I have a question about power factor. I know that compact fluorescent bulbs are more efficient that normal bulbs, but I also know that they have a power factor of .5 or .6. What does this mean? Am I only getting billed for .5 times the power it is actually using? Does this mean it is not as efficient as it appears but you save money anyway?

A simple resistive load such as a heating element obeys Ohms law I=V/R instantaneously, ie throughout the AC voltage waveform the current is always proportional to the instantaneous voltage. This will also hold true for a conventional lightbulb. The power factor is 1.

Any load which contains an inductor (such as an electric motor) or capacitor on the mains will draw current "out of phase" with the voltage. A pure capacitance will draw current proportional to the rate of change of voltage, so the current will be maximised as the voltage crosses zero (in the sinewave this is when it's changing fastest). Summing up the instantaneous power (I.V) you would find that over one cycle, no "power" is drawn by the capacitive load (and no power would be metered by a domestic meter). However, you have still been pulling current through the wires which means that there will be real energy losses (costs incurred) in the distribution network. Capacitive loads draw current "ahead" of the voltage whereas inductive loads draw current "behind" the voltage - consequently an inductive load cna be balenced by applying a capacitor to it.

Modern devices with switch-mode power supplies including computers and low-energy bulbs typically feed the mains straight into a bridge rectifier with a capacitor on the DC side. This means that they draw essentially zero current for most of the AC cycle, and just a brief burst of current at the peak of the voltage waveform. This means that the instantaneous peak current is much higher than the average current of a resistive load of the same nominal power. If a large fraction of the load on the grid (or even just one premesis) starts loading the network in this fashion it can distort the waveform of the mains and cause other issues (I'm not too sure what exactly). It's not really very easy to correct or compensate for loads which only draw currents at the peak of the waveform.

See also

http://en.wikipedia.org/wiki/Power_factorOne consideration is that mains electricity is distributed as three phases 120degrees apart. Assuming all loads are resistive, the current in the neutral wire to the substation will sum to zero (both instantaneously and over time). Consequently no power is dissipated in the neutral wire, and it can be relatively light gauge.

If all the loads only drew current at the peak of the voltage waveform (so there's no instantaneous overlap of loads) the average absolute current in the neutral would be the sum of the currents in the phases, and the neutral cable would need to have three times the cross-section of the phases (for power-loss / cable self-heating considerations). This is an extreme example, but illustrates problems that can arise.