Car battery amps?

Averagesupernova
Gold Member
I don't know how big the factor is from cold vs hot on a filament russ. The cold resistance is alot lower than most people think. It is usually assumed (mistakenly) that it is closer to the hot resistance. Measure it with an ohmeter cold and compute it hot (full voltage) and you'll have it.
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On the inverter, don't forget to figure in the inefficiencies of the inverter.

NoTime
Homework Helper
The cold resistance of a 100w 120v lamp is around 10 ohms.
Quite small.
The hot resistance is 144 ohms @ 120v.
P=EI and E=IR.
No idea what the resistance might be at 12v, but probably fairly close to the cold resistance.

IIRC the old sealed beam automobile headlights were around 100W at 12v.
Don't know about the halogen variety.

Danger
Gold Member
IIRC the old sealed beam automobile headlights were around 100W at 12v.

Where I live, the legal limit for high-beams is 85W; for low-beams it's 55. Those laws were based upon incandescent bulbs, and have never been changed. (Needless to say, my lows are 85 and my highs 100, with 2 x 100W driving lights and a couple of 1,000,000 cp spots in reserve. )

russ_watters
Mentor
I don't know how big the factor is from cold vs hot on a filament russ. The cold resistance is alot lower than most people think. It is usually assumed (mistakenly) that it is closer to the hot resistance. Measure it with an ohmeter cold and compute it hot (full voltage) and you'll have it.
Why didn't I think of that...?

...[...30 seconds later....]...

The resistance of a cold 100w lamp is about 10.3 ohms. So that means at start-up at 120V, it draws 11.6 A and would dissipate just under 1400w, if only for a tiny fraction of a second.

At 12V, it would draw 1.16A and dissipate 14w.

Averagesupernova
Gold Member
I know that the very common sealed beam lights used for 30 or more years will cause a 20 amp power supply to go into current limit. I've used them in various series/parallel to test power supplies at different loads. I can make and break contact enough on the power lead to cause the filament to heat up enough so that the power supply does not go into current limit mode. Then the power supply will light the bulb with ease. Gives an idea of how a filament can act. I also have a variac that I have used as a desk lamp light dimmer on many occasions but have never plotted the voltage/current relationship throughout the voltage range.

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http://www.arrl.org/catalog/?item=9531#top

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led lights would be a smarter choice way less draw

sophiecentaur
Gold Member
2020 Award
This thread is a real muddle from beginning to end. The "45 Amps" marked on a battery must be 45 Amp Hours (not Amps per hour or anything else). That means you can get 1A for 45 hours or 5A for 9 hours or 0.5A for 90 hours. It doesn't actually imply that you could take things beyond 45A for 1 hour, though (the Amps times Hours doesn't extend beyond a 'reasonable' value of current drain). It is a single figure that tells you quite a lot about a battery, though.

Also, you can say nothing about how a light bulb's resistance will change as you alter the supply volts - it may vary by a factor of 10:1. Applying an appropriate voltage to a bulb, you can get any power you like (until it blows) but, except at its design voltage, you can't use it as a reliable measuring device. You need a 'heater' resistor, which will not change resistance over a large range of powers because it doesn't get white hot.

The answer to the original question is 45Ah divided by 8A =5.5 hours approx. But remember - taking more than half the charge out of a battery will compromise its lifespan. So you should really think in terms of 3 hours in practice.

Averagesupernova
Gold Member
Do you realize how old this thread is and that you have simply repeated what has already been said?
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And Taz, for the sake of resistors and light bulbs AC (using RMS voltage and current) is pretty much the same as DC. 100 volts RMS through a resistor will dissipate the same watt as 100 volts DC.

CCA= the amperage a battery can deliver at -18*C (or 0*F) for 30 seconds and maintain a terminal voltage above 7.2V.

Ah= Amp Hour Rating, which is the number of amps a battery delivers at 27*C (80F) for 20 hours and maintains a voltage above 10.5V.*

*The Ah indicates the battery's ability to supply power when, say, the alternator has failed.

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CCA= the amperage a battery can deliver at -18*C (or 0*F) for 30 seconds and maintain a terminal voltage above 7.2V.

Ah= Amp Hour Rating, which is the number of amps a battery delivers at 27*C (80F) for 20 hours and maintains a voltage above 10.5V.*

*The Ah indicates the battery's ability to supply power when, say, the alternator has failed.

Well Done Mate! Thanks for ending this discussion! It really gave me a headache.

I actually wont take on anything anyone has stated as true. Best to obtain the manufactures Specs. on the battery. Rather than fool around with potentially deadly CAR Batterys.

