= = (600.2) (1110.8889 μA²) 225 V2 + 300 Ω = 0.666 W + 0.75 W Total power absorbed = 1.416 W
= = (600.2) (1110.8889 μA²) 225 V2 + 300 Ω = 0.666 W + 0.75 W Total power absorbed = 1.416 W
Chapter29: Service-entrance Calculations
Section: Chapter Questions
Problem 6R: What load may be used for an electric range rated at not over 12 kW? (Table 220.55) _____
Question
how do we get .666 and .75 watts from this?
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