When working on your boat’s electrical system – whether it’s a modification or a new installation – one of the most important things to consider is calculating the wire gauge correctly. Underestimating this aspect can lead to various problems: malfunctions, abnormal consumption, voltage drops, and – in the worst cases – dangerous overheating that can even cause fires on board. That’s why it’s worth taking a few minutes to do the right calculations.
First of all: What do you need to calculate wire gauge?
Before we start with the theory, a small piece of advice: an essential tool to have on board your sailboat is a multimeter to measure battery voltage, check electrical circuits, and solve any problems quickly and efficiently. Nowadays, they are very affordable and offer various features, such as the KAIWEETS HT118E Professional Digital Multimeter that accurately measures voltage, current, resistance, diodes, capacitance, and even temperature.
Calculating Wire Gauge: Where Do We Start?
And now let’s start with the basic concepts that will allow us to calculate wire gauge, where Ohm’s law describes the linear relationship between the three fundamental quantities: voltage (Volts), current intensity (Amperes), and resistance (Ohms).
This law is represented by the equation: U = I*R, where U is voltage, I is current, and R is resistance.
Read differently, Ohm’s law tells us that, given a certain current intensity I, a conductor produces a voltage drop that is directly proportional to its resistance R.
In practice: The greater the resistance of the conductor, the greater the voltage drop caused by a given current flowing through it. Conversely, the lower the resistance of the conductor, the lower the voltage drop.
It can therefore be deduced that the resistance of the conductor, in our case the cable, should be minimized.
But what does the cable’s resistance depend on?
It should be noted that the electrical resistance of a cable increases with its length, while it decreases as its cross-section increases.
The resistance of the cable is given by the following formula: R = K*L/S
K is the specific resistance of the conductor, in our case copper cables, which is 0.02 Ohm per meter.
L is the length of the cable in meters.
S is its cross-section expressed in square millimeters (mm2)
For a practical example, using this formula, we discover that a copper cable 10 meters long with a cross-section of 1 mm2 will have a resistance of 0.2 Ohm (0.02 * 10 / 1 = 0.2 Ohm).
How do you calculate the voltage drop on the cable?
Let’s suppose we use these 10 meters of copper cable to carry current from our battery to the refrigerator compressor. Remember that the cable length includes both the outgoing and return paths, i.e., from the positive pole of the battery to the compressor and back to the negative pole of the battery.
We now need to calculate the current consumed by the refrigerator compressor: let’s say it consumes 90W and is powered at 12V, a current of 7.5 A would flow through the cable (90W /12V = 7.5A).
Knowing the current (7.5 A) and the cable resistance (0.2 Ohm), we can calculate the voltage drop on the cable by applying the formula seen earlier.
U = I * R = 7.5A * 0.2 Ohm = 1.5 V
Let’s assume we power our refrigerator compressor with a fully charged 12.8V battery. After accounting for the 1.5V voltage drop in the cable, we’ll only have 11.3V at the appliance. The compressor will likely go into protection mode to preserve the battery and will be deactivated.
So, how do we calculate the right cable cross-section?
There are two solutions to resolve this undesired effect: decrease the cable length or increase its cross-section.
The first solution probably won’t be possible because each appliance will have a specific place where it needs to be installed, but we can work on increasing the cable cross-section.
Here’s the formula that allows us to calculate the cable cross-section, knowing the length, the current flowing through it, and setting a desired voltage drop.
S = 0.02 * I * L / V
S is the cable cross-section in mm2
I is the current in Amperes flowing through it (normally calculated by dividing the appliance power by the battery voltage I = W / V)
L is the cable length in meters (round trip)
V is the desired voltage drop (do not exceed 0.5V)
Let’s see how to calculate the cable cross-section using the refrigerator example again:
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Current: 7.5 A
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Length: 10 meters
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Acceptable drop: 0.5V
S = 0.02 × 7.5 × 10 / 0.5 = 3 mm²
Watch out for overheating
Caution: The higher the cable resistance, the greater the voltage drop on the cable, and there will also be overheating of the cable. It is also known that the specific resistance of the conductor increases as its temperature rises, thus entering a chain reaction: the resistance increases further, causing the temperature to rise, and so on until it reaches a dangerous overheating that could lead to a fire.
Some other rules to keep in mind
A note on the presence of different voltage lines: if there are 12 – 24V DC and 220V single-phase voltages on board, it is important that they are placed in appropriate flame-retardant conduits or pipes, separating the three lines. The materials of the conduits must not generate toxic fumes and flames in case of fire. The important thing is that these conduits ensure the separation of the three voltage lines, as well as easy identification of the same, for the purpose of correct and quick subsequent maintenance and in the presence of faults that could occur over the years during operation.
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