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E1201_HW3_SOLN_part2

E1201_HW3_SOLN_part2 - P2.43 To minimize the number of...

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Unformatted text preview: P2 .43 To minimize the number of unknowns, we select the reference node at one end of the voltage source. Then we define the node voltages and write a KCL equation at each node. 5 2 V2 —V1 +v2 —20 :_3 2 10 0 Solving, we find v1 = 18.24 Vand 9" v2 = 13.53 V . . 20 — v2 'U' Then, we have /1 = 10 = 0.647 A. 2 P271 The equivalent circuit with a load Rt attached is: we For a load of 100 (2, we have i, = 10/100 = 0.1 A ,and we can write v, = V, — R1,. Substituting values this becomes 10 = v, — 0.1g (1) Similarly for the ZOO-fl load we obtain 12 = V, — 0.06R, (2) Solving Equations (1) and (2), we find V, = 15 V and R, = 50 Q P2.85 (a) With only the Z-A source activated, we have I; = 2 andl/2 = 20;)3 = 16V (b) With only the l-A source activated, we have I; =1Aandv1 =2(/;)3 =2V (c) With both sources activated, we have /'=3Aandv=£Z(/')3 =54V Superposition does not apply because device A has a nonlinear relationship between vand II ...
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