Ignition switch/barrel

[MG]

Richard,
You have to see it in a different way.
Let's use your example of a 60W bulb drawing 5A @ 12V.
Using Ohm's law this bulb has a resistance of R=U/I=12/5=2.4Ohm.
Now let's say that in this circuit there is another resistor, a bad joint or partly broken cable for example.
Let's for example consider a very thin and rather long wire. Here the resistance is calculated by R=rho*l/A, rho (the greek symbol) being the specific resistance, depending on the material. For copper it has a value of 0.01678Ohm*mm^2/m, l stands for the length of the wire in m, A is the cross sectional area in mm^2. You have to imagine this as a hosepipe. The thinner and longer, the less water will go through it, given that the water pressure is constant, or like the artery example Bill proposed.
If we now consider a wire with a cross section of 0.2mm^2 and a length of 20m this will give a resistance of approx. 1.7Ohms (R=0.01678*20/0.2)
This resistance of the wire is connected in series to the bulb, so we can model this whole system as two resistors in series connected to a 12V supply. The two resistors add up, being connected in series, so the overall resistance equals to 2.4+1.7=4.1Ohms.
If we now calculate the current with I=U/R=12/4.1 we get 2.93A. You can see that the added resistance caused by the wiring reduces the overall current through the circuit and therefore the power consumption of the bulb, which will be rather dark now.
If we have the 2.93A through our 1.7Ohms wiring, we will get a voltage drop of U=R*I=1.7*2.93=4.981V, meaning that the bulb will only "see" 12V-4.981V=approx. 7V now.

You have to regard the bulb as a constant resistor (it is not really, as it changes it's resistance with temperature, but we will neglect this for our modelling). You can't say the bulb draws 7.5A at 8V, because this would equal a resistance of R=8/7.5=1.07Ohms, while it really has 2.4Ohms, like calculated before.
A 6V/60W bulb has just a quarter of the resistance of a 12V/60W bulb just for this very reason (R=6V/10A=0.6Ohms), so at 12V it would draw I=12/0.6=20A, that's what causes it to burn in a 12V circuit. Following you logic it would simply draw 5A and shine along happily at 60W, which we all know it doesn't.

Hope these little examples helped a bit to foster understanding.

***smart-idiot mode off***

Cheers, Markus

[Richard]

I got lost on the 1st paragraph,
All I know is when I was doing electrical work many moons ago I was taught that if you used to small a cable for a given circuit then the cable got hot the resistance increased so therefore the cable got even hotter hence fire hazard, the same thing applied if there was a loose or poor connection which is one cause of a plug melting when people do not tighten the connections very well,a as the pins get hot, all the while whatever is being powered up is trying but not succeding and the fuse tends to blow.
I must confess I am a hands on bloke and have never been good on theory its a wonder I ever made an electrician, so I have always endeverd  tried to make good connections, use the right size cable and fuses.

[chaterlea25]

Ok Richard,

In simpler terms

Volts X amps = watts
12    X 5      =  60

                                        Volts =12
Resistance is equal to, (I) Current  = 5    = 2.4 ohms   

voltage is constant, so is the resistance of our mythical bulb, a bad connection, narrow arterie or kink in the hose
is just like adding another bulb in series with the first

so now resistance is going to be 2.4 + 2.4 =4.8 ohms

                        Volts         12
Current is = to, resistance     4.8   = 2.5 amps

So now volts  X amps = 30 watts,      each bulb giving 15 watts

Your point about a bad connection in a plug top is like the second bulb, it generates heat which will burn the cable and plug, remember with something like an electric kettle we are talking about 2000 watts,!!!

If a fuse is rated at 5 amps this means it will carry 5 amps happily, then comes the science bit!! the "fusing factor"
which is a product of current and inverse time, ie the more current the less time it takes to blow,
further characteristics are due to the materials and manufacturing technique of the fuse

HTH
John O R

[trevinoz]

I feel like I am back at Tech in the first year of my apprenticeship getting Ohm's Law bashed into my thick skull!
It's still there!
                         Trev.

[olev]

strewth, my turn i think,

Amps = power/volts
so at 12 volts
a 24watt load pulls 2amps (24/12)
a 36watt load pulls 3amps (36/12)
a 48watt load pulls 4amps (48/12)
a 60watt load pulls 5amps (60/12)

as the wattage goes up so does the current.

Resistance = Voltage/amps

the 24watt load has a resistance of 6 ohms (12/2)
the 36watt load has a resistance of 4 ohms (12/3)
the 48watt load has a resistance of 3 ohms (12/4)
the 60watt load has a resistance of 2.4 ohms (12/5)

as the wattage goes up the resistance goes down

so a high resistance joint will never blow a fuse as it increases the resistance which reduces the current.

a short or partial short or a high wattage load (something that reduces the resistance)
is the only thing that will increase current and blow a fuse.

cheers

[trevinoz]

Olev,
          That is the reason houses burn down.
Hot joints can eventually catch fire and the fuse happily allows the current to flow.
      trev.

