This is why I never became an EE

[Devin]

Measuring stuff that stands still is easy. Velocity isn't very hard either. But I'll never, ever get wattage, voltage and amperage straight (or at least it seems), especially not when considering they are always in motion.

So in an effort for "concisity", I'll just ask: if I have a device taking 60W, and it's drawing it from a car battery (at 12V obviously), how long would the battery last before draining completely?

For reference this should be the specs from my Die Hard:

Power Configuration: Cell Type: Lead-Acid Number of Cells: 6 Voltage: 12.0 Power Ratings: Cold Cranking Amps (CCA at 0 deg.F): 575 Reserve Capacity (RC): 84 min.

 

[Off Road SHO]

Paging SHOAZ, phone call for Mr SHOAZ.

Tom

 

[hawkeye18]

The RC of a battery is defined as the amount of time a battery can maintain a useful voltage (probably ~>11v, I don't have an exact figure) at a constant 25A draw. Since your draw is 5A (60w/12v=5a), you would simply multiply 84 minutes by 5 (25/5=5) to obtain a figure of 420 minutes (duuuuuuuuuuuude), or 7 hours.

Hope this answers your question.

 

[Eric VerValin]

The reaction between the acid and lead plates makes that problem kind of hard to do.

This is how I tell people who don't have a clue about electricity....

Wattage is simply this... (Power)Watts = Voltage x Amperage .... always has, always will.

Now... think about a 60 watt light bulb in a house... 60w = 120v x ? Simple division will tell you its right at .5 Amps. Put a 60 Watt lightbulb in a car with a 12 volt system, it changes 10 fold. Now your drawing 50 amps. 60Watt = 12v x ? So obviously as voltage decreases, current (amperage) increases....

Think of it like a highway full of traffic.. bumper to bumper, nice and evenly spaced. Think of each lane of traffic as your "amperage", and the speed of the cars as being your voltage...

Now back to the thing about using a lead acid battery... there will be a rate of recharge.. and I have no clue what it is.... I deal with higher voltages usually, and I just don't have that much experience with batteries... maybe someone else can chime in with a forumla or two... I'm kind of curious myself... I was trying to figure something like this out a while ago, and gave up after about 10 seconds of realizing I had to go look it up.. lol

 

[hawkeye18]

The reaction between the acid and lead plates makes that problem kind of hard to do.
(snip)

Now... think about a 60 watt light bulb in a house... 60w = 120v x ? Simple division will tell you its right at .5 Amps. Put a 60 Watt lightbulb in a car with a 12 volt system, it changes 10 fold. Now your drawing 50 amps. 60Watt = 12v x ? So obviously as voltage decreases, current (amperage) increases....
(snip)

Now back to the thing about using a lead acid battery... there will be a rate of recharge.. and I have no clue what it is.... I deal with higher voltages usually, and I just don't have that much experience with batteries... maybe someone else can chime in with a forumla or two... I'm kind of curious myself... I was trying to figure something like this out a while ago, and gave up after about 10 seconds of realizing I had to go look it up.. lol

Dude... what in the F are you talking about? 60/12 is not 50! it's 5. lol stop drinking so many red bulls. You make no sense...

 

[jonheese]

Dude... what in the F are you talking about? 60/12 is not 50! it's 5. lol stop drinking so many red bulls. You make no sense...

He almost had it... He did say that it changes 10-fold, which was right, except that 0.5 x 10 = 5, not 50.

Maybe just a typo?

Regards,
Jon Heese

 

[gmorrell]

There might be a capacity rating on the battery, it will be rated in Amp/hours. Most Group F car batteries are about 40 Amp/hours with a full charge.

What this means is that it will supply 40A @ 12V for 1 hour, or 20A @ 12V for 2 hours, or 10A @ 12V for 4 hours, and so on. All of this is sort of simplifying things and assuming a flat discharge curve, so it's approximate, but damn close.

60W at 12V is 5 Amps, so a 40 Amp/hour battery would survive 40/5 or about 8 hours on a 60W load.

 

[hawkeye18]

This is also assuming that your battery is healthy and fully charged, it has not sulfated or shed plate material, that its fluid has not boiled, etc... all of these things will lower your reserve capacity by quite some bit.

I'm not an EE, but knowing some, I wouldn't want to be .

 

[Devin]

Wow, thanks for all the info. The hardest part for me to get is when introducing loads over time, but that simplifies it quite a bit. Of course I will throw in larger numbers than what is needed just to be certain that I'm within a margin of error as well.

 

[Off Road SHO]

So, why are you wanting to light up a 60 watt bulb for? Hmmmm?


Tom

 

[Eric VerValin]

He almost had it... He did say that it changes 10-fold, which was right, except that 0.5 x 10 = 5, not 50.

Maybe just a typo?

Regards,
Jon Heese

LoL yea it was.. I was on the phone and had a nagging woman in my other ear... I've only done electrical work for about 10 years now.. I'm not guessing..


And hawk... Red Bulls are nasty.. unless you add some Jagermeister..

 

[Devin]

That's about the wattage of a low-powered computer...

 

[Eric VerValin]

www.mp3car.com lotta neat stuff there.. if you were wanting to put a computer in your car..

 

[SHO_DOODmorrris]

LoL yea it was.. I was on the phone and had a nagging woman in my other ear... I've only done electrical work for about 10 years now.. I'm not guessing..


And hawk... Red Bulls are nasty.. unless you add some Jagermeister..

first of all red bulls are not nasty and vodka is better 50:50 to be exact

 

[firebat45]

That's about the wattage of a low-powered computer...

If this is indeed what it is for, get a proper shutdown controller. I wouldn't be okay with any amount of drain that's going to let the battery die within 5-7 days, which means you have to be down in the 200mA range, there's no computer that is that low.

I have an M4-ATX in mine, the PC is up and running about the same time I get onto the street.

If this isn't for a CarPC, then disregard my whole post.