How long will an inverter run?

Submitted: Saturday, May 24, 2008 at 07:50
ThreadID: 57939 Views:6697 Replies:5 FollowUps:5
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Learned ones (suck, grovel)
As per last post, preparing for life on the road.
I'm going to put an inverter in the car. I'm an absolute dunce on electrical stuff, but good at maths. Trying to determine what size inverter and what I should promise my bride can be powered.

Is there a formula to calculate how long my deep cycle aux battery will power a particular appliance through the inverter without the battery being charged?
eg How long will a 50 AmpHr battery run a 400 watt appliance?
I know there will be lots of variables, but I'm just looking for an indication.
Thanks in advance.
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Reply By: Col_and_Jan - Saturday, May 24, 2008 at 08:01

Saturday, May 24, 2008 at 08:01
I am not a learned one, but a 400W appliance, without any inefficiencies would be a 400W/12V = 33 A drain on your battery.
50AH at 33A drain would only take 1.5 hours to completely drain the battery.

Probably less than that at such a high current draw, leaving your battery feeling pretty sorry for itself (ie stuffed!).

AnswerID: 305563

Follow Up By: tuck 01 - Saturday, May 24, 2008 at 08:24

Saturday, May 24, 2008 at 08:24
Thanks Col.
Looks the formula is simply:
Watts divided by Volts equals Amps.

If so, from that I can make decisions on inverter size, battery replacement, promises to my bride etc.
FollowupID: 571598

Follow Up By: Col_and_Jan - Saturday, May 24, 2008 at 10:31

Saturday, May 24, 2008 at 10:31
Sorry, I was probably being a little bit over the top.

Another thing to remember is, the capacity of the battery is Amp.Hours and not Amps per Hour, so 50AH has in rough terms, 50A continuous supply for 1 hour, 25A for 2 hours, and 2.5A for 20 hours. A compressor fridge uses probably an average of 2.5A.

It doesnt work out exactly that as a high current draw uses up the battery more quickly, and you should limit the drain on the battery to go down to 50% if you want it to last.

A lot of caravanners limit the inverter to about 300W, and use a pure sine wave inverter for electronic equipment.
FollowupID: 571611

Reply By: k1w1 - Saturday, May 24, 2008 at 10:22

Saturday, May 24, 2008 at 10:22
I have found the following helpful at times. Sorry its a bit long winded.
Calculating the right sized battery for your needs.

Item of equipment Loading Watts x Estimated hours use = Watt Hours
x =
x =
x =
x =
Total Watt Hours =

Watt Hrs ÷ by volts (system voltage) = Ampere Hours

Ampere Hours already estimated =

Plus 10% allowance for cable loss =

Plus 25% over-capacity allowance =

Plus 50% vehicle starting margin =
(if applicable)

Total Ampere Hours Required =

Calculate the ‘Watt Hours’ by multiplying the loading of each piece of equipment (this is expressed in watts and is stamped into the compliance plate attached to the item) by the number of hours you intend to use it between charges.

Determine the Ampere Hour (AH) requirement that the battery must accommodate.
A cable loss margin of 10% is usually appropriate. In addition, it is nice to have a little more capacity than you need. Hence an over-capacity margin of 25%.
If the battery will be needed for vehicle starting you will need to increase your Ampere Hour estimate by 50% to ensure you still have starting power when the battery has been partially discharged.

How long will the Battery last running my Inverter?

AC Watts = figure written on AC (240 volt) appliance

AH = Ampere Hours capacity of the battery

(AC Watts÷10*) = TIME IN HOURS

Example: a 16” TV (67 watts) running on a 130AH battery

(130÷2) 65
(67÷10*) = 6.7 = 65 ÷ 6.7 = 9.7 HOURS

Note: *Use a figure of 20 if your battery voltage is 24V DC.
Only 50% of the batteries capacity will be used to allow it to recover.
Cheers Alan
AnswerID: 305579

Follow Up By: Member - barry F (NSW) - Saturday, May 24, 2008 at 12:38

Saturday, May 24, 2008 at 12:38
Thanks Jeff for putting this question up as I have been wondering how to calculate how long the battery would power a given appliance.
Alan, am I right in saying that if I have a 95 amp hour deep cycle battery and wanted to calculate how long it would power a 15 Watt light globe you would divide the 15W by 12V = 1.25 AMPS divided into the 95 equalling 76 hours of which you would only use 50% now giving an effective use of 38 hours for one 15 Watt globe.

Likewise, a TV that has 2.6A stamped on the back : 2.6 divided into 95 = 36 hours of which would give an effective use of 18 hours?
FollowupID: 571624

Reply By: k1w1 - Saturday, May 24, 2008 at 10:26

Saturday, May 24, 2008 at 10:26
Sorry the first part of the table didint copy. It is basically, loading in watts x the estimated hours of use = watt hours
AnswerID: 305580

Reply By: oldtrack123 - Saturday, May 24, 2008 at 10:29

Saturday, May 24, 2008 at 10:29
Hi Jeff
Also remember for long battery life they should not be discharged below 50%of capacity ie 25amp hrs for a 50amp/hr dc battery[do not take below 11.6v]
You can get low voltage cutouts [operate @ about 11.6v]to give some protection to battery [most only rated @10 amps]
AnswerID: 305582

Follow Up By: k1w1 - Saturday, May 24, 2008 at 14:36

Saturday, May 24, 2008 at 14:36
barry F
Just about spot on with the globe. For the TV calc you need to use the AC watts figure not the amps of it. Then follow the formula.
Cheers Alan
FollowupID: 571643

Reply By: tuck 01 - Saturday, May 24, 2008 at 17:24

Saturday, May 24, 2008 at 17:24
Thanks all for the info and advice.
I didn't think your first answer was too abrupt Col.
Hope this helped you as much as it helped me Barry.
Gotta love the collective wisdom of the forum.
AnswerID: 305629

Follow Up By: Member - barry F (NSW) - Saturday, May 24, 2008 at 17:33

Saturday, May 24, 2008 at 17:33
Yep, it certainly was a help & you are dead right about the good info from a wide variety of people willing to help no matter what the subject, they are a good bunch. Many thanks to all & cheers.
FollowupID: 571682

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