In part 1,
I described current as being the motion of electrons
from one atom to the next within a material. A material’s
ability to conduct current is a function of its ability to
pass on electrons. All materials conduct current to some
degree. Materials that resist the passing on of electrons are
called insulators. Materials that put up very little
resistance to the passing on of electrons are called conductors.
Ohm’s Law and
Electrical Resistance
When we speak of the electrical resistance of a material, we
are referring to the resistance of the material to pass
on electrons. Electrical resistance is symbolized by the
letter R for calculation purposes and is measured in
ohms (abbreviated with the Greek letter ). To measure the
resistance of a material, we apply a voltage to the material
and measure how much current flows as a result of the voltage.
The resistance is then calculated by dividing the voltage V by the current I.
For calculation purposes, the letter I is commonly used to represent the current: R = V / I.
This relationship between current, voltage and resistance is
known as Ohm’s Law because it was discovered experimentally
by the German physicist Georg Simon Ohm around 1840. If we
multiply both sides of the equation by I, we change the
form of the equation to V = I x R. If we then
divide both sides of the equation by R, we change the
form of the equation to I = V / R. Now we
have Ohm’s Law in three forms. I numbered each equation for
future reference.
V = I x R (1)
I = V / R (2)
R = V / I (3)
If we know the value of any two of the
three elements—current, voltage or resistance—we can
calculate the third. Later, I will crank through a few simple
examples and talk more about resistance, but I first want to
touch on power.
Power
In mechanical devices like internal combustion engines, power is the force the engine pushes on an object times the distance
the object moves divided by the time it takes the object to
move that distance. Power is symbolized by the letter P for calculation purposes. Mechanical power is often measured
in horsepower (abbreviated hp). For example, if the force
necessary to keep a car moving at 30 miles per hour (44
ft./second) is 300 pounds, then the amount of power necessary
is 300 lb. x 44 ft./sec.= 13,200 ft.-lb./sec. To convert to
horsepower, we divide by 550. The power necessary to keep the
car moving is 13,200 / 550 = 24 horsepower. In electrical
devices, power is measured in watts (abbreviated W).
One horsepower equals 746 watts. If a one horsepower electric
motor were 100 percent efficient, it would use 746 watts of
power when delivering one horsepower of power. When the
electrons move from one atom to another atom within a
conductor, i.e., when current flows in a wire, heat is
generated within the wire because of the passing on of the
electrons. The power generated by the wire in the form of heat
is the voltage across the wire times the current flowing in
the wire: P = V x I. Within an
incandescent light bulb, there is a small piece of wire made
out of tungsten. Because of its very high melting temperature
(3380°C), tungsten works very well as a filament in light
bulbs. If a light bulb is rated at sixty watts, that is the
amount of power it uses when operated at its rated voltage.
How much current does a sixty-watt light bulb draw when
operating at 120 volts? If we divide both sides of the above
equation by V, we change the form of the equation to I = P / V. Using this equation, we can calculate
the current flowing in the light bulb. I = 60 W / 120 V = 0.5 A. To calculate the resistance
of the filament, we combine the equations P = V x I and V = I x R, we get P = I 2 x R. If we divide both sides of
the equation by I 2 we change the form of
the equation to R = P / I 2.
Using this equation, the resistance of the bulb filament is 60 W / (0.5 A x 0.5 A) = 240 W. Likewise, if
we combine equations P = V x I and I = V / R, we get P = V 2 / R. If we multiply both sides of the equation
by R and divide both sides of the equation by P we change the form of the equation to R = V 2 / P. Using this equation, the resistance of the bulb
filament is (120 V x 120 V) / 60 W = 240 W. Solving for V and I in the above
equations, we create two additional equations. Now we have
nine power equations relating power, voltage, current and
resistance. Using these equations and Ohm’s Law, if we know
any two elements, we can calculate the other two.
P = V x I (4)
V = P / I (5)
I = P / V (6)
P = I 2 x R (7)
R = P / I 2 (8)
I = square root (P / R)
(9)
P = V 2 / R (10)
R = V 2 / P (11)
V = square root (R x P)
(12)
Standard Wire
Sizes
Single strand wire sizes are identified by American Wire Gauge
(AWG). The higher the gauge, the smaller the wire. Typical
house wiring for receptacles is 14 AWG copper. 14 AWG has a
diameter of 0.06408 inches. The next larger size after 1 AWG
is 0 AWG referred to as "one aught"1 and
written as 1/0 AWG. The next larger size above 1/0 AWG is 00
AWG referred to as "two aught" and written as 2/0
AWG. The largest wire is 4/0 AWG, 0.4600 inches in diameter. I
will talk about larger and multiple strand wires later.
Example
Calculations
Now that we have a series of equations relating power,
voltage, current and resistance, I want to crank through some
simple example calculations.
Problem 1: I want to operate a 60-watt
light bulb temporarily in a chicken coop 100 feet from my
house to keep some chicks warm at night. We will assume the
outside air temperature is constant at 20°C
(68°F).
I have a 100-foot extension cord with 18 AWG copper wire. What
voltage do I need at the house receptacle to insure the
voltage at the bulb is 120 V? The circuit diagram looks like figure
1.
The resistance of the 18 AWG copper wire at
20°C
is 6.39 W / 1000 ft per manufacturer’s data. Since the
cord is only 100 feet long, the resistance of the wire is one
tenth of 6.39 or 0.639 W. The resistance of the 60-watt bulb
operating at 120 volts is 240 W as calculated previously. Note
that the amount of power the bulb uses and the light output of
bulb vary with operating voltage. If the bulb voltage is below
120 V, the bulb uses less power, the light output is reduced
and the life of the bulb increases. Likewise, if the voltage
is greater than 120 volts, the bulb uses more power, the light
output is greater and life of the bulb decreases.
