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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