Date:         Thu, 24 Jun 2010 23:03:54 -0700
Reply-To:     Alistair Bell <albell@SHAW.CA>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Alistair Bell <albell@SHAW.CA>
Subject:      the importance of good electrical connections
Content-Type: text/plain; charset=US-ASCII; delsp=yes; format=flowed

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.