= = (600.2) (1110.8889 μA²) 225 V2 + 300 Ω = 0.666 W + 0.75 W Total power absorbed = 1.416 W

EBK ELECTRICAL WIRING RESIDENTIAL
19th Edition
ISBN:9781337516549
Author:Simmons
Publisher:Simmons
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?

=
= (600.2) (1110.8889 μA²)
225 V2
+
300 Ω
= 0.666 W + 0.75 W
Total power absorbed
= 1.416 W
Transcribed Image Text:= = (600.2) (1110.8889 μA²) 225 V2 + 300 Ω = 0.666 W + 0.75 W Total power absorbed = 1.416 W
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