Consider the circuit shown in the figure. Suppose that vs = 50 V.?

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KCL at VΔ
(VΔ-50)/500+ VΔ/1000+( VΔ-Vo)/2000 = 0 Multiply equation by 2000
4VΔ-200+2VΔ+VΔ-Vo = 0 -----> Vo = 7VΔ - 200
KCL at Vo
(Vo-VΔ)/2000 + 10Vo/2000 - VΔ/750 = 0  Multiply equation by 2000
11Vo - VΔ - 8VΔ/3 = 0
Vo = VΔ/3
But Vo = 7VΔ - 200 So VΔ/3 = 7VΔ - 200 Multiply equation by 3
600 = 20VΔ so VΔ = 600/20 = 30V and Vo = VΔ/3 = 10V
Vo = 10V a) <<<<<
b) Vs delivers 50V * 40mA = 2W
The current source delivers 10*30/750 = 0.4W so 2.4W are delivered. b)
Consumed by resistance:
V1 = VΔ = 30, Vo = V2 = 10
20²/500 + 30²/1000 + 20²/2000 + 10²/200 = 2.4W Checks...Show more