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