Question
EET 102 K边尤"de 102 A CT し) CA In 24 SD几 OAC oon 2550 3.
EET 102 K边尤"de
102 A CT し) CA In 24 SD几 OAC oon 2550 3.
Answers
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For a given content at node v1 2 = I2 + I1 2 = v1 - v2/1/4 + v1/1/2 2 = 4(v1 - v2) + 2v1 2 = 4(v1 - v2) + 2v1 2 = 4v1 - 4v2 + 2v1 2 = 6v1 - 4v2 (1) at node 2 I2 = I3 + 6 v1 - v2/1/4 = v2/1 + 6 4(v1 -v2) = v2 + 6 4v1 - 4v2 -v2 == 6 4v1 - 5v2 = 6 (2) solving equation(1) and (2) we get 6v1 - 4v2 = 2 4v1 - 5v2 = 6 multiply equation (1) and (2) = 2 4v1 - 5v2 = 6 multiply equation (1) with 4 and equation (2) with 6 therefore 24v1 - 16 v2 = 8 24v1 - 30 v2 = 36/14v2 = -28 therefore v2 = -28/14 = -2v and 6v1 = 2 + 4 + v2 = 2 + 4(- 2) v1 = 2 - 8/6 = -1 therefore v1 = -1v and v2 = -2v (2) cued can
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