## How to calculate battery discharge time

1. The problem statement, all variables and given/known data

Discharge time from battery if you have a pump drawing 4.4 amps with 12 volts dc and a light drawing 0.0545 amps with 110 volts (ac).

Battery at max has specs: 12 volts, 144 watt hour, 12 amp hour.

2. Relevant equations

I'm not sure if you're able to add up the two amps that are being drawn (pump and light) then do the total current divided by what's being drawn.

3. The attempt at a solution

4.4+0.0545 = 4.4545 amps

12/4.4545 = 2.69 hours
I'm not sure if you're able to do what I did...can anyone help?

Thanks

 Quote by thejoshuaa 1. The problem statement, all variables and given/known data Discharge time from battery if you have a pump drawing 4.4 amps with 12 volts dc and a light drawing 0.0545 amps with 110 volts (ac). Battery at max has specs: 12 volts, 144 watt hour, 12 amp hour. 2. Relevant equations I'm not sure if you're able to add up the two amps that are being drawn (pump and light) then do the total current divided by what's being drawn. 3. The attempt at a solution 4.4+0.0545 = 4.4545 amps 12/4.4545 = 2.69 hours I'm not sure if you're able to do what I did...can anyone help? Thanks
well you have two appliances drawing current, one draws 4.4 coloumb of charge per second and another draws 0.0545 coloumb of charge per second. ( that is the definition of current )

only difference is that, the 4.4 is drawn at a steady rate and the 0.0545 is drawn on average, but since the frequency of the ac current is much higher than a second, the amount of charge drawn in one second is still 0.0545.

So I see no problem with what you have done.
 Recognitions: Gold Member I see a problem. You need to look at this in terms of the fully charged amount of energy in the battery (Watt-hours) and the rate at which energy is supplied to the loading devices (Watt) For this problem, you should note the assumption that the AC device is being supplied by a zero-loss DC-to-AC converter.

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## How to calculate battery discharge time

Note that you're given the battery specs in terms of both amp hours and watt hours. You may find the latter to be more convenient since the devices are running at different voltages. The light, requiring 110V ac, must have a device associated with it that will take the 12V of the battery and produce a 110V ac supply for it. This sort of device is usually called an inverter.

If the inverter is 100% efficient then it will draw from the battery the same power (watts) as it supplies to the light.
 Recognitions: Gold Member Its amazing how one's mind can snap to a potentially invalid situation. I should have said that there is nothing that would prevent attaching the AC device to a 12V battery directly except sanity, perhaps (no magic converter needed--in this case). I can't tell from the question if this is indeed the case.

 Quote by lewando I see a problem. You need to look at this in terms of the fully charged amount of energy in the battery (Watt-hours) and the rate at which energy is supplied to the loading devices (Watt) For this problem, you should note the assumption that the AC device is being supplied by a zero-loss DC-to-AC converter.
I'm sorry I don't understand, doesn't the battery spec of 12 amp hours mean that it can give a 1 amp supply for 12 hours?

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 Quote by Idoubt doesn't the battery spec of 12 amp hours mean that it can give a 1 amp supply for 12 hours?
It can. At 12V. 12 amps for 1 hour at 12V.

 Quote by lewando It can. At 12V. 12 amps for 1 hour at 12V.

then does this not allow us to calculate the total charge inside the battery? And after that isn't it just a matter of finding out how long it takes to use up that charge?

why is the the 12V important?

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 Quote by Idoubt then does this not allow us to calculate the total charge inside the battery? And after that isn't it just a matter of finding out how long it takes to use up that charge?
That's the idea. I prefer the term energy instead of charge because the analysis becomes simpler.

 why is the the 12V important?
Voltage is important because the "amp hour" rating is only defined for a battery's nominal output voltage. For example, would you want to power your circuit with a 12 amp-hour battery with a nominal output voltage of 12V or 5000V? Which one would be easier to lift? Which one would make your pump act as a temporary source of light? ;)
 Still not getting it even if a 12V battery is supplying a 110V ac device ( presumably through some transformer network) as long as we know how much current leaves the battery shouldn't the voltages be irrelevant ?
 Recognitions: Homework Help What's delivered to the device is Energy, and more specifically, the rate of energy delivery is in Watts. Energy is conserved. In order to deliver 0.0545 Amps at 110V to the device which comes to about 6W, you need to draw the same amount of power from the 12V battery. Otherwise you're getting something for nothing, and you've invented perpetual motion!

 Quote by gneill What's delivered to the device is Energy, and more specifically, the rate of energy delivery is in Watts. Energy is conserved. In order to deliver 0.0545 Amps at 110V to the device which comes to about 6W, you need to draw the same amount of power from the 12V battery. Otherwise you're getting something for nothing, and you've invented perpetual motion!
Agreed.

But can we not approach the problem with conservation of charge? I mean both energy and charge are conserved right?

If I find the solution with one, shouldn't it also be the solution with the other? ( assuming the data given is realistic )

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 Quote by Idoubt even if a 12V battery is supplying a 110V ac device ( presumably through some transformer network) as long as we know how much current leaves the battery shouldn't the voltages be irrelevant ?

As long as the voltages are the same. Take a look at that converter for a moment. As gniell pointed out, the converter must provide 0.0545A at 110V = 6W to the light. For the battery to provide 6W to the converter, it must provide 0.5A at 12V = 6W. So the current coming out of the battery (at 12V) is...

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 Quote by Idoubt Agreed. But can we not approach the problem with conservation of charge? I mean both energy and charge are conserved right? If I find the solution with one, shouldn't it also be the solution with the other? ( assuming the data given is realistic )
In one case the charges are being moved in a potential of 12V, in the other a potential of 110V (rms). The work done is not the same for the same amount of charge moved.

 Quote by lewando As long as the voltages are the same. Take a look at that converter for a moment. As gniell pointed out, the converter must provide 0.0545A at 110V = 6W to the light. For the battery to provide 6W to the converter, it must provide 0.5A at 12V = 6W. So the current coming out of the battery (at 12V) is...
I see , so the current that leaves the battery is actually,

4.4 +0.5 = 4.9A ?
 Recognitions: Gold Member Yes.