known. If any two of the three variables (watts, amperes, volts) are known, the third can be 

 computed as follows: 



To find watts when volts and amperes are known, multiply the variables. 

 15 amperes X 120 volts = 1,800 watts 



To find amperes when watts and volts are known, divide watts by volts. 

 1,800 watts -J- 120 volts = 15 amperes 



To find volts when watts and amperes are known, divide watts by amperes. 

 1,800 watts ■=- 15 amperes = 120 volts 



The longer the circuit for a constant load and conductor size, the greater the power loss in 

 the circuit due to resistance of the wire. The resistance in the wire dissipates some of the 

 power by converting it to heat. For example, the input voltage may be 120 volts, but the 

 voltage may be only 114 volts at the output end, which is a drop of 6 volts or 5 percent. To 

 reduce this loss to 3 percent, a larger wire or cable with a larger cross-sectional area (and 

 therefore a lower resistance) would be required. To determine the cable or wire to be used 

 for a 3-percent voltage drop instead of 5 percent, the following method may be used. This 

 method is based on the International-Ohm-Mil-Foot (I-O-M-F), which has been estabhshed 

 as a standard constant of 10.8 ohms per foot of wire length. The ohm is defined as a unit of 

 resistance. An example of this method follows: 



EXAMPLE PROBLEM 



Problem: What size wire should be used to serve a load of 50 amperes on a single-phase 

 2-wire circuit 300 feet from the power source? The supply voltage is 120 volts 

 and the resistance loss is to be kept to about 3 percent. 



Solution: Find 3 percent of 120 volts = 120 volts X 0.03 = 3.6 volts; I-O-M-F constant for 

 two wires = 2 X 10.8 = 21.6; Distance = 300 feet, one way; Current = 50 

 amperes. 



Find wire size in circular-mils: 



constant X length X ampheres _ 21.6 X 300 X 50 

 acce 



324,000 



Circular mils , , , , o ^ i j 



acceptable voltage drop d.o-volt drop 



3.6 



= 90,000 circular mils 



164 



