## Marine Electrical Systems, Part 1c: WIRING NETWORK (Wiring Circuit)

by Cameron Clarke

Ok, now lets examine a typical circuit, a simple wiring diagram, in a boat. How does this affect the bilge pump? A typical 12-volt pump draws about 5.4 amps when pumping water. What comprises the wiring circuit? Lets see, there is a battery, a battery switch, a + terminal distribution block, a fuse, an on/off switch, a float switch, the pump, a - terminal distribution block, back to the battery. Agree? Oh, plus some wire. Now I will examine this again and indicate in parenthesis (#) the number of connections.

Refer to figure 2 as we trace the electrical path. The positive post of the battery is connected (1) to a battery lug, which is connected (1) to wire, connected to battery switch (1 + 1 for switch contacts too), connected (1) to a wire connected to pos. distribution post (1). Continuing from post to fuse (2 + 2 for removable fuse) via wire, to on/off switch (3, each wire to lug plus switch itself), connected to float switch (3), then wire to +lead of bilge pump (1), -lead of pump (1) negative distribution post (1), and negative distribution post (1) to negative battery post (1) via wire and terminals. Did you follow that? That's 21 connections! Therefore, 21 connections times 0.03 ohms each results in 0.63 ohms over all. That does not account for any wire losses, just connections. How did we get so many connections in the first place? Can we eliminate any?

Figure 2, Typical Electrical Circuit

Let's cheat a little to make our example in figure 2 easier to understand, at least the math, OK? Let's say there were only 20 connections, measuring 0.03 ohms each. The result, 20 x 0.03 is 0.6 ohms. We will also assume no wire resistance loss, which of course is impossible. A typical battery voltage in use is 12.5 volts to run our pump. What then is the voltage the pump actually sees? Does it receive the full 12.5 volts? No. The voltage is reduced at the pump in proportion to the resistance of the wire path (the sum of connection resistance and wire resistance) and the current draw. In other words, more resistance or more current draw reduces voltage. Ohm's law states current in amperes times resistance in ohms equals the voltage drop in volts. So 5.4 amperes times 0.6 ohms equals 3.24 volts. What this means is that the wiring and connections, (the wiring network) consumes 3.24 volts, so that the bilge pump receives only 9.26 of the original 12.5 battery volts. Power measurement is the product of amperes times voltage and is expressed in watts. In summary:

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Battery Wire Network Bilge Pump Volts 12.5 3.24 9.26 Amps x 5.4 x 5.4 x 5.4 Watts 67.5 17.5 50.0

Table 2, Wiring Network LossesNote: Amps remain the same in each item as the system is wired totally in series. The full amount of current (5.4 amperes) must flow through each component, wire, terminal, and connection.

In this example, the Battery must provide 67.5 watts for the pump to receive 50 watts. A full 17.5 watts are consumed in the wiring network and lost as heat. That is a 26% Loss of power!

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