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Were the load purely resistive, voltage and current would be in phase, watts would equal volts times amps, and the power factor would be 1.0. Since the load has an inductive component, however, voltage and current are never simultaneously at their maximum values, so the true power (instantaneous product of voltage and current) is less than the maximum voltage times the maximum current, and the power factor is less than 1.0. As shown in Figure 7.6, PF equals cos , the cosine of the phase angle between voltage and current. For purely resistive loads such as incandescent lamps, = 0 , so PF = 1.0. In the example shown, current lags behind voltage by 60 ( = 60 ), so PF = cos 60 = 0.5. The de nition of PF can also be written as Amps = Watts/(Volts PF) The above equation indicates the problem with small PF. For the same wattage of useful power, halving PF doubles amps. Greater amps means larger supply conductors, greater voltage drops, greater resistive heating, and decreased ef ciency. The power factor of an electric motor can be very small when the rotor is stalled. As a result, locked-rotor and start-up currents can be three to ve times normal running current, a factor which must be considered when protecting motor circuits.
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