martin Gales Nov 1 '11 at 6:53
@VineetMenon: Martin Gales explicitly wrote that the ammeter is connected in parallel and of course you can do so. You are right of course that this would not be how an ammeter should be used but nobody prevents you from doing that. –  Alexander Nov 1 '11 at 13:40
@MartinGales: If you add an ammeter shunt with a known resistance you can use the formula I gave to calculate the internal resistance. –  Alexander Nov 1 '11 at 13:44
@Alexander:I didn't read the comment of Martin.... :( –  Vineet Menon Nov 1 '11 at 14:50

I will attempt to answer the question,

We will need three things,

1. Open circuit voltage across the battery $v_{oc}$
2. Closed circuit voltage accross the battery using a known resistance $v_{cs}$
3. Closed circuit current $I_{cs}$
|
------| |-----/\/\/\----
|  |               |
|                  |
|<----$v_{oc}$---->|

|
|-----| |-----/\/\/\-----------|
|   | |               |        |
|   |                 |        |
|   |<----$v_{cs}$--->|        |
|                              |
|                              |
(A)                             |
|                              |
|                              |
|--------/\/\/\/\/\/\/\--------|

Now,

$v$ is the emf generated by battery, $R$ is the known resistance, $r$ is the internal resistance

$$I_{ol}=\frac{v}{R+r}$$ $$v_{cs}=v-I_{ol} r$$ $$v_{cs}=v-\frac{v}{R+r}r$$ $$v_{cs}=v(1-\frac{r}{R+r})$$

Now, the unknown here are $v$ and $r$. But we know $v$ since, the open circuit volatage is nothing but emf i.e. $$v = v_{oc}$$

Hence, $$v_{cs}=v_{oc}(1-\frac{r}{R+r})$$

which yields, $$\frac{v_{cs}}{v_{oc}-v_{cs}}=\frac{R}{r}$$

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