Quote:
Originally Posted by AlecS
The voltage the battery outputs is proportional to the amperage being drawn, and the resistance of the circuit. This is also known as ohm's law.
V=IR
Although batteries are not always perfect voltage sources, you can use ohm's law to determine the voltage drop in your robot fairly accurately. If you were drawing 300 amps, and the total circuit resistance was .020 ohms, you would lose 6 volts. This does not fluctuate over time, it is directly related to the current being drawn. (ignoring temperature and other factors). Hopefully this clears things up.
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I'm not an engineer (yet), but doesn't this calculation have to be related back to the capacity of the battery somehow? When drawing, say, 5 amps, intuition tells me some tiny lead-acid battery would have a much larger drop in voltage than a huge deep-cell marine battery would. Maybe I'm rusty with my basic electrical physics, but with V=IR, if you had a 300 amp draw with .02 ohms, that would mean that you must be drawing from a 6 volt source (or be in series such to draw 6 volts), not "losing" 6 volts. Our applications are of course done in parallel however on the robot, so series is not an option. If we know the current draw and resistance of the whole circuit like you said, then we solve for our voltage across the whole circuit. I'm thinking the theoretical voltage drop of the battery under load is another calc entirely. I'm simply raising a concern, not saying this it absolutely correct. I'd love to be educated otherwise if someone has a good explanation.
Here, they relate a voltage drop on current draw to a capacity fraction.