If a battery gives me a 5V terminal and a 0V terminal I calculate power by multiplying total current of the circuit by 5V. What if the battery has a +12V and -12V terminal I'm using to power my circuit? What value do I multiply current by to give me Watts?
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asked Apr 29, 2016 at 2:43
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1\$\begingroup\$ Does it also have a 0V terminal? \$\endgroup\$– Ignacio Vazquez-AbramsCommented Apr 29, 2016 at 2:45
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\$\begingroup\$ Yes, my power supply has +12V and -12V I'm using to power an op amp comparator, and +5V and 0V I'm using for everything else \$\endgroup\$– eddiewastakenCommented Apr 29, 2016 at 2:45
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\$\begingroup\$ Measure +12V current (Iplus) and -12V current (Iminus). Power is 12 * Iplus + 12 * Iminus. If Iplus = Iminus in all cases, then you can just multiply Iplus * 24V and get the same answer. \$\endgroup\$– user57037Commented Apr 29, 2016 at 2:48
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\$\begingroup\$ Oh, and use the actual exact battery voltage if you know it instead of 12V. \$\endgroup\$– user57037Commented Apr 29, 2016 at 2:49
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1 Answer
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Yes, if you connect your load between the \$+12V\$ and \$-12V\$ terminals, you will have \$24V\$ across it. That load will draw a current which is sourced from the positive terminal and sinked (?) by the negative terminal. The power will be the current multiplied by the total supply voltage.
The negative terminal can be considered effect a virtual ground which you can use as a reference point for all of your calculations. To convert voltages referenced to the \$0V\$ to be relative to the virtual ground you simply add \$12V\$ on to them.
Furthermore, if there is also a \$0V\$ terminal as well, and your load is also connected to that in some way, you have to measure the power for each rail individually (as @mkeith mentions in the comments) as some current may be flowing into/from the \$0V\$ terminal. You would treat the two supplies as if they were not even connected to each other and each has a separate load.
answered Apr 29, 2016 at 2:47