I'm assuming you're in the U.S. and by "#14" wire you mean 14 AWG.

A power factor of 30% means that only 30% of the product of V and I is
power being delivered to a load. The rest is simply being moved back and
forth in the power system. To get 1500 watts of power at a power factor
of 100%, you need P / V amperes. For V = 120 volts, that works out to
12.5 amps. But to get 1500 watts of power with a power factor of 30%,
you need P / V / 0.30 = 12.5 / 0.30 = 41.67 amps. Heating in wires is
due to the current going through them. Ordinary #14 wire in houses is
rated for 15 amps. So you'd be grossly exceeding the rating of the #14
wire in getting 1500 watts with a 30% power factor. [All voltages and
currents in this posting are in RMS.]

When the power factor is less than 100%, you need more current for the
same amount of power, because it requires current to move the "reactive
power" -- power moved back and forth without being used up in a load.

If by "1500 watt capacity" you actually mean "1500 volt-ampere
capacity", then you're ok. In this context, watts are real power, but
volt-amperes are the product of V and I, which is the sum of the real
power (watts) and "reactive power" (VARs, for "Volt Ampere Reactive").
If you limit your volt-ampere product to 1500 volt-amperes, then your
maximum current is 12.5 amps. But if your power factor is 30%, you'll
only be delivering 0.30 * 1500 = 450 watts to the load.

This is why you'll generally see transformers and other power devices
rated by volt-amperes rather than watts.

Roy Lewallen, W7EL

clifto wrote:
A bit off topic, as it has to do with house wiring. But how does power factor
affect dissipation in the conductors? I'm wondering, for example, if a 1500
watt capacity and a 30% power factor would overload #14 wire.