131223-0927 EST
102 Inspector:
Can you provide more details?
I agree with kwired.
Consider the following assumptions and analysis:
Your tub has a 5000 W heater at 240 V. That is a resistance of 11.5 ohms. The resistance in the bad connection is 11.5 ohms to obtain the maximum power point. Now each resistance dissipates 1250 W. This a lot of power in one small location. Will it melt the wire? Probably not. Since the tub power is 1/4 its normal value, then it will take 4 times as long to heat the tub, or 4 times as long during the heating part of a keep warm cycle, but the off time will remain the same.
Input energy to heat the tub normally is 5*Hours in kWh. With the equal series resistance the heating time is 4 times longer, but at 2500 W input, or 2 times the normal energy required. Half of the available energy was lost in the unwanted series resistance.
Did you use the power company meter to measure energy or power? Was the problem you discovered simply a series resistance problem, or some sort of shunt resistance?
.