Test a battery pack. How can i easily create a high amp load?
- Thread starter eikido
- Start date
Frank
100 W
What voltage? Electric heater controlled via contactor is one possibility.
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Frank
100 W
Maybe try some incandescent light bulb(s) but a small heater might be better (depends on how much load you want to create).
A 1500 watt heater (or toaster) draws about 12-13 A at 120V (mental arithmetic) so at 60V would be half that.
Someone else want to check my numbers please?
A 1500 watt heater (or toaster) draws about 12-13 A at 120V (mental arithmetic) so at 60V would be half that.
Someone else want to check my numbers please?
That's not a bad idea.
Here is a 2000w heater at 230v.
https://www.biltema.se/en-se/construction/fans/heating-fans/heating-fan-2000-w230-v-2000034997
But how would it behave if i connected it to a 36v source?
It is specified to 230/2000w
->230v source
R = VxV / P = 230x230/2000= 26,45 ohms
->36v source
I = V / R = 36 / 26,45 = 1,36 amps?
So it would only consume 1,36 amps? I might need more
Here is a 2000w heater at 230v.
https://www.biltema.se/en-se/construction/fans/heating-fans/heating-fan-2000-w230-v-2000034997
But how would it behave if i connected it to a 36v source?
It is specified to 230/2000w
->230v source
R = VxV / P = 230x230/2000= 26,45 ohms
->36v source
I = V / R = 36 / 26,45 = 1,36 amps?
So it would only consume 1,36 amps? I might need more
Frank
100 W
^^^
Great idea if it will accept the input voltage.
Great idea if it will accept the input voltage.
Define easily
And note that it's best for cap testing if possible to do individual cells rather getting a pack average - depends on the batt type of course.
I don't know the math myself, but can just use resistors.
They make aluminum and ceramic wire-wrap ones, seen 1000W and up in different ohms.
People toss them in a container of water if air cooling is too slow.
And note that it's best for cap testing if possible to do individual cells rather getting a pack average - depends on the batt type of course.
I don't know the math myself, but can just use resistors.
They make aluminum and ceramic wire-wrap ones, seen 1000W and up in different ohms.
People toss them in a container of water if air cooling is too slow.
ginekolog
10 W
Buy 4 50W 12V halogent bulbs, connect in series and you get 200W 48V 5A load. Thats how I test my batt.
https://endless-sphere.com/forums/search.php?keywords=batter*+load*+test*&terms=all&author=&sc=1&sf=titleonly&sr=topics&sk=t&sd=d&st=0&ch=300&t=0&submit=Search
more results, not as many proportionally relevant
https://endless-sphere.com/forums/search.php?keywords=load*+test*&terms=all&author=&sc=1&sf=titleonly&sr=topics&sk=t&sd=d&st=0&ch=300&t=0&submit=Search
more results, not as many proportionally relevant
https://endless-sphere.com/forums/search.php?keywords=load*+test*&terms=all&author=&sc=1&sf=titleonly&sr=topics&sk=t&sd=d&st=0&ch=300&t=0&submit=Search
Buy 4 50W 12V halogent bulbs, connect in series and you get 200W 48V 5A load. Thats how I test my batt.
[/quote]
Awesome! I will try this It will be a cheap test. Thanks.
So the resistance over one 12v, 50w halogen bulb is really low?
Something like 2.5 ohms (2.5x4=10 ohms)?
I=48v/10ohms=4.8A
[/quote]
Use of incandescent bulbs (including halogen) is a poor means of load testing if there is any need for measurement, like capacity testing. Filament resistance of an incandescent bulb varies extremely with filament temperature. When I tested one common automotive bulb (194), the resistance was about 5 ohms cold and 60 ohms hot, a factor of 12. Resistance measured cold with a meter is low. Resistance with power applied to the bulb rises with temperature, which may be roughly estimated by color. It is calculated by measuring current through the bulb and voltage across it and applying Ohm's Law. As voltage is raised, the filament gets hotter/brighter, resistance rises, and the resistance calculated thus should be close to that calculated from the rated wattage of the bulb at its rated operating voltage.
In the example given of a 12V 50W bulb, current (amps) is 50/12 = just over 4A. R is V/A = 12/4 = 3 ohms. Power is 12 x 4(and a tad) = 50 watts. At 6V, resistance might be 1.5 ohms, thus 6/1.5 = 4A; 6V x 4A = only 25W.
Connecting 4 50W bulbs in series will not result in a 200W load as stated above because the resistances are additive. Neither will connecting them in parallel, except when they are operated at rated voltage.
Use of suitable resistors will give much greater accuracy. Their resistance too will vary some with temperature, but if they are of suitable wattage, they will not get very hot and it will not vary much.
Feel free to check my arithmetic. I think youtube has better tutorials on Ohm's Law than I can provide.
[/quote]
Awesome! I will try this
It will be a cheap test. Thanks.So the resistance over one 12v, 50w halogen bulb is really low?
Something like 2.5 ohms (2.5x4=10 ohms)?
I=48v/10ohms=4.8A
[/quote]
Use of incandescent bulbs (including halogen) is a poor means of load testing if there is any need for measurement, like capacity testing. Filament resistance of an incandescent bulb varies extremely with filament temperature. When I tested one common automotive bulb (194), the resistance was about 5 ohms cold and 60 ohms hot, a factor of 12. Resistance measured cold with a meter is low. Resistance with power applied to the bulb rises with temperature, which may be roughly estimated by color. It is calculated by measuring current through the bulb and voltage across it and applying Ohm's Law. As voltage is raised, the filament gets hotter/brighter, resistance rises, and the resistance calculated thus should be close to that calculated from the rated wattage of the bulb at its rated operating voltage.
In the example given of a 12V 50W bulb, current (amps) is 50/12 = just over 4A. R is V/A = 12/4 = 3 ohms. Power is 12 x 4(and a tad) = 50 watts. At 6V, resistance might be 1.5 ohms, thus 6/1.5 = 4A; 6V x 4A = only 25W.
Connecting 4 50W bulbs in series will not result in a 200W load as stated above because the resistances are additive. Neither will connecting them in parallel, except when they are operated at rated voltage.
Use of suitable resistors will give much greater accuracy. Their resistance too will vary some with temperature, but if they are of suitable wattage, they will not get very hot and it will not vary much.
Feel free to check my arithmetic. I think youtube has better tutorials on Ohm's Law than I can provide.
markz
100 TW
Get some power resistors and put it in a cup of water to keep them ultra cool.
angusinalberta
1 mW
An old electric kettle will do the job.
Mine is rated 1500 W on 120 V
Since power is proportional to V squared we can readily find the power at any other voltage.
For a 48 V battery, for example, the power will be
1500 W * 48^2 / 120^2 = 240 W
The current will be 240 W / 48 V = 5 A That's probably close to what you want.
There is a glitch of course. The kettle will have a thermal switch that shuts it off when it starts to boil. You might want to disable the switch and then be _very_ careful to keep the kettle topped up. Otherwise you might fry your kettle. (or burn down your house)
Mine is rated 1500 W on 120 V
Since power is proportional to V squared we can readily find the power at any other voltage.
For a 48 V battery, for example, the power will be
1500 W * 48^2 / 120^2 = 240 W
The current will be 240 W / 48 V = 5 A That's probably close to what you want.
There is a glitch of course. The kettle will have a thermal switch that shuts it off when it starts to boil. You might want to disable the switch and then be _very_ careful to keep the kettle topped up. Otherwise you might fry your kettle. (or burn down your house)
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