# How Is Ohm's Law Tested Here?

1. Feb 13, 2012

### DocZaius

I have trouble understanding how the following experiment tests Ohm's Law. Imagine a simple DC circuit with a 6V battery connected to a resistance that is in parallel with a voltmeter and in series with an ammeter.

Switch on the circuit, and measure the resistance, voltage and current. Plot that as a point on a voltage (y axis) vs current (x axis) graph. Now replace that resistance with another resistance. Plot that as a point on the graph. Do this for a few resistances.

You find that these points (each corresponding to a resistance) make a downward sloping line. The line is not horizontal as you might expect, but the voltage seems to change based on the resistance. There is an interesting linear relationship there, sure. But I don't see how Ohm's Law is tested there.

Ohm's Law is a statement about how when temperature and resistance is kept constant, the voltage applied across a resistor is directly proportional to the current passing through that resistor. It seems to be a statement involving a single resistor.

Once you start measuring different resistances, you don't seem to be testing Ohm's Law but some other linear relationship (which to be honest I haven't been able to clarify to myself).

The way I would test Ohm's Law is I would pick a single resistor, and change the voltage across it, and see if that plot of voltage vs current is a line.

I would love some help on figuring out what I'm apparently missing.

Thanks.

Last edited: Feb 13, 2012
2. Feb 13, 2012

### jambaugh

Ultimately you are correct, this isn't either a test of, nor demonstration of Ohm's law.
Rather you are checking the resistance values given for the resistors which are calibrated using the same Ohm's law. It is at best, practice using the Ohm's law formula... in short a math exercise (as well as an exercise in using the meters).

You are right, a better test would be to vary the applied voltage, measure the change in current, and graph THAT relationship for a single resistor kept at a constant temperature.

 To do this would require a variable supply voltage which costs more than a 6V battery and a bag of resistors. That and sloppy thinking on the part of the supplier of lab experiments is probably why such lame experiments get taught in school.

3. Feb 13, 2012

### Antiphon

I disagree. It's an interesting measurement because you're also measuring the internal resistance of the battery. It's not a strait line because of the equation for two resistors in series. Its quite interesting.

4. Feb 13, 2012

### DocZaius

Antiphonal, could you explain how Ohm's Law contributes to the interest that you are expressing? Please elaborate on your point about the internal resistance of the battery.

5. Feb 13, 2012

### gsal

Yes and No :)

(A) Yes, it seems that a clearer Ohm's Law experiment would consist of keeping the resistance constant and change the voltage...but even that experiment may fail to provide a straight line.

On the other hand, if you had an ideal variable voltage source (no internal resistance) and capable of supplying as much or little current as demanded by the resistance under the given voltage, experiment (A) above would produce a straight line of points in the V vs I plane that passes through the origin.

And if using the ideal voltage source, once again, but this time keeping the voltage constant and changing the resistor (OP experiment)....THEN, the data points would be in straight horizontal line.

6. Feb 13, 2012

### Antiphon

The battery can be modeled as a perfect voltage source in series with an internal resistor. This is why you can short the terminals of a (real) battery but get a finite current. The internal resistance of the battery will limit the current.

In your experiment the voltmeter will droop as you decrease the external resistor. This allows you to determine the effective internal resistance in the battery without shorting it and possibly damaging it. Maybe this is boring to some, but I consider it a powerful diagnostic to be able to see inside a battery with a resistor and a voltmeter. It is no less interesting than aiming a radio telescope at a star and listening to the radio emissions or carbon dating a fossil.

7. Feb 14, 2012

Extending this a bit further,from Ohm's law the emf (E) of the battery is given by:

V=E-Ir

(r= internal resistance of battery and V=p.d. across load for a current I)

A graph of V against I should be a downward sloping straight line as described by the OP.The intercept on the y axis gives the value of E and the slope of the graph gives the value of r.

8. Feb 14, 2012