posted 04-08-2006 09:57 AM ET (US)

To explain further:The voltage drop across a resistance is directly proportional to the current flow:

E = IR

where

E = electromotive force in volts

I = current in amperers

R = resistance in ohms

i.e., Ohm's law.

If a bad connection exists in your 12-volt distribution wiring, it adds resistance. Assume it adds 5-ohms of resistance.

The current drawn by the electronic devices is small. Let us say 0.2-ampere. The voltage drop across the bad connection will be

E = IR

E = 5 x 0.2

E = 1.0 volt

Thus the 12-volt supply voltage to the electronic devices will drop to 11-volts. This is still enough for them to operate.

When the navigation lighting circuit is energized, the lamps draw about 2-amperes of current. The voltage drop across the high resistance will now be:

E = IR

E = 5 x 2.2

E = 11 volts

Now the supply voltage drops to 1-volt. The electronics shut off.

This example is a bit simplistic, but it demonstrates the problem.

Also, if the above situation were to occur, consider the power being dissipated across the bad connection.

Power is proportional to current and voltage:

P = IE

By substitution and algebra, we can combine the two formulas to solve for power in terms of voltage and resistance:

P = IE and E=IR, or I=E/R

thus

P = E(E/R)

P = E^2/R

If 11-volts drop across a 5-ohm load, the power dissipated will be

P = 11^2/5

P = 96.8 watts

This is a great deal of heat and will soon make itself known.

In actual practice this won't be quite as described above. The current drawn by the lamps will be reduce by the drop in voltage supplied to them. For example, if each lamps is rated as a 10-watt lamp at 12-volts, this implies that three of them will have a resistance of

R = E^2/P

R = 12^2/(10 X 3)

R = 144/30

R = 4.8 ohms

If there is a 5-ohm resistance in the distribution, the 12-volt supply will divide across them in proportion to their resistances, thus across the lamps will be

E = 12 x [ 4.8/(4.8 + 5) ]

E = 5.87 volts

and the voltage across the 5-ohm resistance will be

E = 12 x [ 5/(4.8+5)]

E = 6.12 volts