mcg
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| posted on 23/6/11 at 09:18 PM |
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Fuse help!!!?!?
As you can tell from my previous post I am doing a bit of wiring at the mo!
Can people tell me if the following need fuses, and if so, what rating of fuse will I need:
- Starter solinoid?
- Coil?
- Fan?
- Alternator? (sorry if this is a stupid question!)
- Lights: Front (12v 55W bulb)
- Lights: Indicators (LED)
- Lights: Rear (21W and 5W bulb)
- Lights: Rear middle (LED)
- Fuel Pump. (Facet)
- Digital Speedo
Many thankyouverymuches.
Matt
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JoelP
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| posted on 23/6/11 at 10:07 PM |
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You need some background reading, which sadly i cant point you to! However, remember the equation:
P=VA
or
A=P/V, which in this case is A=P/12
You work out what your circuit will draw in watts, divide by 12 and round up to the next fuse size, and then select a wire size capable of carrying
the fuses rated current.
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Chippy
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| posted on 23/6/11 at 10:32 PM |
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- Starter solinoid?.......................................................No!
- Coil?.........................................................................No!
- Fan?.........................................................................Yes! Depends on amperage
- Alternator? (sorry if this is a stupid question!).........No!
- Lights: Front (12v 55W bulb)....................................Yes!
- Lights: Indicators (LED)............................................Yes!
- Lights: Rear (21W and 5W bulb)..............................Yes!
- Lights: Rear middle (LED).........................................Yes!
- Fuel Pump. (Facet)....................................................Yes for me, No for some people
- Digital Speedo...........................................................Yes!
Can't help with the fuse ratings, as that depend on the Wattage of each item, and some items can be set to the same fuse. But hope that may
help. Cheers Ray
To make a car go faster, just add lightness. Colin Chapman - OR - fit a bigger engine. Chippy
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rayward
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| posted on 23/6/11 at 11:03 PM |
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one thing to consider is the fuse is there to protect the cable as much as the components,
i.e in the event of a fault (short circuit) you don;t want your cable melting before the fuse blows !!!!
hth
Ray
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Macbeast
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| posted on 24/6/11 at 03:58 AM |
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As a guide, look at any Haynes manual. They give functions of each fuse ( N/S headlights, brakelights etc ) and the corresponding fuse rating.
I'm addicted to brake fluid, but I can stop anytime.
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John Bonnett
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| posted on 24/6/11 at 06:42 AM |
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I would suggest that all main circuits are fused individually and the components which draw high currents are relayed as well. Many modern switches
are only rated for low currents and to avoid short life relays are essential. Fuse the relay coils and main feeds separately.
You can Google colour code convention for the wiring so that you, and others later on, can identify what wire goes to where. I've found
Autosparks excellent to deal with and offer the best prices on cable.
On a general note, bearing in mind that most electrical problems stem from a bad earth, run an earth from each component back to the battery and make
sure the earth cables are rated, as their respective feeds, to suit the loading. It's fairly easy at the build stage to put in earth posts as
terminals for a number of earths and then run back a single heavier duty cable.
Both relays and fuses can be mounted in modular fittings for easy access. Attached picture shows the installation, in its early stages, of fuses and
relays in my Trials car.
John
[img]
 
Wiring started
[/img]
[Edited on 24/6/11 by John Bonnett]
[Edited on 24/6/11 by John Bonnett]
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02GF74
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| posted on 24/6/11 at 06:47 AM |
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quote:
Originally posted by JoelP
You need some background reading, which sadly i cant point you to! However, remember the equation:
P=VA
or
A=P/V, which in this case is A=P/12
You work out what your circuit will draw in watts, divide by 12 and round up to the next fuse size, and then select a wire size capable of
carrying the fuses rated current.
Common misconception that car electrics are 12 V; when the alternator is running, the voltage is 13.5 V to 14 V. Use 14 in your sums.
Also when supplying current to a bulb, there will be an initial surge current until the bulb warms up (resistance of the filament increases with
temperature) - you need to take this into account when fitting a fuse.
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daviep
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| posted on 24/6/11 at 08:10 AM |
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quote:
Originally posted by 02GF74
quote:
Originally posted by JoelP
You need some background reading, which sadly i cant point you to! However, remember the equation:
P=VA
or
A=P/V, which in this case is A=P/12
You work out what your circuit will draw in watts, divide by 12 and round up to the next fuse size, and then select a wire size capable of
carrying the fuses rated current.
Common misconception that car electrics are 12 V; when the alternator is running, the voltage is 13.5 V to 14 V. Use 14 in your sums.
I'm not entireley sure that is good advice, by using 14v instead of 12v the current will appear to be lower than it may actually be. If you were
to use cable and fuses which were very close to your calculated load then you may find them inadequate.
Example: 100watts at 12v = 8.3amps whereas 100watts at 14v = 7.1 amps
Davie
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ReMan
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| posted on 24/6/11 at 08:26 AM |
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quote:
Originally posted by daviep
quote:
Originally posted by 02GF74
quote:
Originally posted by JoelP
You need some background reading, which sadly i cant point you to! However, remember the equation:
P=VA
or
A=P/V, which in this case is A=P/12
You work out what your circuit will draw in watts, divide by 12 and round up to the next fuse size, and then select a wire size capable of
carrying the fuses rated current.
Common misconception that car electrics are 12 V; when the alternator is running, the voltage is 13.5 V to 14 V. Use 14 in your sums.
I'm not entireley sure that is good advice, by using 14v instead of 12v the current will appear to be lower than it may actually be. If you were
to use cable and fuses which were very close to your calculated load then you may find them inadequate.
