10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/ − 45◦ V, I = 20/15◦ A Therefore 1 P = (100)(20) cos[−45 − (15)] = 500 W, 2 Q = 1000 sin −60◦ = −866.03 VAR, [b] V = 100/ − 45◦ , B→A I = 20/165◦ P = 1000 cos(−210◦ ) = −866.03 W, B→A Q = 1000 sin(−210◦ ) = 500 VAR, [c] V = 100/ − 45◦ , A→B I = 20/ − 105◦ P = 1000 cos(60◦ ) = 500 W, A→B Q = 1000 sin(60◦ ) = 866.03 VAR, [d] V = 100/0◦ , A→B A→B I = 20/120◦ P = 1000 cos(−120◦ ) = −500 W, B→A Q = 1000…
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