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July 1997 Willie's Workbench

Posted in Features on July 1, 1997 Comment (0)
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The subject of electrical current in a vehicle seems to be a confusing issue to a lot of readers, judging from the mail I receive. One way to better explain how amps, volts and watts relate to each other is to compare them to a water system. Consider, for instance, the well pump as an alternator that keeps filling the water tank, which is the battery. Let's start out with a short conversion dictionary.


Gallons available = amps
Pump = alternator
Pipe = wire
Valve = switch
Water tank = battery
Friction loss = resistance
Sprinkler = use (lights, winch, etc.)
Gallons per minute = watts (total use)
Pressure = volts
Total use = watts

Therefore one can express total use as pressure x gallons available or volts x amps = watts.

To equal a constant wattage (gpm), voltage and amps must be in equal proportion. Remember volts x amps = watts. Example: 12 volts x 10 amps = 120 watts. Keep in mind that voltage is pressure and amps are gallons available. To maintain the same flow if the voltage drops, then the amps must increase. Ten volts then would require a 12-amp draw to equal 120 watts.

Pressure (voltage) can drop because too small a water pipe (electrical wire) causes excessive friction (resistance). Because of the increased friction (resistance), heat will build up. In both instances, heat causes the molecules to expand and separate, causing vapor lock (reduced electrical conductivity).

If the pump (alternator) is not keeping the tank (battery) full, (which acts as an excessive use reservoir), once the tank (battery) becomes empty, total demand is then placed on the pump (alternator). Example: The well pump puts out 10 gpm, the tank holds 100 gallons, and the sprinkler uses 20 gpm. In five minutes, the sprinkler could deplete the tank. However, the running pump resupplies the tank at 10 gpm.

In 7 1/2 minutes, the tank will be dry and maximum output will be only 10 gpm or the capacity of the pump. Not only will gpm be less, but pressure will be about half of original.

Now compare that to a winch: The alternator puts out 50 amps, the winch at half load draws 200 amps, the battery holds 1,000 amps (these are actually amp/hour ratings, but for sake of simplicity we'll pretend they're minutes). Five minutes of winching depletes the battery if the alternator was not recharging. However, even with the alternator charging at 50 amps it cannot keep up with watts (volts x amps) draw, so voltage drops in the battery as more amps are needed, just like the pressure drops in a water line as the tank gets lower. (Remember, this is all theoretical. We didn't factor in resistance in the wires and battery, or consider battery type or a multitude of other factors that affect real operation, and is used for an example only.)

A similar situation could happen with auxiliary lighting. Four 100- watt driving lamps draw about 30 amps (or more, depending on wiring resistance). With the headlights on and the engine running, that just about uses up all of our alternator's 50-amp output. The lights would stay bright until the reserve capacity of the battery was used up, then progressively dimming as voltage dropped.

The solution? The first that comes to mind is a dual battery installation. This can really help, but it's only a short-term solution to the problem of current depletion. Remember, a battery is only a storage tank and is refilled by the pump-in this case, the alternator. The more storage, the longer you can draw from it, but eventually if the draw exceeds the refill capacity, the tank will go dry. This is where larger alternators come in handy. Something like a 160-amp alternator can just about double the output of the alternators generally installed in newer vehicles, or more than four times the output of early alternators. While 160 amps won't keep up with prolonged winching, it will more that double the winch operation time.


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