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**Unformatted text preview: **2. Mesh Analysis Problem : Find I 1 , I 2 , I 3 and I 4 by mesh analysis. ANS: Define the currents I 5 and I 6 as in the diagram. I A = I 6- 5. KVL around loop of I 5 : 5I A = 5(I 6- 5) = 3I 5 + 12(I 5 +2) + 8 (I 5-I 6 ) (1) KVL around loop of I 6 : 0 = 5I 6 + 8 (I 6-I 5 ) + 9 + 4(I 6-5) (2) Simplify: 23I 5- 13I 6 = -49 (1)' -8I 5 +17I 6 = +11 (2)' Solve: I 5 = -2.404 A I 6 = -0.484 A Now convert back to original currents: I 1 = I 5 = -2.404 A I 2 = -2 A I 3 = I 6- 5 = -5.484 I 4 = I 6 = -0.484 A 5A 2A I 5 I 6...

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