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Discussion Starter #1
Hi All,

I have a question about some lights. Question is if I have 2 lights that are rated at 9 - 60 Volts, have them hooked up to one 12v/30 watt relay. Am I getting the full output of these lights? Would I be better having 2 12v/40 watt relays, one for each light?

Thanks for the responses.

Steve
 

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Discussion Starter #2
Bump... wondering if there are any electrical pros here?
 

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Hi All,

I have a question about some lights. Question is if I have 2 lights that are rated at 9 - 60 Volts, have them hooked up to one 12v/30 watt relay. Am I getting the full output of these lights? Would I be better having 2 12v/40 watt relays, one for each light?

Thanks for the responses.

Steve
Usualy when you have a selectable input like you have described, you will find one of two methods for supply connections. 1) Multiple wires to choose from depending on the supply voltage. 2) Autoranging electronics built into the equipment that will automaticaly adjust to what ever voltage that you select. Either way, Your output SHOULD be consistant accross the entire input range. Now, as far as how to wire these thing up? You have not supplied enough information to determin the load. The basic formula is:Amps multiply by Volts = Watts. You have only supplied "Volts". You need to supply one more variable to be able to calculate. Once that you have determined the load. Then you can determine relay size, wire size and fuse size. Adding extra relays will not help.
:beer:
 

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Discussion Starter #4
Usualy when you have a selectable input like you have described, you will find one of two methods for supply connections. 1) Multiple wires to choose from depending on the supply voltage. 2) Autoranging electronics built into the equipment that will automaticaly adjust to what ever voltage that you select. Either way, Your output SHOULD be consistant accross the entire input range. Now, as far as how to wire these thing up? You have not supplied enough information to determin the load. The basic formula is:Amps multiply by Volts = Watts. You have only supplied "Volts". You need to supply one more variable to be able to calculate. Once that you have determined the load. Then you can determine relay size, wire size and fuse size. Adding extra relays will not help.
:beer:
Thanks, that's all I was looking for. :beer:
 

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Formula

Thanks, that's all I was looking for. :beer:
The info needed is there to figure what load his relay will handle. He makes no mention of the watts of the lamps. If it is say a 100 watt light at 12 volts it pulls 8.5 amps. One of these lamps on a 30 amp relay are no load for it at all. We need to confirm if the relay is 30 amps or 30 watts. My guess his light does not put out full output until the highest voltage allowed. Divide watts by volts = amps. I think he means he has it hooked to one 12 volt 30 amp relay. I have never seen a relay rated by watts. If it is truely a 30 watt relay, then 30/12 = approximately 2.5 amps rating on the relay. Not likely the relay is rated this small.
 

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Discussion Starter #6
The info needed is there to figure what load his relay will handle. He makes no mention of the watts of the lamps. If it is say a 100 watt light at 12 volts it pulls 8.5 amps. One of these lamps on a 30 amp relay are no load for it at all. We need to confirm if the relay is 30 amps or 30 watts. My guess his light does not put out full output until the highest voltage allowed. Divide watts by volts = amps. I think he means he has it hooked to one 12 volt 30 amp relay. I have never seen a relay rated by watts. If it is truely a 30 watt relay, then 30/12 = approximately 2.5 amps rating on the relay. Not likely the relay is rated this small.
I'm running LED's... the maximum draw with the 2 lights will be 4 amps per light. Would this make a difference?

Thanks
 

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The info needed is there to figure what load his relay will handle. He makes no mention of the watts of the lamps. If it is say a 100 watt light at 12 volts it pulls 8.5 amps. One of these lamps on a 30 amp relay are no load for it at all. We need to confirm if the relay is 30 amps or 30 watts. My guess his light does not put out full output until the highest voltage allowed. Divide watts by volts = amps. I think he means he has it hooked to one 12 volt 30 amp relay. I have never seen a relay rated by watts. If it is truely a 30 watt relay, then 30/12 = approximately 2.5 amps rating on the relay. Not likely the relay is rated this small.
If you are considering a standard lamp, then your assumption would be correct. But then the output would be very low at the small end of the scale. And very high at the large end of the scale. Why would anyone want that. ?:shaking: As I stated earlier, Most ( not all but most ) gear with selectible inputs have some type of package to maintain a constant output no mater what input is selected. Since the OP has only provided us with ( 1 ) specific output. My GUESS! would be that you may be partially correct. Since he only states "Max Amps". That would mean that his wattage would vary accross the input range. But, without this new information. There would be no reason to consider such things as: wire size, fuse size or RELAY SIZE!

I'm running LED's... the maximum draw with the 2 lights will be 4 amps per light. Would this make a difference?

Thanks
Yes. Now you have all the information to design an instalation for this gear. And since they gave you the information in total amps. This means that you do not need to do any of the load calculations to find this information.
So, Total Load = 2 lights @ 4 amps / light = 8 amps total load
Fuse size - 8 to 10 amps
Wire size - 18 awg ( or larger if you prefer )
Relay size - anything larger than 10 amps.
Have Fun! :beer:
 

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Discussion Starter #8
Thanks Guys... all hooked and nice and bright
 
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