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Author Topic:   Sizing conductors
jwestwood posted 01-31-2014 10:39 AM ET (US)   Profile for jwestwood   Send Email to jwestwood  
I have read many of the posts here concerning sizing conductor wiring when batteries are moved forward to a console location. This is what I plan to do to help clean up the aft space. Do you, Jim, or others here have a suggestion for sizing starter conductors for an Outrage 21 with twin 1988 Yamaha 90's. As always, thanks for the help.
jimh posted 02-02-2014 10:23 AM ET (US)     Profile for jimh  Send Email to jimh     
As I have mentioned many times, in distribution of 12-Volt power the conductor will almost always have to be selected on the basis of voltage drop, not on the basis of current capacity. One chooses the conductor size so, for the particular length of conductor, the voltage drop is limited to the desired tolerance.

The tolerance for voltage drop seems to be three-percent in the most rigorous case, and as much as 10-percent in more lenient cases. The three-percent tolerance would be used for loads that are sensitive to voltage, such as a navigation lighting circuit. (The light output from an incandescent lightbulb varies greatly with voltage, and low voltage will produce a significant dimming of the light output.) For loads that can tolerate a greater voltage drop, the 10-percent tolerance can be used. Some modern electronic devices are happy to operate over a wide range of input voltage, as they contain internal voltage convertors and regulations for their power circuits.

The voltage drop is a function of the resistance of the conductor and the current being carried. If we consider the nominal voltage of our system to be the voltage of a fully-charged battery, we can use 12.6-Volts as the system voltage. A three-precent voltage drop would then be

12.6-Volts x 3/100 = 0.378-Volts

The voltage drop in a conductor is equal to the current times the resistance. I demonstrate by example:

A circuit is to carry 3-Amperes of current to a load that is 15-feet from the battery. What is the maximum resistance in the conductor for a three-percent voltage drop in a 12.6-Volt system?

The maximum voltage drop will be 0.378-Volts, as previously calculated. From Ohm's Law we can write

E = IR

and rearrange to

R = E / I

Solving for R and substituting our known values

R = 0.378-Volts/ 3-Amperes

R = 0.126-Ohm

This resistance is for a circuit of 15-feet out and 15-feet back, or a total wire length of 30-feet.

0.126-Ohms/30-feet = 0.0042/foot

Most tables of wire resistance give the resistance of 1,000-feet of the wire, so we have to scale this up by a factor of 1,000 which gives us a value of


as the maximum wire resistance we could use. Now we consult a table to find a suitable conductor. A good table is

Inspecting the table, we see:

16-AWG = 4.016-Ohms/1,000-feet

Thus a total conductor length of 30-feet of 16-AWG will distribute 12.6-Volts DC to a load of 3-Amperes without incurring more than a three-percent voltage drop.

Note also that in boat wiring a conductor of 16-AWG is the smallest recommended conductor to be used for any electrical wiring (with some exceptions) according to the recommendations of industry groups like the ABYC and 14-AWG is mandatory according to certain federal regulations (and forgive me for being too lazy to look up and cite the appropriate section of our government's voluminous Code of Federal Regulations).

Note that 16-AWG is listed as having a current capacity of 22-Amperes for short chassis wiring use. This example demonstrates how voltage drop is more significant in selecting a conductor size, as the current in this example was only 3-Amperes, or 13-percent of the rated maximum current.

jimh posted 02-02-2014 11:47 AM ET (US)     Profile for jimh  Send Email to jimh     
For many cites of regulations regarding wire gauge and type per federal regulations, see an earlier discussion at

which has a number of hyperlinks to primary sources.

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