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Please excuse me if my terminology is not accurate, or if I'm not asking the right question. I will explain my goal first.

I am building a circuit/device to drain a battery to a specific voltage. I am monitoring the voltage with an arduino and that is opening/closing a relay depending on the voltage.

I need to select a resistor to drain a 3.2v cell and draw 500 milliamps.

EDIT:

I am aware that my cells are 3.2v NOMINAL (LifePo4 26650 cells)

Also aware of safety concerns, etc. I'm draining very slowly and under supervision, and in a safe environment.

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  • \$\begingroup\$ added a serious edit into my answer \$\endgroup\$
    – user136077
    Commented Mar 5, 2017 at 21:22
  • \$\begingroup\$ Note that a 3.2V cell will not deliver exactly 3.2V, the voltage will vary over time. So whatever resistor you use, the current drawn will vary (somewhat). \$\endgroup\$
    – Wouter van Ooijen
    Commented Mar 5, 2017 at 21:53
  • \$\begingroup\$ Please do note that (besides charge) battery voltage is also dependent on temperature and load current. I'm not really sure what you're trying to achieve here but take care and remember that some batteries will explode if drained to fast. \$\endgroup\$
    – user1890202
    Commented Mar 5, 2017 at 22:01

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a half ampere from 3.2 volts - that's 6.4 Ohm resistor. 6.8 Ohm is a standard part and only about 5% too high. You dissipate 1.6 watts. Take a 2W or 4W capable model. 0.25W model smokes, 0.5W model gets brown and 1W model burns your fingers.

Maybe you should learn Ohm's law. NOTE playing with high capacity batteries is dangerous. A short circuit is very easy to generate by accident. That can at very high probability burn your home and blow out your eyes.

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  • \$\begingroup\$ Can you share how you got to those numbers? \$\endgroup\$
    – AaronJAnderson
    Commented Mar 5, 2017 at 21:25
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    \$\begingroup\$ @AaronJAnderson Ohm's law. It's really simple, U=R·I, and you should really head user287001's advice and read up on it, if you're doing anything with electricity. \$\endgroup\$
    – Marcus Müller
    Commented Mar 5, 2017 at 21:26
  • \$\begingroup\$ R as ohms is voltage divided by the current in amperes. 500mA is 0.5 amperes. Power as watts is the voltage multiplied by the current in amperes. The voltage and the current for the resistance or power calculation must be taken from the same 2 wire part at the same time. \$\endgroup\$
    – user136077
    Commented Mar 5, 2017 at 21:33

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