ASK AN EDUCATOR! “How do I verify Ohm’s law when using a battery?”

Mustafa asks:

I connected a 1 ohm resistor to a 9V battery.. the resistor rating was 1/4W.. I know by common sense that the battery won’t deliver 9A.. the resistor didn’t burn.. when I did it theoretically I = V/R.. = 9A but practically this does not apply.. can you please explain..

You are correct with Ohm’s Law stating at 9V you will get 9A across your 1Ohm resistor, but this is in a perfect world. Because you are working with a battery and not a regulated power supply, the battery’s voltage is going to start at ~9V then quickly drop, resulting in significantly less power being dissapated. There is a really cool article by Richard Friedrich and Rogelio Ramirez from Friedrich Engineering that uses a series of tests to determine the output capability of a PP3 9V battery. From their tests, it appears that your battery might be either a bit low or with a chemistry that does not allow for high current loads.

My advice to you would be to perform the same experiment with the same battery, but this time have a multimeter attached to the + and – on the battery to monitor voltage. Record your voltage drop every 5s or so, then repeat the experiment using the multimeter to measure the current flowing through the circuit. You can then analyze your data and determine the cause of your resistor not igniting 🙂

Good luck with your experiment, but MAKE SURE YOU ARE CONDUCTING YOURSELF IN A SAFE MANNER! Wear safety glasses, work in a ventilated area, and don’t touch hot things…because they are hot! We do not condone unsafe activities, and hold no responsibility for injury of any kind if any happens to occur.

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As the “really cool” article says, (after all the cool bits about stuff exploding) the real take-away here is calculating the internal resistance of the battery. You can model the battery as a perfect 9V source, in series with a “virtual” mystery resistor. So lets say that when you put your 10 ohm resistor across the terminals, you measure 6v across the 10 ohm resistor. You then know, from Ohm’s law, that the current in the resister is 6v/10ohms, =.6A
Then, from Kirchoff’s law, you know that the voltage across the mystery resistor and the voltage across the real resistor have to add up to 9v. So the voltage across the mystery resistor must be 3v, when the current through it is .6 amps, so the internal resistance is 3v/.6A = 5 ohms. Once you calculate the internal resistance, try other load resistors, and see if your results are consistent. But be careful of short-circuiting the battery! They can and do explode!

Another 2 things that can happen, even if you use a voltage source with less internal resistance than a 9V battery.

1) As the resistor heats up, its resistance increases from 1 ohm.

2) Your ammeter might have a finite resistance when trying to make this type of measurement.

All of these non-idealistic become insignificant at lower currents. I think rather than a bench supply, I think you should try the experiment with a 1k or higher resistor.

As the “really cool” article says, (after all the cool bits about stuff exploding) the real take-away here is calculating the internal resistance of the battery. You can model the battery as a perfect 9V source, in series with a “virtual” mystery resistor. So lets say that when you put your 10 ohm resistor across the terminals, you measure 6v across the 10 ohm resistor. You then know, from Ohm’s law, that the current in the resister is 6v/10ohms, =.6A

Then, from Kirchoff’s law, you know that the voltage across the mystery resistor and the voltage across the real resistor have to add up to 9v. So the voltage across the mystery resistor must be 3v, when the current through it is .6 amps, so the internal resistance is 3v/.6A = 5 ohms. Once you calculate the internal resistance, try other load resistors, and see if your results are consistent. But be careful of short-circuiting the battery! They can and do explode!

Another 2 things that can happen, even if you use a voltage source with less internal resistance than a 9V battery.

1) As the resistor heats up, its resistance increases from 1 ohm.

2) Your ammeter might have a finite resistance when trying to make this type of measurement.

All of these non-idealistic become insignificant at lower currents. I think rather than a bench supply, I think you should try the experiment with a 1k or higher resistor.

Thevenins theorem anyone? then mention that the battery is a bit dynamic due to chemistry