[Richard]

Well there you have it Mosin put a 30 amp fuse in circuit or a 40 amp as it seems it does not matter what size it is so long as it is big enough to run the total wattage of the lamps and horn you have fitted as the only time it will blow is if you have a short, however from the debate I read that you must beware of a bad joint ir a loose connection as two things will happen, 1)  your headlamp will go to half its brightness and 2) the wiring may catch fire as the fuse would just not blow so I am now going to carry a 6" nail to replace my 15amp fuse when it blows as that will save me having to bother with it again.


Joking apart you need to fit a fuse that is going to protect your regulator so find out what the manufacturer rates it at and fit a fuse at that rating.

Gentlemen thank you for the lessons I did not really get Ohms law when I was an apprentice and I still do not fully understand it now but I think I see the error of my ways, a bad joint is in series which reduces demand additional lamps are in parallel which would increase the current demand
Richard

[trevinoz]

Richard,
               Nails are very useful for finding exactly where a short circuit is!
       Trev.

[olev]

Richard, Your part 2 is not right.

heres another example (more maths - sorry)

using 12 volts a 36watt load draws 3amps and the circuit resistance is 4ohms.

if a connection in series with the load went crook and developed a resistance of 2ohms the following occurs;

Total circuit resistance = load resistance + joint resistance
4 + 2 = 6ohms
The current in the circuit = Voltage/Resistance
so the current in the circuit with the crook connection is 12/6 = 2amps
The current in the circuit has dropped!!

But check this out
Power (watts) = amps squared X resistance (ohms)
So the wattage of the old 36watt load is now 2amps X 2amps X 4ohms = 16watts
the wattage dissapated by the joint is 2amps X 2amps X 2ohms = 8watts
that 8watts is dissapated as heat hence a 'hot' joint

The circuit is only running 2amps. You aren't going to burn out wiring.
The fire is going to start at the hot joint.
cheers

[muskrat]

OK you lot, who's got the blue face ? I can see the cows coming home !! Bl@@dy bright sparkz. LOL

[Richard]

Olev,
Part 1 and 2 were tongue in cheek and not meant to be serious, I may not know my theory but that does not seem to stop all my mates from getting me to wire their bikes or sort the problems out, I have never had anything catch fire,get hot or not work after I have done it.
It is alright for all these experts who can roll of formulas from the armchair but it all means jack sh*t as when it comes to doing a job its the ability to use hands as well as brains, not get the wallets out and pay firms to do the jobs then belittle others .
I may have been incorrect and bowed to my mistakes but enough is enough and it is such a shame as this was, is a good forum.
I wish to thank those on here that have helped me and conversed but I am not bothering any more.
Richard
PS I will not be answering any emails or posts

[alanp]

Olev, thanks for the info. Most of us understand Ohm's law but I for one had not used it to understand why bad joints get warm/hot. Just assumed you needed good joints to keep your lights shiny, without thinking about it any deeper. Cheers.
Alan

[a101960]

Richard

Calm down dear, calm down. I know what you are saying, and I can personally endorse your practical expertise. What all of this amounts to is the fact that any resistance within the circuit will compromise the efficiency of its operation. It matters not weather it is a corroded joint or an incorrect cable rating. Any resistance will generate heat and steal power. I once new a bloke that wired up a pair of Halogen spot lights on a Mini with telephone cable. I kid you not!!! Elektriky things can behave very strangely sometimes for no good reason. I once had a car that suffered from a bad earth on the rear lights, the odd thing was it happened out of the blue for no apparent reason. Every time the brakes were applied the tail lights went out. On test continuity seemed O.K. but the problem was only resolved by fitting earth straps. Your starter for 10? At the end of the day common sense must prevail. By trade I am an aircraft electrician and well versed in the theory of electrical practice but, and it is a big but, no matter what your depth of knowledge might be, sometimes things happen and the solution defies all conventional theoretical protocols. But hey ho, that is what makes life interesting. A little anecdote. I remember once an aircraft heavy duty winch (used for hauling cargo aboard) being sent back to the repair bay twice for being U/S on fit. The fault turned out to be the exciter conductor had broken. The problem was that this was a large heavy duty cable and in the bay the two severed halves always made contact because of the angle that the cable was connected to the power supply.On the aircraft the cable had a different orientation and the conductor failure revealed itself. No matter what we did in the bay no fault was ever found. Anyway we replaced the cable and all was well, and it was only when as a matter of curiosity we stripped the cable down that we found the fault.

John

[trevinoz]

We should all be flogged with a stick of rhubarb and hang our heads in shame.
How dare we attempt to point out and try to explain the reason why someone is thinking erroneously!
Bring back the stocks and transportation!
Seriously, all of the replies in this topic have been intended to help, not cause offense, in my view.
 Trev.

[RichardL]

All,

I would hate to lose a friend in Richard from Minety over this Ohm's Law nonsense. The nonsense being flaunting the colorful plumage of theory to show who has the biggest fuse.  We definately want Richard amongst us and need to know when to chill (borrowing current slang with clear meaning) when it comes to demonstrating how smart we are with technology and how dumb we can be with diplomacy. I'm not pointing any fingers, because I didn't really try to track the offense, since offense is in the sense of the offended. Anyway, I'm only preaching enough to have hope that Richard will reconsider. 

Richard L.