# Mistake in calculating the last time of battery (Ampere hour problem)

I've a question about the mAh rating of batteries. At the Dutch Wiki page of Ampère-uur which is the Dutch word for Ampere hour, they say for example a $100$ Ah battery can supply 5 A for 20 hours by a temperature of 20 degrees Celsius. So, in my case I have a battery of 1 Ah and it can deliver $\frac{1}{20}$ A for 20 hours by a temperature of 20 degrees Celsius. So when my battery has to deliver $0.088$ A it will last for:

$$\frac{20\space\text{h}\times0.088\space\text{A}}{\frac{1}{20}\space\text{A}}\approx35.4\space\text{h}\tag1$$

But this seems wrong, because when I decreas the current comsumption to $0.2$ A the battery will last for $80$ hours which is higher then when I use the $0.088$ A.

Where is my mistake?

• Not sure how you've arrived at the above but it should just be 1 Ah / 0.088 A = ~ 11 hours. – PeterJ Dec 15 '17 at 11:30
• @PeterJ Is the Dutch definition of Amphour wrong? – adaa Dec 15 '17 at 11:31
• Try write down the numbers together with units – Gregory Kornblum Dec 15 '17 at 11:33
• I can see that, but i can't see Ah in the i initial calculation. – Gregory Kornblum Dec 15 '17 at 11:40
• @adaa unfortunately condensing units does not help with this error since the error is the inversion in the ratio so the units end up the same. – Trevor_G Dec 15 '17 at 12:28

You made the common mistake we all make now and again and inverted the ratio.

It should be

$$\frac{20\space\text{h}\times\frac{1}{20}\space\text{A}}{0.088\space\text{A}}\approx11.36\space\text{h}\tag1$$

Or you could simply have divided $1\text{Ah }/\text{ } 0.088\text{ A} = 11.36\text{ h}$

(Note the top part of the corrected equation just undoes what you did to calculate the 20hrs in the first place.)

Whether you actually get those numbers is however dependent on a number of other factors that are way beyond the simple equations and out of scope for an answer here. Unfortunately those "other factors" are rather less well defined mathematically and include things that have more to do with chemistry, and rates of exothermal reactions, than electronics. Suffice it to say, unless you are using extremely low currents or extremely large currents, the formula above is close enough.

Variance battery to battery can be large though. It really depends on the quality of the manufacturer and the history of what happened to the thing before you received it. If it's from a fully automated ISO company and fresh from the factory and not made on a Friday afternoon, the Ah number is probably within about 10% correct. If it's some outfit in Lower Bungholia that job lots manufacture out to local "house-wives" and ships them around in donkey carts... the numbers mean nothing.

• First of all thanks, can you edit the other factors into a new equation? – adaa Dec 15 '17 at 12:57
• @adaa unfortunatly the "other factors" are rather less well defined mathematically and include things that have more to do with chemistry, and rates of exothermal reactions, than electronics. Suffice it to say, unless you are using extremely low currents or extremely large currents, the formula above is close enough. – Trevor_G Dec 15 '17 at 13:03

Your calculation is wrong. Capacity (ampere hours) = current (Amperes) * time(hours). Thus a battery of 1Ah can supply 50mA for 20 hours, or 200mA for 5 hours, and so on.

• FYI the equivalence shown here is not valid with real batteries (only with theoretical "ideal" batteries). A real battery's capacity varies with the discharge current. For example, manufacturers often quote SLA battery capacity at the 20h rate. So if a battery was quoted as 1Ah at the 20h rate (so 50mA for 20h) it would have less capacity (say 0.9Ah) when discharged at 200mA = 4.5h duration, not 5h duration. See p2 of this datasheet from Yuasa "NP Discharge Characteristics Curves" shows 4x higher current means << 1/4 duration. HTH! – SamGibson Dec 15 '17 at 14:11