Basic Electrical Units
A volt is a unit of electrical potential. Electrical potential in turn is the force
which a power source (such as a battery) applies to a connected device in order to push
electrons through it. Electrical potential (voltage) is comperable to the presure which
a pump applies to a plumbing system in order to push water through the pipes.
An ampere is a unit describing the volume of an electrical current. (One ampere is
equal to approximately 6,241,000,000,000,000,000 electrons passing a barrier per second.)
Using the water analogy, it is comparable to gallons per minute.
The Ohm is the unit of electrical resistance. Resistance is the ability of a substance
to resist an electrical potential applied across it and pass less current than an
theoretical ideal conductor. Again using the water analogy, a pipe resists the passage of water
by having a small diameter or a rough internal surface.
The watt is the unit of electrical power expended.
What is Electrical Power
When we refer to electrical power we mean exerted power as opposed to the mere ability to act.
To illustrate, imagine a loaded railway boxcar and a muscular man standing beside it.
We might call him powerful in view of his likely ability to unload the railway car. But in
electrical terminology this is not power since he is not doing anything, it is potential which
is measured in volts.
If the man now starts to unload the railway car, then there is starts to exert power. If the power were electrical, we could measure it in watts. The watt is a unit of work done. This work might be accomplished by a motor turning a shaft or an electric heater warming a room.
Electrical power is exerted when pressure is applied to electrons to make them move. The
exerted power rises if either the pressure (in volts) is increased or the number of
electrons so moved (in amperes) increases. (As we will see later, increasing the pressure
will also increase the number of electrons moved.)
Computing Power Consumption
In order to compute the power consumed by a device, we must know the voltage supplied at
its input terminals and the current drawn. The voltage can be measured by connecting
a voltmeter to the input terminals. In order to compute the current drawn in a DC system
it is necessary to disconnect one of the power leads from the input terminal and then connect an ammeter in series with it.
The power in watts is equal to the applied potential (in volts) multiplied by the drawn
current (in amperes):
watts = volts * amps
For example, if the supplied voltage is 48 and the current is half
an ampere, then the power drawn is 24 watts.
This formula can be transformed algebraically so that we can find any of the three parameters if we know the other two:
volts = —————
amps = —————
An analogy with a water wheel is helpful here. If potential (voltage) is compared to the height from which the water falls and current (amperage) is compared to the water’s quantity, then power (wattage) is the size of the machinery inside the mill which it can turn. If we install more machinery, we must either open the sluice wider (increase the amperage) or arrange for the water the come in higher on the wheel (increase the voltage).
Ohm’s law explains the relationship between the voltage applied to a device, its resistance, and the amount of current which will flow through it. Specifically, it states that a potential of one volt applied to a device with a resistance of one Ohm will cause a current of one Ampere to flow. This can be expressed as the formula:
amps = ———
This formula can also be transformed algebraically so that we can find any of the three parameters if we know the other two:
volts = amps * ohms
ohms = ———–
Returning to our analogy with a pump and a water pipe, if we want to get more water through the pipe, we can do one of two things:
- Install a more powerful pump (increase the voltage)
- Clean the pipe or install larger one (decrease the resistance)
Ohm’s Law and Series Connected Resistors
If we connect a battery to two resistors connected in series, their total resistance
will determine the amount of current that will flow. For example, if we apply twelve volts to a one Ohm and a two Ohm resistor, Ohm’s law says the current will be:
———— = 4 amps
1 + 2 ohms
But what if we connect a voltmeter across one of the resistors? Ohm’s law can answer this question too:
1 ohm * 4 amps = 4 volts
2 ohms * 4 amps = 8 volts
Note that four volts plus eight volts is 12 volts which is what we applied. Each resistor ‘sees’ a fraction of the total voltage proportionate to its contribution to the total resistance.
Everything is a Resistor
When we say resistor we sometimes mean an electrical component specifically designed to consume power without producing a useful result. But in reality anything that will pass a current in a manner which conforms to Ohms law can be analyzed as a resistor. This includes electric heaters, incandescent lamps, and the wires used to connect them.
Wires Acting as Series Connected Resistors
Though the resistance of copper wire is low, it is not zero. For example, 1000 feet of
number 10 AWG copper wire (such as might be used to connect an electric dryer) has a
resistance of about one ohm. Much thinner number 24 AWG wire (such as might be used to
connect a telephone) has a resistance of about 25 ohms per 1000 feet.
When we connect a device (say an electric lamp) to a battery we must remember that wires
are resistors. The lamp is a resistor too. They are two resistors connected in series.
Remember that each resistor in a series sees only part of the applied voltage. If the
wire-resistor gets too large a share, then a voltage significantly lower than intended
will be applied to the lamp. Consequently it will receive less power than intended and
will glow only dimly.
For example, suppose that we want to connect a 25 watt 12 volt lamp to an automobile
using a 50 foot telephone cord. How much of the 12 volts will the cable get and how
much will the lamp get?
First of all, what is the resistance of the lamp? The lamp is designed to consume
25 watts at 12 volts. Therefor the intended current is:
————- = 2.083 amps
We can now use Ohm’s law to determine what the lamp’s resistance must be:
————- = 5.761 ohms
Second, what is the resistance of the wire? A 50 foot cable contains 100 feet
of wire. If 1000 feet of number 24 AWG wire has a resistance of 25 ohms, then
100 feet has a resistance of 2.5 ohms. That is the resistance of our cable.
