Voltage Drop By Mike Holt for EC&M Magazine Electrical equipment shall be installed so that it operates within its voltage rating as specified by the manufacturer. Because of circuit conductor voltage drop, the operating voltage at electrical equipment will naturally be less than the output voltage of the power supply. UL does not have any specific requirement that equipment manufactures specify the voltage range of equipment but, typically manufactures recommend that the minimum circuit voltage not be less than 10 percent of the equipment voltage rating. We must be careful to understand the difference between nominal circuit voltage, equipment voltage rating, and actual operating voltage. For example, an 115V rated motor is designed to be installed on a 120 nominal voltage circuit, but the actual operating voltage should not be less than 104V. Inductive loads (e.g., motors, ballasts, etc.) can overheat, resulting in shorter equipment operating life and increased cost if they operate at voltage below their rating. In addition, under voltage can cause sensitive electronic equipment such as computers, laser printers, copy machines, etc., to lock up or suddenly power down. This can result in data loss, increased production cost, and possible equipment failure. Resistive loads that operate at under voltage simply will not provide the expected rated power output. For example, a 10 kW heater rated 230V will provide less than 8.2 kW of power (work) at 208V (P = E^{2}/R). This (under voltage and under power) might not be a hazard, but it could cause production delays as well as increased production cost because the process is not function as it was intended. Reduced circuit voltage can cause incandescent lighting to flicker when other appliances, office equipment, or heating and cooling systems are cycled on. Though this might be annoying for some, it’s not dangerous and does not violate the NEC. The actual operating voltage for a load can be determined by subtracting the conductor voltage drop from the nominal voltage rating. The voltage drop of the circuit conductors for a single-phase load can be determined by multiplying the current of the circuit by the total resistance of the circuit conductors: VD Single Phase = I x Z x 2 VD Single Phase = I x Z x 1.732 I = The load in amperes Z = The impedance of the conductor as listed in the National Electrical Code Chapter 9, Table 9. Question: Can a 16A, 115V motor be connected to a 120V circuit, if the 12 AWG circuit conductors are no more than 100 ft each? Answer: Yes this installation will be fine, because the NEC does not limit the voltage drop to this type of load. The voltage drop of the circuit conductors is equal to the current of the load times the impedance of the circuit conductors. The load is equal to 16A and the resistance of 12 AWG conductors in accordance to Chapter 8, Table 9 is equal to: 0.4 ohms [(2 ohm/1,000 ft) x 200 ft]. The circuit voltage drop is equal to 6.4V (16A x 0.4 ohms), therefore, the load will be operating at 113.6V, assuming the circuit voltage is 120V nominal. Other factors that need to be taken into consideration include the power factor of the load as well as some of the NEC requirements. But remember, we need to be sure the equipment is supplied with a voltage that meets the equipment manufacturer’s requirements. Mike Holt's Comment: If you have any comments or feedback, please let me know at Mike@MikeHolt.com |
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