New Branch Circuit for Deck Lighting

[rcapriola]

[MOVED TO SMALL BOAT ELECTRICAL for discussion]

I'm installing two electric deck illumination lamps on a boat. Each lamp is rated as consuming 1.3-Amperes of electrical current. I am wiring the two lamps in parallel.

Q1: what value of fuse should be used to protect this circuit?

The wiring at the lamps is AWG-18. A conductor of AWG-18 is rated for a current of something like 14-Amperes.

Therefore, I think I will use a fuse rated for 3-Amperes or 5-Amperes.

Q2: does branch circuit wired with AWG-18 conductors and having a current draw of two 1.3-Ampere lamps being fused at 3-Amperes or at 5-Amperes sound right?

[jimh]

I'm installing two electric lamps with floodlight patterns on the aft deck of a boat. Each lamp is rated as consuming 1.3-Amperes of electrical current. I am wiring the two lamps in parallel.

Q1: what value of fuse should be used to protect this circuit?

The total current for the two lamps will be 2.6-Amperes. Depending on the type of lightbulb in the lamps, there may be a higher in-rush current when the circuit is energized. Using a 3-Ampere fuse may work if there is not a high in-rush current. Otherwise a 5-Ampere fuse may be more useful.

The type of fuse to be used will determine what values of fuses are available. Typically with boats an ATC fuse will be used. The ATC fuses are available in the following Ampere values: 1, 2, 3, 4, 5, 7.5, 10, 15, 20, 25, 30, 35, and 40. Fuses of 1, 2, or 4-Amperes may be harder to find than the other values.

The wiring at the lamps is AWG-18. A conductor of AWG-18 is rated for a current of something like 14-Amperes.

The maximum current rating of any conductor depends on the insulation used and if the wire is in free air or in a bundle with other wires and on the ambient temperature. The best reference for current rating of a wire is the table of American Wire Gauge ratings. See

Engineering Toolbox
https://www.engineeringtoolbox.com/wire ... d_419.html

The values in the table are for a single solid round copper conductor with PVC insulation. The table specifies an 18-AWG conductor as rated for 8.5-Amperes in an ambient temperature of up to 30-degrees-C or 86-degrees-F.

Wiring the entire branch circuit with 18-AWG wiring would be possible, but I do not recommend that. The American Boat and Yacht Council (ABYC) standard of marine electrical wiring specifies the minimum conductor size as 16-AWG.

Q2: does branch circuit wired with AWG-18 conductors and having a current draw of two 1.3-Ampere lamps being fused at 3-Amperes or at 5-Amperes sound right?

As I suggest above, you should use 16-AWG conductors to bring the power to the lamps. The circuit can be fused at 3-Amperes if there is not a large in-rush current; otherwise a 5-Ampere fuse will be suitable.

The reasons for using a minimum conductor size of 16-AWG is to provide a larger physical conductor that will be less likely to be damaged.

Another factor in choosing conductor size is the amount of voltage drop that will occur in the circuit and reduce the actual voltage at the load. This is important for lighting circuits, as often the intensity of the light produced by a lamp varies rapidly with change in voltage. This is especially true with incandescent lightbulbs.

A general design parameter is to limit the voltage drop in conductors in a branch circuit to not more than three-percent of the system voltage. The voltage drop in a power distribution circuit is a function of the conductor size, the length of the conductors, the current flowing in the circuit, and the system voltage.

I have developed a simple method for analysis of these four influence, which I call the Ampere-Feet method. See my article at

Ampere-Feet
A Method for Selection of Proper Conductor Size

https://continuouswave.com/whaler/refer ... tion1.html

A pre-computed value of Ampere-feet is given in Table 4 in the above article for power distribution in nominal 13.2-Volt circuits. The value for 18-AWG is only 31-Ampere-feet.

In the case of the circuit under discussion in this thread, the current is 2.6-Amperes. This suggests that the maximum length of 18-AWG conductors in the circuit will be 31-Ampere-feet / 2.6-Amperes = 11.9-feet. This means the distance between the power source and the load for the proposed circuit cannot be longer than 11.9-feet if you want to prevent a voltage drop of less than three percent in the conductors.

If the circuit were wired with 16-AWG conductors, then the value of Ampere-feet would rise to 49.3-Ampere-feet. With a load of 2.6-Amperes the maximum length of the circuit would then increase to 18.9-feet.