Basic Electricity, Part 4

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.

Photo 1. Test setup



Test results:
in voltsI
in ampsResistance
R= V / I
in ohms
P=V x I
in watts 1461
115.511.769.821 1358 1235
105.110.729.804 1126 1021
95.49.769.774 931 840

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:


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)

Figure 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

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.