Pauluk, a million thanks.
So...in the case of the DC battery showing 12.8 volts, can I assume that the battery is capable of providing up to 31.5 amps? I=E/R = 12.8/.4 = 31.5 amps.
Maybe the voltage will drop and change all the numbers as the amp load is applied. Do I work from the full, unladen voltage to calculate amps and ohms, or do I have to let the amp load bring the voltage down before I'll have good numbers to work with?
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"The voltage being lost across the internal resistance of the battery is therefore
12.6 - 11.8 = 0.8V
You know that 2A is flowing through the battery, so the internal resistance by Ohm's Law is then
R = E / I = 0.8 / 2 = 0.4 ohm."
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So it looks like you're saying that we're not working off the full EMF when we calculate the resistance through a battery, but instead, we work off the voltage drop and the full amp reading. Seems like the formula would be this:
R=EMF lost/I
It would be easy to make the mistake of working off the full voltage divided by the amps drawn, which would be 11.8 volts/2 amps = 5.9 ohms.
Please correct me if I'm wrong on any of this, I appreciate it.
[This message has been edited by Spark Master Flash (edited 08-03-2003).]