*It is good for people to seek knowledge, but it's much safer to google the facts than try re-discovering the wheel.

*This wheel was spuare - not round. Thats why it took 3 years work to make it round, from many so called experts. And few with knowledge.

Ok folks, this is freshman electronics. CCA is the measurement of the amount of amperage a battery can produce under extreme load. Amp Hours is the amount of amperage a battery can produce over a longer period of time at a lower load, say for instance a car radio.

RC is equal to approximately 1/5 of CCA.

Amp/Hours = (Reserve Capacity / 2) plus 16

I know this because I work for a data company and I have to test and certify every battery back up, every quarter, in every security panel and ups we have installed in the midwest. I learned it at ITT Tech.

sophiecentaur
Gold Member
2020 Award
That, above, will be a rule of thumb for a particular battery. The "16" in the formula gives it away; it would not be the same for a large battery as for a small one.

I just read all the posts in this four-year thread and feel much more knowledgeable with all the back and forth. My reason for following this thread is that I was hoping to find out how long my 5 24-V deep cycle batteries in series (for example) could power a 1,500-Watt SkilSaw after inversion to 120-V AC. Assuming no loss in the inversion process, I'm figuring the Saw will be drawing 12.5 Amps (1,500 / 120); but I'm still not sure how to figure out how long these fully-charged batteries can generate 12.5 amps without a recharge (from a solar panel or wind turbine).

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i googled and learned to use the formula Watts x volts = amps, how are you getting 10 watts from a 100 watt bulb if you don't mind, from what i've read i've learned to do it like this.

100 watt bulb at 120 volts draws 100/120 = .8 amps
100 watt bulb at 12 volts draws 100/12 = 8.3 amps

or you could do watts like this for example:

.8 amps from 120 volts draws .8 * 120 = 96 watts per hour
8 amps from 12 volts draws 8 * 12 = around 96 watts per hour

the cold cranking amps on the battery is labeld 650 CCA or cold cranking amps, i would assume that would mean the battery could take a load of 6 100 watt bulbs for a very short period of time.

now that the battery is 650 CCA that would mean 650 * 12 = 7800 watt hours, which they are made to take loads like that for small amounts of time? if 45 amp hours was right then 45 * 12 = 540 watt hour, or 45 amp hour, same thing in my book.

so the question is whats the amp hours of the battery, i read on the internet the other day that the basic car battery holds about 45 amp hours, and i also read somewhere else last night that a basic car battery holds around 60 amp hours or 60 * 12v = 720 watt hours, meaning it could produce 720 watts in one hour and no more. i think marine batteries start at 100 amp hours and go up from there?

i need to get me a ohm or amp meter and learn to measure the ohms or amps to see more about this. by the way, i like this thread, im learning lots about amps watts and volts.

if my 650 CCA battery in my truck was 650 amp ours, that would mean it was 7800 watt hours, if the battery produced that over time then that would mean it would run a 100 watt light bulb 7800 watts/100 watts = 78 hours, or 100 watt bulb on 12 volts uses 100/12 = 8.3 amps, so
650 amps / 8.3 amps = 78 hours, same formula as above giving the same hours, which i dont think 650 CCA is what you go by? funny how that works out.
650 CCA means it can deliver 650 amps at peak capacity, which is only a few seconds. It does not mean you have 7800 watt hours. 650 doesn't say anything about how much watt-hour or amp-hour. You have to get that from the manufacturer directly because often the battery doesn't label them. I had this problem and had to go to the manufacturer's website to look up the detail specification. They list the amp-hour (AH).

A 650 CCA battery usually has about 10-14 Amp-hours.

There is no simple mathematical equation that relates the amp-hour to the Watt/Volt/Amp because the amp-hour is determined by a combination of the physical size of the battery as well as its chemical internal structure. In general, a bigger battery will have more amp-hour than a smaller one, but that relationship involves too many variables to calculate.

Once again check the manufacturer's website for the detail spec. You can't get the amp-hour from calculations or explanation of physics. The manufacturer determine the amp-hour by measuring the battery, not by calculation.

You can measure it too by turning on your car radio and see how long it lasts :)p
If you know the watt. of the car radio, the voltage of the battery, you can determine the amp. Multiply the amp with the hours that your car radio lasts, then you get the amp-hour. ;)p

I just read all the posts in this four-year thread and feel much more knowledgeable with all the back and forth. My reason for following this thread is that I was hoping to find out how long my 5 24-V deep cycle batteries in series (for example) could power a 1,500-Watt SkilSaw after inversion to 120-V AC. Assuming no loss in the inversion process, I'm figuring the Saw will be drawing 12.5 Amps (1,500 / 120); but I'm still not sure how to figure out how long these fully-charged batteries can generate 12.5 amps without a recharge (from a solar panel or wind turbine).
what's the make and model of the battery? google for that model and go to manufacturer's site.

