Resistor Discharging Batteries

I have 3.3 ohm 20w resistors 2 of them and I wanna know how to hook stuff up and what I can discharge from 1s lifepo4 180ah and maybe 2s

Ohm's Law:
V/R = I

For 1S Discharge:

3.7V / 3.3Ω = 1.12A

P = V * I

3.7V * 1.12A = 4.14W - Well within limits of your resistors for 1S. You could put both in parallel, and drain at about 8.28W. For a 180Ah battery, that's gonna take a while

For 2S Discharge

7.4v / 3.3Ω = 2.24A

7.4v * 2.24A = 16.59W - Just under the limits of your resistors. Make sure they're well ventilated and discharge in a cool position (Don't leave them in the sun). If you use both in parallel, you could get just over 33W discharge.

All you have to do is connect the resistors across the + and - terminal.

EDIT, I was typing mine up as you posted.

BTW, he's using LiFePO4, so 3.2v is a better voltage to use than 3.7v, for the calculations.

Fastassmotors said:

I have 3.3 ohm 20w resistors 2 of them and I wanna know how to hook stuff up and what I can discharge from 1s lifepo4 180ah and maybe 2s

Click to expand...

First, make sure you stay with the thing the entire time you're discharging, watching the voltmeter all the time, so you don't overdischarge the cells, which will damage them permanently. You'll be there a long long time if discharging from full to empty, so you might want to buy an alarm in case you doze off.

How you hook them up depends on the current you want to draw. To draw the most current, you parallel both resistors, which cuts the resistance in half (see http://allaboutcircuits.com in the learning section for how this works). To hook them to the batteries, one end of the resistor (or pair) goes to the + of the battery, the other goes to the -.


One formula you may use a lot is Ohms Law. The Vulture that sees the Coyote next to the Rock (V = C * R); the Coyote sees the Vulture *over* the Rock (C = V / R), and the Rock sees the Vulture over the Coyote (R = V / C). (this is not the usual version, but if you're not familiar with the terms / abbreviations used, it may be easier to remember)

In this case, you're using the C = V / R version, to find the rate you'd be discharging the batteries at.

So if you have 1 LiFePo4 cell of about 3.2 Volts (average) and two parallel resistors of 1.65ohms Resistance, then 3.2 / 1.65 is 1.94 Amps of Current.

While the battery has more charge on it, the current will be higher by a bit, and will drop as the voltage drops as the battery discharges.

If you have two cells in series, the voltage is twice as high so the current will be, too.

FWIW, the single 180Ah cell discharged at 1.94A is going to take about 93 hours, almost 4 days, to discharge from full to empty.


To make sure you don't melt your resistors, you also need to know how many Watts they'll be making of heat. That's Current times Volts, so 3.2 * 1.94 is 6.208 Watts. Since both resistors are in parallel, half the current goes thru each one, so half the watts too. Well under 20. Even if you double it due to double voltage, it's still fine.

But it's probably enough heat you won't want to hold the resistors in your hand.

Thanks guys you both made it easy to understand the algorithm which I'll remember and now I know what's safe and not safe for my resistors just what I needed to discharge these suckers thanks. Also yes they're right around 3.2 nominal so that's a little more appropriate.

What voltage are the cells now ? Are they the same voltage ? Only parallel and then drain cells if matched voltage. 180ah is a lot of energy.

Fastassmotors said:

. Also yes they're right around 3.2 nominal so that's a little more appropriate.

Click to expand...

I tend to over engineer for worst case scenario. LiFePo4 has almost no capacity between 3.7v and 3.2v, so for small pack, thats not a problem. You'll burn off the extra voltage in minutes if not seconds. It won't kill your resistors...

But for a 180Ah? That 3-4% of capacity could last hours, hence better safe than sorry. Either way though, you were a good 20% under spec, so you're safe.