Good question!

If we assume a wind turbine efficiency of 50% and that a household can demand up to about 13 kW, we can calculate the wind turbine diameter.   Let us assume a dry air density of 1.225 kg/m³, which corresponds to a temperature of 15^oC and pressure of 101.325 kPa and wind velocity of 5 m/s.

An equation from this page will be used:

That is a large turbine!  A little good news might be that if the wind velocity were assumed to be 10 m/s, the turbine diameter drops to 7.35 m.

I undertook to estimate my peak power consumption to compare with the 13 kW assumption above.

 So the 13  kW rule of thumb is not bad, assuming that 3,000 W of total demand is not active at any given time.  Note that this is a winter time list - the oil furnace and truck engine block heater would lower the peak demand to 12,785 W in warm weather.  The oven, clothes washer and clothes dryer could be scheduled to run at different times further reducing the peak demand another 2,500 W to 10,285 W.  The refrigerator and coffee maker could also be made to not run simultaneously - another 725 W peak demand reduction to 9,560 W.

Assuming a 10,000 W peak demand for my house and redoing the calculation above, I would still need an 18 m diameter turbine at wind speed of 5 m/s or a 5 m turbine at wind speed of 10 m/s.  Wind power is not looking so attractive .... unless we could store the continuous output.  Realistically, the furnace might run 10% of the time, the water heater maybe 2% of the time the refrigerator maybe 5% of the time, the laundry appliances maybe 0.7% of the time.... There is actually only about 2,540 W that might be considered continuous but that would still require a nine metre turbine just to power the continuous loads.