The
receptacle at the house is the source of EMF (voltage). How
much current is flowing in the circuit? We know that the
voltage across the bulb is 120 volts and the power the bulb is
using is 60 watts. From this information, we can calculate the
current in the circuit using equation (6) I = P /
V = 60 W / 120 V = 0.5 A. If
the bulb were connected at the source receptacle, the source
voltage and the bulb voltage would be the same. In this
example, the bulb is connected to the source receptacle by a
100-foot extension cord. The resistance of the extension cord
reduces the voltage that gets to the bulb. This reduction in
voltage is called voltage drop.
Since we know the
resistance of the extension cord and how much current is
flowing in the extension cord, we can calculate the voltage
drop using equation (1) V = I circuit x R extension cord = 0.5 A x 0.639 W = 0.3195 V. This
is the voltage drop in just one wire of the extension cord.
There is also voltage drop in the other wire so we have to
multiply the voltage drop times two. The total voltage drop in
the extension cord is 2 x 0.3195 V = 0.639 V.
This voltage drop is quite low because the current is low. If
we were trying to use a 1500 W electric heater at the end of
the same extension cord to heat the entire chicken coop, the
voltage drop would be so high the heater would produce much
less heat. For the light bulb circuit to deliver 120 volts to
the bulb, the source voltage will have to be 120 V plus the
voltage drop in the extension cord or 120.639 V.
Problem 2: Let’s consider the problem of
heating the entire chicken coop with a 1500 W electric heater
at the end of the same extension cord. Note that the nameplate
rating on the heater is 1500 W at 120 volts. When the voltage
is lower than 120 volts, the heat output of the heater goes
down. When the voltage is above 120 volts, the heat output
goes up. What source voltage do we need at the house so that
the heater gets 120 volts? We can calculate the current the
heater draws using equation (6) I = P / V = 1500 W / 120 V = 12.5 A.
We can
calculate the voltage drop in the extension cord using
equation (1) V = I circuit x R extension
cord = 12.5 A. x 0.639 W = 7.9875 V. Again the
voltage drop is in both wires of the extension cord so we have
to multiply by two. The total voltage drop = 2 x 7.9875 =
15.975 V. In this case, the source voltage would have to be
120 V + 15.975 V = 135.975 to maintain the heater voltage at
120 V. I wired up the above example in my lab with the source
voltage at 120 V. With the 100 foot extension cord, the
voltage at the heater was only 105.1 V and the circuit current
was 10.5 A.
Using equation (4) P = V x I,
the amount of power the heater was producing was only 105.1 V
x 10.5 A = 1103.55 W. Note that the reduced voltage at the
heater greatly reduces the heat output of the heater. The
voltage drop in the extension cord was 120 V – 105.1 V =
14.9 V. Using equation (3) R = V / R, the
resistance of the extension cord wire is 7.45 V / 10.5 A =
0.709 W. I used half the voltage drop in the equation because
half the voltage drop is in each wire. Note that this
resistance is higher than the manufacturer’s data. One
difference might be because the manufacturer’s data was at
20°C
(68°F).
When I did the test, the temperature in the room was 80°F
and this test was not the first one I did with the extension
cord. You will recall that the resistance of the wire produces
heat.
That heat gradually raises the temperature of the wire.
As the temperature increases, the resistance of the wire
increases. After several tests or after running one test for
several minutes, the wire within the extension cord may be
considerably hotter than 68°F.
Using equation (7) we can calculate the amount of heat being
produced by each wire of the extension cord. P = I 2 x R = 10.5 A x 10.5 A x 0.709 W= 78.16 W.
The total heat produced by the extension cord is two times
that produced by each wire or 2 x 78.16 W = 156.32 W. This
type of wasted power is called loss. Not only is it
important to use adequate wire size to insure the heater gets
the correct voltage and produces the expected amount of heat,
it is also important to reduce the losses. In this example,
the total power being used by the circuit is 1103.55 W +
156.32 W = 1259.87 W. Only 1103.55 W or 87.6 % is being used
where we want it in the chicken coop. The other 12.4% is
heating up the back yard. If we left the circuit running for
several hours, the heating of the extension cord might
eventually cause the wire to melt and start a fire. Note that
the 15 amp fuse or breaker that protects the 14 AWG copper
house wiring will not protect the 18 AWG extension cord. That’s
another story I will eventually get to. If I had used an
extension cord made with 12 AWG copper wire (1.59 W/ 1000
feet), the voltage drop in the extension cord would only be
3.975 volts and the heat loss associated with the extension
cord would only be 49.68 W.
1Aught is a Middle English word meaning "a cipher; zero."
It is pronounced "ôt."
Please send me your comments on this
series. If you have questions about basic electricity or
general questions about the NESC, please e-mail me at dave.young@conectiv.com .
National Electrical Safety Code and NESC
are registered trademarks of the Institute of Electrical and
Electronics Engineers. National Electrical Code and NEC
are registered trademarks of the National Fire Protection
Association.
Dave is a consulting engineer with Conectiv Power
Delivery of Wilmington, Delaware, where he has been working with and
teaching all aspects of the NESC for over 33 years. He is a member of
the NESC Interpretations Subcommittee and represents the Edison Electric
Institute on the NESC Overhead Line Clearances Subcommittee 4. Dave is
also an inspector member of the IAEI. |
