Those are great examples Alistair!

Makes you just want to get out there and disconnect and clean all your
electrical connections, doesn't it?  ;)

Thanks, Jake

On Thu, Jun 24, 2010 at 11:03 PM, Alistair Bell <albell@shaw.ca> wrote:

> I was reading "The 12 volt Doctor's Practical Handbook for the boats
> electric system" this evening, for the umpteenth time (slow learner),
> and I thought this part would be of interest to list members.
>
> It really brings home how important good electrical connections are
> in high current draw circuits especially headlights and starter motors.
>
>
> alistair
>
>
>
> From "The 12 volt Doctor's Practical Handbook for the boats electric
> system", Edgar J. Beyn.
>
>
> First example: There Is a switch for the compass light. The lamp in
> the compass Is rated 1 W. The switch is slightly corroded and has
> developed contact resistance of 5 Ohm. How will that affect the
> compass light?
>
> Answer: the compass light draws: 1 Watt divided by 12 Volt = 0.083
> Ampere, it has a resistance of: 12 Volt divided by 0.083 Ampere = 144
> Ohm. Adding another resistor of 5 Ohm, the switch, in series to the
> lamp will increase total resistance to 149 Ohm, the effect will
> hardly be noticeable.
>
> Second example: The same type of switch is used to switch a 40 Watt
> cabin light. This switch also has corroded and developed 5 Ohms
> resistance at its contacts. Will that affect the cabin light?
>
> Answer: On 12 Volt, the cabin light draws: 40 Watt divided by 12 Volt
> = 3.3 Ampere, it has: 12 Volt divided by 3.3 Ampere = 3.6 Ohms of
> resistance. Adding the switch resistance In series with this lamp
> will more than double the total resistance, so that the current will
> be reduced to: 12 volt divided by 8.6 total Ohm = 1.4 Ampere which
> flow through switch and lamp. This lamp will therefore hardly glow at
> all. And another problem, at the switch we now have a voltage drop of
> 7 Volt (12 Volt split by the ratio of resistances, or 12 divided by
> 8.6, multiplied by 5), so that 7 Volt times 1.4 Ampere = almost 10
> Watt of heat are generated at the switch contact.
>
> Third example: the starter motor on your engine is rated 3000 Watt. A
> battery cable terminal has developed very slight contact resistance
> at a battery post, the resistance is only 1/100 of an Ohm and now is
> in series with the starter motor. Can you start your engine? Can you
> find the battery connection by hand touch?
>
> Answer: The starter motor uses: 3000 Watt divided by 12 Volt = 250
> Ampere, its resistance is: 12 Volt divided by 250 Ampere = 0.048 Ohm.
> The added resistance of 0.01 Ohm is series brings the total
> resistance to 0.058 which allows only:  12 Volt divided by 0.058 Ohm
> = 207 Ampere, so that power is reduced to: 207 Amps times 12 Volt =
> 2484 Watts instead of 3000 Watt. And of these 2484 Watts, not all are
> applied at the starter motor. Of the full 12 Volt, about 2 Volts are
> dropped at the added resistance at the battery post which corresponds
> to: 2 Volt times 207 Ampere = 408 Watt, so that only 2484 minus 408 =
> 2076 Watt are trying to turn the engine, about two thirds of normal,
> perhaps barely enough. The 408 Watt at the battery post are enough to
> make that post and terminal HOT in seconds. You would easily find it
> by just feeling the cable connections until you find the hot one.
> Much less resistance of almost perfect connections, makes itself
> noticeable by the generated heat during starting.
>

--
Jake

1984 Vanagon GL 1.9 WBX 'The Grey Van'
1986 Westy Weekender/2.5 SOHC Suby 'Dixie'

Crescent Beach, BC

www.thebassspa.com
www.crescentbeachguitar.com
http://subyjake.googlepages.com/mydixiedarlin%27