So, we have a 5.77 ohm resistor (the lamp) and a 2.5 ohm resistor (the wire) connected
in series to our automobile’s battery. What will be the current?
————— = 1.453 amps
2.5 + 5.761 ohms
Now we can compute the actual power drawn from the battery:
12 volts * 1.453 amps == 17.436 watts
The resistance of the thin wire means that only 17.4 watts are drawn from the battery rather than the 25 watts intended. But that is not the worst of it. How much of this reduced power is the lamp getting? First, we figure the voltage across its terminals:
1.453 amps * 5.761 ohms = 8.371 volts
And the power:
8.371 volts * 1.453 amps = 12.16 watts
We can compute the power consumed by the wire in the same way:
1.453 amps * 2.5 ohms = 3.633 volts
3.633 volts * 1.453 amps = 5.279 watts
As you can see, the resistance added by the thin wire has not only reduced the total power drawn from the battery, but it has taken a large chunk of what was left for itself. What, you may ask, has it done with this power? It has turned it into heat. If you coil up the wire and leave it for a few minutes you will find that it has become warm.
Since there was no intention to use the wire as an electric heater, the power which it
consumes is considered to have been wasted. And the fact that it takes 3.633 of the
battery’s 12 volts for it self is termed a voltage loss.
And that brings us to our next important topic: ampacity. Ampacity is the amount of
current which a wire can carry without becoming unreasonably hot. The idea is to
keep the wire well below any temperature at which its insulation might be damaged
or it might start a fire. Obviously how hot is acceptable is a matter of opinion.
Besides the squishy concept of unreasonably hot the ampacity of a wire depends on
two factors. The first is its resistance. All else being equal a thinner wire will
have a lower ampacity because it has a higher resistance and hence for any given
current generates more heat. The second factor is the wire’s ability to dissipate
that heat. This depends on how it is installed. If it is running bare high in the
air on a winter’s day the ampacity will be much higher than if it is inside
insulation inside a cable in a hot attic in July.
Though ampacity is a subjective concept, there are standards which assign
ampacities to particular conductors when they are installed under particular
conditions. For example, number 12 AWG wire with insulation which can stand
a temperature of 75 degrees Celsius is generally considered to have an
ampacity of 20 amps (16 amps in continuous duty).
Note that ampacity does not necessarily tell you what size wire you should
install. It tells you the minimum size you can install without raising
legitamate objections grounds of safety or durability. But on long runs the
minimum safe size will waste too much power.
For example, if you might be able to connect 1900 watts of lamps to the end of a 250
foot number 12 AWG cable without exceeding its ampacity. But, you would lose
about 12 volts (which would make the lamps dimmer) and burn up about
170 watts in the cable. The 170 watts will be spread out over the whole length of
the cable, so it won’t get hot, but it is still a substantial amount of
The commonly used rule of thumb is that you should make the wire big enough that
you do not waste more than 5% of your power heating up the wire. You can make a
rough determination of the required wire size in the following manner.
- Determine the system voltage. Compute 5%. This is the maximum acceptable voltage loss.
- Measure the wire run length in feet from power source to powered device. Multiply by two (to reflect that fact that there will be one wire in each direction).
- Figure out how much power the connected devices actually draws. Do not rely on information which may be printed on them. These figures are often heavily padded. Instead hook one up with am ammeter in series. If the devices can supply power to other devices, hook them up too. Do not let the device idle (where it may draw less power than normal). Put it into the normal operational mode.
- Use Ohm’s law to determine the the wire resistance which would cause the maximum acceptable voltage drop determined in step one. In other words, divide the maximum acceptable loss in volts by the maximum current drawn in amperes. (Convert mill-amperes to amperes first by multiplying by 1000.) The answer will be in Ohms.
- Divide 1000 by the number of feet of wire from step two and multiple by the number of Ohms from the previous step. This is the maximum acceptable resistance of the wire per 1000 feet.
- Open http://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes and run your eye down the “Copper Resistance, (Ω/kFT)” column until you find a number which is lower than the answer you got in the previous step. Look in the far left column of the same row to find the cooresponding American Wire Gauge size.
- If the American Wire Gauge size obtained in the previous step is odd, subtract one from it. (Only even size are generally available.) This is the smallest size you should use.
Long Runs at Low Voltage
Are careful reading of the above section will show that long runs at low voltage require surprisingly large diameter wire. If we halve the voltage in a system while keeping the power the same, we must quadruple the cross-sectional area of the wire in order to keep the losses the same. Why is this?
The first reason is that if we halve the voltage we must double the current
if we want to keep the same power. At double the current we must double the
cross sectional area of the wire if we want to keep the same voltage drop
measured in volts. But we actually want to keep the same voltage drop as a
percentage of the supply voltage. Since we have halved the supply voltage
we must halve the acceptable voltage drop which requires us to double the
cross sectional area of the wire again.
This effect can be dramatic. For example, we might be able to power a 48 volt device using a 50 foot run of number 24 AWG telephone wire. But if we replaced it with a 12 volt device and substituted a 12 volt power supply we would have to replace the wire with number 14 AWG which is like a fairly heavy-duty extension cord.