Watts (Power) = Volts x Amps
Amps = Watts ÷ Volts
Volts = Watts ÷ Amps
Electrical Study Hall:
And an excerpt from a Good Old Boat article
Ohms Law & You ~ by Gord May
Ohm's law provides the fundamental basis for all electrical
and electronic circuit design. It states:
"Current flow is directly proportional to the applied voltage and inversely proportional to the circuit resistance."
Simply put, any increase in resistance will cause a commensurate decrease in current
The descriptive term "voltage drop" denotes the power losses in the wiring
and other circuit elements and also denotes the power actually delivered to the intended load, such as a bilge pump
. This is a useful concept
because the voltage drop across the entire circuit is always equal to the battery
voltage that powers the circuit.
Every section of wire and each termination has resistance and, therefore, voltage drop. Too bad. We'd like all the voltage to reach the load and "drop" there. The typical boat only has about 12.5 volts available to begin with, so even very small increases in circuit resistance will cause a serious loss of power at the intended load.
The fundamental expression of Ohm's law is:
One expression for power (watts) is W = E x I. This is read as power equals volts times amps.
There are actually three useful mathematical expressions for each of the four terms (current, voltage, resistance, and power). These may be arranged as shown below.
The four fundamental terms are in the inside circle. Next to them in the sectors of the outside circle are the three expressions that equal each term. For example, if you need expressions that equal resistance, you find the R (inner circle, lower left) and see that resistance is equal to:
Voltage squared, divided by power
Voltage divided by current
Power divided by the square of current.
Note that each of the three outer expressions is significantly different because each expresses resistance in a different pair of the other terms. The 12 expressions you will need to understand Ohm's law and calculate aspects of the circuit are arranged in this figure in a way that makes them easy to find and apply.
There is one other expression related to Ohm's law that is critical for doing good wiring
. It describes the fact that resistances of components in a series circuit add up to make the total circuit resistance. This expression is:
The point here is that every circuit element adds resistance to the circuit. Voltage drops across each and every resistance in the circuit and therefore robs the circuit of power. We want to minimize this effect, but we can't avoid it.
The American Boat and Yacht Council's (ABYC) standard E9 specifies the maximum voltage drop permitted in the wires of a DC circuit as follows:
A 3-percent maximum drop for panel feeders, navigation
pumps and blowers, electronic equipment, and other essential equipment. This category encompasses most situations.
A 10-percent maximum drop for non-essential equipment such as cabin lighting
(Be aware that the ABYC develops standards by consensus. According to the ABYC board rules, these standards are advisory only. Compliance with ABYC standards is -- according to ABYC literature -- entirely voluntary. The standards are not law but, as a practical matter, parts
or all of them may be adopted into law by various authorities. Some surveyors and representatives of insurance
companies use them as guidelines. Unfortunately, access to these standards is restricted to members of ABYC. Short of going to the substantial expense of becoming a member
, there is no easy way to know what ABYC standards require. Much of the information and many of the needed tables, however, are published in Charlie Wing's book, Boat Owner's Illustrated Handbook of Wiring. Ed.)
You can find tables in marine catalogs that will suggest what wire size to use to meet these 3-percent and 10-percent requirements, but the correct wire size can also be calculated as well. The industry standard for this calculation involves the concept
of circular mil area, or CM.
Circular mil area is the effective cross-sectional area of the wire expressed in circular mils. One circular mil is the area of a wire that is 0.001 inches in diameter. The area calculation is easy for solid wire, but for stranded wire it is the sum of the areas of all the strands expressed in circular mils. Don't get hung up on calculating the circular mil area of wire by gauge. It is always given in tables.
What you want to know is how heavy a wire you need to use to stay within the guideline in the specific circuit you are wiring.
The equation for the required circular mil area is:
You may assume that battery voltage is 12.5 for this calculation, and so a 3-percent drop will be 0.375 volts.
This expression does not address the resistance added to the circuit by the terminations (joints and splices). The reality is that each "old" termination could, after a time in service
, add a resistance of between 0.01 and 0.03 ohms to the circuit. This additional resistance is, in many installations, greater than the resistance of the wire.
A practical example
You are going to wire a bilge pump
with a capacity of 2,000 gallons per hour (GPH). It is rated at 110 watts when connected to a nominal 12-volt circuit. The wire run is 20 feet to the pump, so the two-way length of wire is 40 feet. Assume a moderately charged battery is at 12.5 volts.
First we find the current in the circuit. Since we know watts and volts and want to find current we select the expression
The minimum wire size required is found with the circular mil calculation:
So you need wire with at least an area of 10,090 circular mils. You go to the table below and find that #10 wire has an area of 10,380 circular mils. That is what you need. Don't skimp on the wire. Always pick a wire with an area as large as, or larger than, calculated.
Note: The table is for American Wire Gauge (AWG). It is not for SAE wire, which has a smaller area for the same gauge. It is best not to use SAE wire at all, but if you are in a pinch, always bump the gauge up by one, so if you needed #10 AWG you will need #8 SAE.
The voltage drop with #10 wire, according to Ohm's law, is I x R, or 8.8 amps x 40 feet x 0.00102 ohms/ foot = 0.359 volts. (You get wire resistance in ohms/foot from the table.) Ignoring the connections for a moment, we get 12.5 volts less 0.359 volts, or 12.141 volts for the pump.
If we'd chosen #12 wire, the calculation would show that the wire voltage drop was 0.570 and only 11.93 volts was left to power the pump. This would severely reduce the output of the pump.
It gets worse
But remember that our calculations did not include any termination resistance. If we assume only 6 "old" splices in our circuit, two each at the panel, switch, and pump, and the old splices have as much as 0.03 ohms, we find that if voltage equals current divided by resistance (E = I x R), then 8.8 amps x (0.03 x 6) = 1.584 volts. And that is just the splices.
Now add in the voltage drop of the wire (0.359 volts), and the pump gets 12.5 (1.584 + 0.359) = 10.56 volts, which approaches its minimum operating voltage. Any further deterioration, such as a low battery, and the pump may not operate at all.
From these calculations we can see the implications of Ohm's law and how it affects our electrical system
. Obviously we don't want "old" dirty connections and undersized wire powering that bilge
pump. Resistance is a fact of life in these circuit elements, but we need to minimize it as much as possible if we want the pump to work.
In fact, the major causes of poor performance and electrical
failures in a boat are:
High resistance from loose, corroded terminals caused by the corrosive effects of moisture penetration.
Undersized wire, often factory installed.
Vibration and mechanical damage to wires.
To avoid these electrical problems:
Plan your wiring to eliminate all unnecessary joints and splices (terminations).
Make high-quality terminations that are clean, tight, and water/moisture resistant.
Install wiring and terminations in high, dry locations.
Use high-quality marine-related materials, including Type III tinned copper marine wire.
Calculate and select adequate wire size.