From my preliminary google for "24v deep cycle battery" I've found that they range as much from 14AH to 100AH, depending on the size.

since you connected the batteries in series in order to achieve 120 volts, the AH capacity of the entire system will be the same as the capacity of a single battery. So find out that AH from the manufacturer and divide by the 12.5 amps that you've already figured out. That will give you the hours it last.

davenn
Gold Member
i googled and learned to use the formula Watts x volts = amps, how are you getting 10 watts from a 100 watt bulb if you don't mind, from what i've read i've learned to do it like this.
100 watt bulb at 120 volts draws 100/120 = .8 amps
100 watt bulb at 12 volts draws 100/12 = 8.3 amps
or you could do watts like this for example:
.8 amps from 120 volts draws .8 * 120 = 96 watts per hour
8 amps from 12 volts draws 8 * 12 = around 96 watts per hour

You are missing the point that averagesupernova made a few msgs back

a 100W bulb is designed to draw/use 100W AT 120V NOT 12V, at 12V its resistance is going to be somewhat higher that its gonna draw next to nothing and its probably not even going to glow.

a 100W bulb for 12V is going to have a very different cold and hot resistance than a 100W bulb at 120V

ok hunted around the house for a couple of globes closest in wattage I could find .....

40W 240VAC globe (we have 240V mains here in Oz) and a
20W 12V globe

cold resistance of 40W globe = 100 Ohms

cold resistance of 20W globe = 5 Ohms

On a 13V supply the ...

40W 240V globe drew 53mA (no Glow)
20W 12V globe drew 320mA (normal bright glow)

there's my 2 cents worth

cheers
Dave

davenn
Gold Member
interesting....
that 20W globe ....

13V x 320mA = 4.1W maybe its been mislabelled will have to try and find another 12V rated globe for a test :)

Dave

Averagesupernova
Gold Member
You cannot expect cold filament ohmeter readings to reflect wattage.

sophiecentaur
Gold Member
2020 Award
Why not use auto bulbs (say 100W worth) and just measure current? If it's that vital to know the actual capacity of the battery then test it under real conditions. But mind you don't let it dip much below 12V or you will start to damage the battery.

hello, i am going to do my project work on automatic hydraulic jack which will be driven by an electric wiper motor.for this wiper motor it requires of 100 watt to operate and requires a torque of 40 nm so can the vehicles battery power can provide that sufficient amount of power to this wiper motor???your help will be very grateful to me.

sophiecentaur
Gold Member
2020 Award
hello, i am going to do my project work on automatic hydraulic jack which will be driven by an electric wiper motor.for this wiper motor it requires of 100 watt to operate and requires a torque of 40 nm so can the vehicles battery power can provide that sufficient amount of power to this wiper motor???your help will be very grateful to me.

The torque is not directly linked to the power. With appropriate gearing, you can get any torque you want because Power is Torque times Speed. If your motor uses 100W (when not stalled), does it produce 40nm at normal speed as opposed to requiring 40nm? How fast do you need the jack to operate? If the speed of the jack operation is not critical then you can use pulleys and a coupling belt between motor and pump to get the gearing you need to match the motor torque..

The torque is not directly linked to the power. With appropriate gearing, you can get any torque you want because Power is Torque times Speed. If your motor uses 100W (when not stalled), does it produce 40nm at normal speed as opposed to requiring 40nm? How fast do you need the jack to operate? If the speed of the jack operation is not critical then you can use pulleys and a coupling belt between motor and pump to get the gearing you need to match the motor torque..

just to lift thevehicle of around 4 to 5 tons from the hydraulic jack which is now a days generally operated by manually but i m just trying to transfer the rotation of wiper motor to the to and fro motion that is linear motion by linkages such as quick return mechanism,slider crank mechanism and by the calculation we did we just need the wiper motor which can provide the torque to the linkage arm of aroubd 50nm with the speed of 38 rpm that is at lowest speed which can obtaiin that power is torque*speed which forms the value of around 100 watt and this motor has to be driven from the vehicles internal power so i would want to know that can the vehicle internal power or vehicles battery can provide that sufficient amount of power to the vehicle??i hope you have understood what i n trying to say so please ur answer or suggesion will be very grateful n helpful to me thank you

russ_watters
Mentor
Yes, a car battery can provide 100W for a decent amount of time.