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4. CH 10 In the circuit below determine giving the units: The total impedance of the circuit (20+j5 )2 b. The current in the

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202 ilt) 115-2 120kms (~ HA -J102 a) Total imbedance (7) = 20 + 115-110 [Z= 20 +152 1. b) Current W = 120 20+J5 i = 5:825/-14- Power dissipated in R (PR) = i²R = (5.82 L)* x 20 TPR = 677.68 W] 1 Reactive power Keactive power in cap (Qc) = 1² te = (5.OX Q = 10,1-10) Q = 169.4 VAR S =698.52 Zo P=677.65W 0-14.0362 8) IZ = 20+35 120 Voor GoHZ

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