EE 240 - HOMEWORK - Solution 7

EE 240 - HOMEWORK - Solution 7 - EE240-A2-2010 Assignment#7...

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EE240-A2-2010 Assignment #7 SOLUTION 1. In this industrial situation, find the power factor of the load accounting for power lost in the line. The power lost in the line is 4 kW = (0.1 Ω )(|I L,RMS |) 2 from which we find |I L,RMS | = 200 A RMS . The load power is P LOAD = |I L,RMS | |V L,RMS | pf = 40,000 W pf = (40,000W)/( |I L,RMS | |V L,RMS |) = (40,000)/(200 x 240) pf =0.833 2. In this industrial situation, the load power factor is 0.8 lagging. Find the amount of power absorbed by the line. Calculate the load power: P LOAD = |I L,RMS | |V L,RMS | pf = (200)(240)(0.8) = 38,400 W The amount lost in the line is therefore P LINE = (40kW - 38.4kW) = 1.60 kW. 3. The complex power of the V 1 source is S 1 = 2000VA/-30 O . If V 1 = 200V RMS /10 O , find the rms current phasor I and then find the phasor V 2 . Using the passive sign convention, we find the current I from S 1 = 2000VA/-30 O = V 1 (-I)* = 200V RMS /10 O (-I)* Therefore (-I)* = 10 A RMS /-40 O . (-I) = 10 A RMS /+40 O . and I = 10 A RMS /-140 O . The RMS voltage drop across (5+j10) in the direction of I is V 1 - V 2 = (5+j10)I V 1 - V 2 = (5+j10)(10 A RMS /-140 O ) = 111.8 V RMS /-76.6 O V 2 = V 1 -111.8 V RMS /-76.6 O = 223.2V RMS /40.0 O .
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