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I was advised by several people here to use a relay to relocate a bad power lead to another circuit on my fuse box. However, the problem now is that the two fuses on the original bad circuit total 45 amps (a 30 for the ac/heat and a 15 for the other). The only relays that I can find are 30 amps, and I've called everywhere. What if I just used a 30 amp relay? Would this cut the load down enough? I would imagine that it would only draw 45 amps if everything was going at once. On the 15 amp fuse are the turn signals, backup lights, od on/off switch and rear defroster (which sucks and never gets used).
If I can't make this work, I'm going to have to take the dashboard apart and I don't feel like doing that.
I don't know why all the auto parts stores and auto electric repair shops around here acted like asking for a relay that handles 45 amps was totally impossible. I found one at jcwhitney.com in about 2 seconds.
Just use a 30A relay for the 30A circuit & a 30A relay for the 15A circuit! The fuse will still protect the circuit & 30A relays are cheap & EASY to find. Just use the same trigger wire for both so they come on together.
http://www.partsexpress.com/webpage.cfm?&Webpage_ID=3&DID=7&CAT_ID=47&ObjectGr oup_ID=31&SO=2
A good idea. I thought of that too. But both of the fuses are on one metal "strip" of the fusebox and share a common power supply. To do what you suggest, I would have to cut two sets of wires away from the fusebox. I am trying to do this as neatly as possible, and would rather leave as much as I can the way it was. I might get motivated one day and trace the defective wire down. Anyway, I did not know that a relay to handle 45 amps would be so hard to find. Like you said, I found 30's all over town. I already ordered the 70 amp relay from JC Whitney, it should be here this week. BTW - for something like the blinkers, will the relay have to click on and off each time the blinker blinks?
70 amps will be sufficient, although I am sure a 300 amp relay would have worked! So, if I understand this correctly, once I add the relay to this circuit, it won't draw 45 amps from the fusebox? I don't understand how this works. Is the relay some sort of soak, and holds the power inside of it or something? So once I do this, I can run up to 70 amps off an already used circuit of my fusebox without worrying about melting anything?
A starter relay is only for intermittent duty - it'll overheat its own coil in that application. But there are continuous-duty relays available that look & work like starter relays. They're used in electric winches & cost about $20ea.
you are gonna trip the relay with switched power . you would take batt voltage anconnect it to normally open side of relay and connect a wire from common terminal to where you want the power. when you turn on the key it will trip the relay and give you continuity from common to normally open. all a relay is is a electically activated switch. low current trips relay to switch high current.
Thanks for the diagram. But now I am more confused. Why do I take a power lead from the switched on power AND the battery? Also you have the word "NO" written on the diagram over the battery power lead connection. What does that mean? Don't use that one?
Wait...maybe I understand after all, after reading your first post. The switched power lead just turns on the relay. Then, it is not used anymore, and the battery supplies the power. So in other words, this would be the same as just taking the power straight from the battery, except the accessories could not be turned on without the key being on. Right?