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How to figure out how much current lights will use.
This isn't just for lights but for all electronics (accept amplifiers).
What you need:
- You need to know the power and voltage of the item. Use 11.5V for a worst case scenerio in a vehicle.
What to do:
- Input the power here (If it's a pair of 150W lights, insert 300W)
Here are the forumulas for figuring out the pieces of ohms law:
| P = Watts |
|
I = Amperes |
| P = |
E * I |
I = |
E / R |
| I2 * R |
P / E |
| E2 / R |
√(P / R) |
| |
| E = Volts |
|
R = Ohms |
| E = |
I * R |
R = |
E / I |
| P / I |
E2 / P |
| √(P * R) |
P / I2 |
Notices:
- This does not work accurately for automotive audio amplifiers so don't use it for that.
- If you are very close to a fuse rating, in cold conditions it could possibly spike over the limit. For example, if you had 220W of light power (110W/light) and figured you needed 20A fuse since it would have a 19.1A draw at worst conditions, you might occasionally still pop a fuse form time to time at startup. If so, bump you're fuse up one level to 25A IF everything in the chain can safely handle it. Remeber the wire and other parts (relays, switches) are what limits how large you can go with a fuse. Do not go larger than the lowest rating of the pieces in that chain. If you use a switch to control a relay as you should, you don't have to take concern with the switch as that part of the circuit because it is a separate low-draw circuit.
All Content Copywright © 2003-2006 Christopher Stiefel. All Rights Reserved.
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