I want to discharge a lead acid battery of 12 V having a capacity of 100 Ah for 20 hours.

100 Ah divided by 20 hours = 5 A

12 V divided by 5 A = 2.4 ohms

Is my formula correct? Should I discharge with 2.4 ohms rheostat?

  • \$\begingroup\$ Have you thought about energy losses? Are your electronics capable of handling that much watts? \$\endgroup\$ – Artūras Jonkus Jul 3 '16 at 8:19
  • \$\begingroup\$ Why discharge it? \$\endgroup\$ – Leon Heller Jul 3 '16 at 8:28
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    \$\begingroup\$ A 2.4 ohm resistor would work, as long as it would survive dissipating 60 watts of heat. \$\endgroup\$ – Mark Jul 3 '16 at 8:34

Your calculations are correct. You need one more calculation - the load power rating.

From \$ P = VI \$ we can calculate \$ P = 12 \times 5 = 60~W \$.

Use a 55 W car headlamp bulb - it's close enough and you probably have one. This has the added advantage of built-in visual indication of state of battery.

I would recommend that you put in some protection to shut off the test load when the battery voltage droops to, say, 10 V. This will reduce likelihood of damage.


simulate this circuit – Schematic created using CircuitLab

Figure 1. Discharge circuit.

If you have a spare car relay you could try the circuit of Figure 1.

  • R1 is added so that the relay drops out at the required voltage. (Relays require less voltage to hold them on that they do to pick-up in the first place.)
  • SW1 provides full 12 V to energise the relay and is then released. When the voltage drops to the drop-out voltage the relay will switch off and the discharge will cease.
  • This circuit is not very good in that the relay drop-out is very slow and you may see some arcing on the contacts as they lose pressure.
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  • \$\begingroup\$ Collided in the doorway there. I'll just upvote yours. \$\endgroup\$ – Neil_UK Jul 3 '16 at 8:35
  • \$\begingroup\$ If the relay coil was placed across the headlamp and the pushbutton start switch across the relay contacts there will be much cleaner switch off at the end point and less arc problems because chatter is gone . \$\endgroup\$ – Autistic Jul 3 '16 at 11:39
  • \$\begingroup\$ @Autistic: I don't follow your scheme. Feel free to post a different version of my answer. SW1 is only used to power the circuit up so it won't affect drop-out. How would your scheme set the drop-out voltage of the relay to 10 V, for example? A typical 12 V relay may stay on until voltage drops to 5 or 6 V which would cause the battery to discharge to that level and, I presume, the OP would wish to avoid that. \$\endgroup\$ – Transistor Jul 3 '16 at 12:22
  • \$\begingroup\$ @transistor Remove R1, Place relay coil across the headlamp, Place the pushbutton across the relay contact . \$\endgroup\$ – Autistic Jul 3 '16 at 20:25
  • \$\begingroup\$ @Autistic: That's what you said the first time. How are you going to drop-out the 12 V relay at 10 V? (What I asked the last time.) \$\endgroup\$ – Transistor Jul 3 '16 at 20:28

On a simplified level your calculation is correct.

However the voltage of a Lead Acid battery is not constant. It follows a curve. So if you use a 2.4ohm resistor the current will vary according to the state of charge.

Depending on what you are trying to achieve this might still be Ok, just be aware of it.

Also be aware that your resistor will have to dissipate a lot of power:


enter image description here Original/Stackexchange post

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    \$\begingroup\$ Worth noting that for 5A and this battery. follow the C/20 curve. \$\endgroup\$ – Brian Drummond Jul 3 '16 at 10:33

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