# Amp Hours

1. Aug 27, 2009

### JD88

I am trying to determine how long a particular battery will last for my application.

These numbers are just for an example.

Battery stats:
12 V 7 Ah

Application
1700 W

Power=Current*Voltage
1700W = Current * 12V
Current=141.7A

Time = 7Ah / 141.7A
Time = 0.05 hours

Is this correct, or are there other factors that I must consider?

2. Aug 27, 2009

### mgb_phys

That the wires would melt and the battery would explode if you could pull 140A from it.

Batteries have an internal resistance due to the chemistry - think of it as a small resistor in series with the battery.
As you draw a current, the resistance creates a voltage difference and so the voltage coming out of the battery drops, the current flowing through the resistance also generates heat in the battery.
For a lead acid battery this resistance is very low so you can get a very large current from a lead acid battery - but this doesn't mean that it is safe.

3. Aug 27, 2009

### JD88

I don't care about the numbers from my example. Just the procedure. Is that how I would go about determining how long the battery would last?

4. Aug 27, 2009

### ank_gl

The battery rating, 12V 7Ah, means that 7 amps can be drawn from the battery at 12 volts for an hour, this is its energy content.

For an appliance rated at 1700W, you need to calculate its resistance at its rated voltage, ie use P=V^2/R. Calculate the current requirement. Now 7 amps for 1 hr = x amps for y hrs!!
Or
Directly use energy balance.

Last edited: Aug 27, 2009
5. Aug 27, 2009

### bm0p700f

Sorry your method has confused me.

An example. A lithium ion battery rated at 4.4Ah powers a 20W bulb at 14.4V

20W/14.4V = 1.39A 4.4Ah/1.39A = 3.17hrs

These are the specifications of my mountain bike lights. They do run for approximatley 3.2 hours before the world goes dark and I am walking back to my car blind. Its dark in Thetford Forest when moon is new.

No resistance needed to work this out. Also if you follow your resistance method the current works out to be 141.7A. The same as JD88's answer. So your method is just a more complicated way of getting to same answer!

JD88 you are right!

6. Aug 27, 2009

### flatmaster

The amp hours is the total energy available in the battery.

7 Ahr *(3600s/hr) = 25200 Asec

power = energy / time

time = energy/power

time = 25200 Asec / 1700 W

time = 14 Sec