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Monaco Executive 2005 - Battery Voltage & Current

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Section 8 --- Electrical Systems - House
The Executive 2005
368
Battery Voltage & Current
Why does the voltage on a discharged battery measure the same as a fully charged battery until the
loads are applied? The simple answer to this might go as follows: A battery creates electrical power by
converting energy from a chemical reaction into electrical energy. As this reaction slows down the
battery voltage will drop. In a lead acid battery the electrolyte conductivity (how well electrical current
can flow through it) changes. The same current may be available but the rate of the reaction decreases,
causing a voltage drop.
Another way of looking at this is to use the analogy of a water pump (a battery is an electric pump).
The pressure in psi (pounds per square inch) that a pump delivers is like a battery’s voltage. The volume
of water in GPM (gallons per minute) is like the electrical current. Look at a 12 psi pump with no loads
(the pump is running but the outflow valve is turned off). The pump will run and the internal pressure
of the pump will build up to some point higher than 12 psi. When the valve is opened, and the water is
free to flow into the loads, the pressure will drop to the rated output pressure of 12 psi, but only if the
load is not too big. If the pump is designed to maintain 12 psi at 15 GPM, and a load demanding 20
GPM is connected, the pump will not be able to keep up and the pressure will get sucked down to a
lower psi. If the load is reduced or removed the pump will catch up and return to its rated 12 psi pres-
sure. If the pump has an infinite source of water, such as a lake or the water utility (this is like the grid,
no battery), the pump will never run out of pressure.
If the pump never runs out of pressure, and is operated at or below its 15 GPM level, it will
hold 12 psi. However, a pump that is connected to a water tank with a finite capacity will start to lose
the ability to hold pressure as the level of water in the tank drops. Think of siphoning water from a buck-
et. As the level of the water drops, the volume of water exiting the siphon slows down.
When the tank is full it is capable of feeding more “pressure” to the pump inlet due to gravity, and
the pump always has enough water available to maintain its rated pressure and volume. However, if
the water tank gets low the pump will not have enough water volume coming in to maintain 12 psi at
15 GPM. If the loads are removed from the pump by closing the valve on the outflow, even with low
pressure in the tank the pump will eventually pressure up to 12 psi. It will just take it longer to get
there. When the valve is opened the pump will sustain 12 psi for a brief period, but since the tank is no
longer feeding the pump as fast as needed the pressure will eventually drop. This analogy can be
restated by replacing the pump with a battery, pressure with voltage, volume with amps, outflow valve
with a switch, water with electricity and the water tank with the battery electrolyte.
The level of the tank could be thought of as the rate of the reaction occurring in the electrolyte. When
the battery is fully charged the electrolyte has an excess of reactions taking place to feed the battery
terminals. This tapers off with time as the electrolyte is spent, so maintaining voltage becomes possi-
ble. With no loads the discharged electrolyte will be capable of producing close to the rated voltage,
but only after a period of time has elapsed for enough of a reaction to take place to bring the voltage
back up. Hopefully, this explanation will clarify why a battery measured at rest can indicate close to its
rated voltage but will not run a load.

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