Example: 100watts at 12v = 8.3amps whereas 100watts at 14v = 7.1 amps
Davie
Spot on, just beat me to it, my thoughts exactly
And Chippys advice is about right too
www.plusnine.co.uk
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paulf
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| posted on 24/6/11 at 09:51 AM |
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This is still not exactly correct, if you supply a bulb or any other component with more than its rated voltage then it will draw more current and
provide a higher wattage output.The correct way if you really wanted to be bothered is to work out the extra current that would be generated by the
increased voltage above the actual rating, however most circuits are fused well above the working current in a car and fuses really just need to
protect against short circuits causing prolonged fault currents above the cable rating.
Paul
quote:
Originally posted by daviep
quote:
Originally posted by 02GF74
quote:
Originally posted by JoelP
You need some background reading, which sadly i cant point you to! However, remember the equation:
P=VA
or
A=P/V, which in this case is A=P/12
You work out what your circuit will draw in watts, divide by 12 and round up to the next fuse size, and then select a wire size capable of
carrying the fuses rated current.
Common misconception that car electrics are 12 V; when the alternator is running, the voltage is 13.5 V to 14 V. Use 14 in your sums.
I'm not entireley sure that is good advice, by using 14v instead of 12v the current will appear to be lower than it may actually be. If you were
to use cable and fuses which were very close to your calculated load then you may find them inadequate.
Example: 100watts at 12v = 8.3amps whereas 100watts at 14v = 7.1 amps
Davie
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ReMan
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| posted on 24/6/11 at 10:28 AM |
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I'm changing camps 
The theory was flawed. I now submit that )ignoring any specific characvhteristics of a light bulb) given a fixed load (RESISTANCE) the current will
increase.
So 12v and a 12 ohm load = 1A
But 14v and a 12ohm load = 1.16A
www.plusnine.co.uk
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mcg
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| posted on 24/6/11 at 11:43 AM |
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okeydokie.
thanks everyone. I'll do a bit of calculating and a little more reading. Its good to know people do things in different ways.
Thanks again
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ReMan
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| posted on 24/6/11 at 02:32 PM |
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Don t get too hung up though!
ATEOTD In this non perfect world I would leave some margin in to the next standard fuse size anyway.
So to requote this example from earlier
Example: 100watts at 12v = 8.3amps whereas 100watts at 14v = 7.1 amps
I would be fitting a 10A fuse and the wiring would be at least 10A capeable
www.plusnine.co.uk
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Chippy
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| posted on 24/6/11 at 04:09 PM |
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If you really want to get into it then there is a very good book called the 12 VOLT BIBLE, although primarily written for wiring boats, the
principal is exactly the same. This goes into wire sizes, lengths of runs, Ohms law, relays and there use, etc, etc. A very good book IMHO. HTH Ray
To make a car go faster, just add lightness. Colin Chapman - OR - fit a bigger engine. Chippy
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mcg
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| posted on 24/6/11 at 08:58 PM |
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cool - I think I am going the Reman route. keep it simple to start and then I can always tweak. Thanks alot
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Angel Acevedo
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| posted on 25/6/11 at 01:34 PM |
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quote:
Originally posted by mcg
cool - I think I am going the Reman route. keep it simple to start and then I can always tweak. Thanks alot
mcg,
I think you´re better off measuring twice cutting once.
Wiring may not be something you want to have near perfect as if you have a problem it may end on an electrical fire.
Check this page, it is a very good step by step wiring guide.
http://home.clear.net.nz/pages/phil.bradshaw/car_wiring.htm
HTH
AA
Beware of what you wish.. for it may come true....
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JoelP
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| posted on 25/6/11 at 08:39 PM |
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quote:
Originally posted by ReMan
I'm changing camps
The theory was flawed. I now submit that )ignoring any specific characvhteristics of a light bulb) given a fixed load (RESISTANCE) the current will
increase.
So 12v and a 12 ohm load = 1A
But 14v and a 12ohm load = 1.16A
i went through that argument on an electricians forum once, with people who assume the power rating is constant irrespective of current. They thought
that a house with a higher voltage would send less current down the cables 
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mcg
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| posted on 27/6/11 at 04:38 PM |
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Thanks alot Angel - Thats a great article. Really good reference.
quote:
Originally posted by JoelP
quote:
Originally posted by ReMan
I'm changing camps
The theory was flawed. I now submit that )ignoring any specific characvhteristics of a light bulb) given a fixed load (RESISTANCE) the current will
increase.
So 12v and a 12 ohm load = 1A
But 14v and a 12ohm load = 1.16A
i went through that argument on an electricians forum once, with people who assume the power rating is constant irrespective of current. They thought
that a house with a higher voltage would send less current down the cables
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MikeRJ
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| posted on 27/6/11 at 05:42 PM |
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quote:
Originally posted by JoelP
i went through that argument on an electricians forum once, with people who assume the power rating is constant irrespective of current. They thought
that a house with a higher voltage would send less current down the cables
For many modern electrical devices the power rating really is constant with voltage (more or less). Switched mode power supplies (used for virtually
all low-medium power mains powered electronic devices these days - TV's, Sat Box, DVD/Blu Ray, PC, Games Consoles etc.) ) actually have an
apparent negative resistance - increase the voltage and the current will reduce proportionally.
Even incandescent lamps don't increase power proportionally with voltage - as the tungsten filament gets hotter the resistance goes up.
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SteveWalker
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| posted on 27/6/11 at 07:17 PM |
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quote:
Originally posted by Chippy
- Starter solinoid?.......................................................No!
- Coil?.........................................................................No!
Cheers Ray
You are correct that the power to the starter solenoid will not have a fuse, but the "start" supply to the solenoid coil is usually
supplied from the ingition switch and the supply to that is generally fused and similarly for the coil.
SteveW
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