In Part
2 of this series in the September/October issue, I worked
through two simple example voltage-drop calculations. In both
examples, I calculated what voltage I would have to have at
the house to insure the voltage of 120 V at the chicken coop.
Let’s call this calculation method A. The reason I used
method A was because I had limited information about the bulb
and heater. In each case, the manufacturer’s information
only gave me the power requirement of each load when operating
at 120 V. The term load relative to electric circuits
is a term commonly used to identify appliances or equipment
that are designed to use electric power. There are two
disadvantages of using method A. In reality, we don’t
usually have the ability to adjust the source voltage and the
calculation method does not tell us what the voltage would be
at the load if the source voltage were 120 V. The alternative
method is to assume a source voltage of 120 V and calculate
the voltage at the load. Let’s call this method B. Though
this method sounds simple, it is not. In order to use method
B, we must know how the power consumed by the load is affected
by the load voltage, the voltage at the bulb or heater. The
only way to acquire this information is to run some tests. In
my lab, I tested a 1500 W space heater to determine its
voltage-current characteristic. I used a variac type 100-Q
variable voltage autotransformer as the voltage source to the
heater. I used a fluke Model 175 multimeter to measure the
voltage and an ideal Model 61-766 clamp-on ammeter to measure
the current. The test setup is pictured in photo
1. Table 1 is a summary of the test results.
Test results: |
V
Voltage
in voltsI
Current
in ampsResistance
R= V / I
in ohms |
Power
P=V x I
in watts |
120.112.179.868 |
1461 |
115.511.769.821 |
1358 |
110.111.229.812 |
1235 |
105.110.729.804 |
1126 |
100.010.219.794 |
1021 |
95.49.769.774 |
931 |
90.59.289.752 |
840 |
Table 1. Test results
Note that if we plot the current versus
voltage on graph paper, the relationship is very close to a
straight line. You may recall from your high school algebra
class that the equation for a straight line is Y = m
X + b where m is the slope of the line and b is the Y intercept. Using the end data points, the
slope of the line m = (12.17 – 9.28) / (120.1 –
90.5) = 0.0976. The equation for the voltage-current
relationship is then I = (0.0976 x V) + b.
Substituting 12.17 for I and 120.1 for V, we can
solve for b.
b = 12.17 – (0.0976 x 120.1) = 0.45.
We now have the equation that defines the
load current as a function of the load voltage for this
particular space heater.
I = (0.0976 x V) + 0.45
(13)
For reference purposes, I will continue to
number the equations. Since we want to determine the load
voltage when the source voltage is 120 V, we must establish
the relationship between the source voltage and the load
voltage. This relationship was first postulated by German
physicist Gustav Robert Kirchhoff around 1880. Kirchhoff
established the basic laws of network analysis. The laws are
now known as Kirchhoff’s Laws. Kirchhoff’s voltage law
states that the sum of the voltages in a series circuit must
equal 0. A series circuit is one where the elements of the
circuit line up in a single loop.
Applying Kirchhoff’s voltage law to the
above circuit diagram, we find that the source voltage plus
the voltage drop in the extension cord wires plus the load
voltage must equal 0. For this circuit, the numerical values
of the voltage drop in the wires and the load voltage will be
negative because the polarity of these voltages is opposite to
that of the source voltage. I will explain polarity and the
differences between direct current (dc) and alternating
current (ac) in Part 5. The equation generated by Kirchhoff’s
voltage law for this circuit then becomes:
Vsource – Vwires – Vload = 0 (14)
We are assuming the source voltage is 120
V. The voltage drop in the extension cord is the circuit
current times the resistance of the wire. As determined in
Part 2, the resistance of 100 feet of 18 AWG copper wire is
0.639 . You will recall that we have to multiply this by two
because the circuit is made up of two wires between the source
and the load. Putting these together we have a second equation
relating the circuit current I to the load voltage V:
120 – ( 2 x 0.639 x I ) – V = 0
(15)
Note that in this case the load current is
the same as the circuit current. Now we have two equations and
two unknowns. By combining equations (13) and (15) together,
we can solve for the load voltage and the current. To do that,
we substitute the right side of equation (13) for I in
equation (15) and solve for V. Equation (15) then
becomes:
120 – (2 x 0.639 x ((0.0976 x V) +
0.45) – V = 0
(16)
Following through with the multiplication,
the equation (16) changes to:
(17)
120 – (2 x 0.639 x 0.0976 x V) –
(2 x 0.639 x 0.45) – V = 0
Equation (17) then reduces to:
120 – (0.1247 x V) – 0.5751 – V = 0
(18)
Combining like values, equation (18)
becomes:
119.4249 – (1.1247 x V) = 0
(19)
Subtracting 119.4249 from both sides of the
equation and dividing both sides of the equation by 1.1247 we
get:
V = 119.4247 / 1.1247 = 106.18 volts
The voltage at the heater is 106.18 volts.
We can now calculate the circuit current by substituting
106.18 for V in equation (13).
I = (0.0976 x 106.18) + 0.45 = 10.81
amps
To check to make sure we did the math
correctly, we substitute 106.18 for V and 10.81 for I in equation (15).
20 – (2 x 0.639 x 10.81) – 106.18 =
0.0048 (20)
Within the accuracy of our calculation and
measurement, the answer is close to 0. The math checks. The
voltage drop in the extension cord is 13.82 volts, the wire
resistance multiplied by the current, the middle section of
equation (20). The power output of the heater is V x I = 106.18 x 10.81 = 1147.8 watts. If that amount of heat is not
enough to heat the chicken coop, we may have to use a larger
wire size to reduce the voltage drop. Another alternative
would be to use a 240-volt heater and use a 240-volts source.
The advantage is that a 240-volt 1500-watt heater draws half
as much current. With half as much current, the voltage drop
in the wire is cut in half. With less voltage drop, the heater
would produce much more heat. In general, the higher the
voltage, the lower the current for the same size load. That is
why electric utilities use very high voltages to distribute
power over great distances.
Figure
1.
Please send me your comments on this
series. If you have any general questions about the NESC,
please e-mail me at dave.young@conectiv.com.
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. |
