topsoil :
LOL if your CPU used 62.3 amps your house power would trip all the time . I mean my houses main is 60 amps
You are confusing power with current. My OC'd Q6600 pulls 9.5 amps from the CPU power connector. Figuring 5% loss in the SMPS on the motherboard, that's 12 volts X 9.5 amps X .95 equals 108 watts. It needs 1.42 volts vcore for around 76 amps.
Your 60 amp house CB is more than likely 240 volts for 14.4 KW.
Noworldorder,
While not exact, the TDP approximates the amount of power the CPU will use at it's stock frequency.
Haserath, all semiconductors are by their nature nonlinear devices. A pretty good approximation of how much more power a highly OC'd CPU will need is:
(higher OC'd voltage)^2/(lower stock voltage)^2. Multiplying that by the TDP will give a very usefull approximation of power needed.
For example - my Q6600, 90 watt TDP, 1.2625 volt VID needs 1.42 volts at 3.6 GHz to be stable. Measured current at stock is 8 amps for 96 watts. Allowing for losses in the SMPS gives something very close to the TDP. At 3.6 GHz, the CPU pulls 9.5 amps.
(1.42)^2/(1.2625)^2 = 1.26. That is almost exactly 10 amps, very close to the measured 9.5 amps.
Or just assume TDP times 1.25 for a highly OC'd CPU. You will be 5% - 10% high.
If you are estimating power needs, you are better off on the high side.