Mathcad - 13-07 - Problem 13-7 Solution: a. Define three...

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S tot 8.922 10 3 × 685.577j + va = S tot S ref S lamp + S oven + = c. The total real and reactive power is the sum ot the three loads, adding P and Q seperately: S ref 672 685.577j + var = S oven 8000 watt = S ref V 1 I 1 = S lamp 250 watt = b. The lamp and the oven are unity power factor loads so they have no reactive power. The refrigerator does have reactive power: arg I line2 ( 29 0deg = I line2 35.417 amp = I line2 I 2 I 3 + = arg I line1 ( 29 8.348 - deg = I line1 39.35 amp = I line1 I 1 I 3 + = I 3 I oven = I oven 33.333 amp = Problem 13-7 Solution: a. Define three loops for the system. Loop 1 includes the top source, line 1, the refrigerator and the neutral. Loop two includes the bottom source, line 2, the lamp, and the neutral. Loop three includes both sources, lines 1 and 2, and the oven. V 1 120 volt = V 2 120 volt = V 3 240 volt = I ref 8 amp = F p 0.7 = θ acos F p ( 29 = I 1 I ref cos θ ( 29 j I ref sin θ ( 29 - = The current is lagging

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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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Mathcad - 13-07 - Problem 13-7 Solution: a. Define three...

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