2007 Electrical Apparatus

2007 Electrical Apparatus

This is a summary of the Electrical Apparatus November 2007 featured technical article,  by Richard L. Nailen, P.E.  

In any electrical system, current flow results in a reduction or drop in voltage from source to load, because of the impedance in the intervening conductors. In the United States, standards say little about voltage drop. The requirements of ANSI C84.1, such as 480 volts at the source of a circuit in which minimum load voltage is 440, allow for voltage drops of 8 to 10 percent. The National Electrical Code recommends (but does not require) maximum voltage drop of 5 percent. Such limits are often far exceeded during motor starting when load current is 5 to 8 times normal. Motor terminal voltage must then be determined accurately to make sure motor torque is adequate for acceleration.

At low values of current, circuit impedance can be considered as resistance only. In a-c circuits, however, inductive reactance is also present. This becomes significant as conductor size increases to 10 mm or more. Because conductor ampacity also varies with size, some electricians mistakenly assume that ampacity and voltage drop calculations are directly related. However, inductance does not affect ampacity, and temperature influences only resistance.

Because voltage drops across resistance and reactance are out of phase with each other, circuit voltages cannot be calculated by simple addition or subtraction. The resistive component of voltage drop will be in phase with current; the reactive component will be 90 degrees out of phase.

Most literature explains voltage drop calculation using a phasor diagram in which current is presumed known. The drop is then calculated and subtracted from the source voltage. In practice, however, the source voltage is the known quantity; current is calculated based on total circuit impedance including the load, and load voltage is derived from that.

Whatever the method, impedance of both sides of the circuit must be accounted for. Therefore, if conductor impedance is determined for the length of circuit from source to load, that figure must be doubled (or multiplied by 1.732 in a three-phase circuit) to calculate voltage drop.

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