HW _5 Solutions - Problem 3.26 Apply mesh analysis to...

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Problem 3.26 Apply mesh analysis to determine the amount of power supplied by the voltage source in Fig. P3.26. 3 Ω 6 Ω 2 Ω 2 Ω 4 Ω 4 A 48 V + _ I 1 I 2 I 3 Figure P3.26: Circuit for Problem 3.26. Solution: Mesh 1: 2 I 1 + 3 ( I 1 I 3 )+ 2 ( I 1 I 2 )+ 48 = 0 Mesh 2: 48 + 2 ( I 2 I 1 )+ 6 ( I 2 I 3 )+ 4 I 2 = 0 Mesh 3: I 3 = 4 A . Solution is: I 1 = 8 . 4 A , I 2 = 0 . 6 A , I 3 = 4 A . Current entering “ + ” terminal of voltage source is: I = I 1 I 2 = 8 . 4 0 . 6 = 9 A . Hence, P = VI = 48 × ( 9 ) = 432 W .
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Problem 3.27 Use the supermesh concept to solve for V x in the circuit of Fig. P3.27. 1 Ω supermesh 2 Ω 4 Ω 2 Ω 3 A I 1 I 2 + _ V x Figure P3.27: Circuit for Problem 3.27. Solution: Supermesh: I 1 ( 1 + 2 )+( 2 + 4 ) I 2 = 0 Auxiliary: I 2 I 1 = 3 A Solution is: I 1 = 2 A , I 2 = 1 A . V x = 4 I 2 = 4 × 1 = 4 V .
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Problem 3.28 Use the supermesh concept to solve for I x in the circuit of Fig. P3.28. supermesh
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This note was uploaded on 06/15/2011 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas at Austin.

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HW _5 Solutions - Problem 3.26 Apply mesh analysis to...

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