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The solution to an electrical engineering assignment involving circuit analysis and power calculation. It includes the simplification of circuits, application of kirchhoff's laws, and determination of power absorbed by each element.
Typology: Exercises
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Branches = 5
v 1 = 60 V v 2 = 60 V i 2 = 60/20 = 3 A i 4 = v 1 /4 = 60/4 = 15 A v 3 = 5i 2 = 7.5 V By KVL, -60 + v 3 + v 5 = 0 v 5 = 60 – 7.5 = 52.5 V v 4 = v 5 = 52. i 5 = v 5 /5 = 52.5/5 = 10.5 A i 3 = i 4 + i 5 = 15 + 10.5 = 25.5 A i 1 = i 2 + i 3 = 3 + 25.5 = 28.5 A (b) It is now a simple matter to compute the power absorbed by each element: p 1 = -v 1 i 1 = -(60)(28.5) = -1.71 kW p 2 = v 2 i 2 = (60)(3) = 180 W p 3 = v 3 i 3 = (7.5)(25.5) = 191.25 W p 4 = v 4 i 4 = (52.5)(15) = 787.5 W p 5 = v 5 i 5 = (52.5)(10.5) = 551.25 W and it is a simple matter to check that these values indeed sum to zero as they should.
150V
