Mesh_Nodal_Analysis

Mesh_Nodal_Analysis - EE201.3 Consider the circuit shown...

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EE201.3 Mesh/Nodal Analysis Denard Lynch Page 1 of 10 Nov., 2008 Consider the circuit shown below: 25 0 0 V 2 -j6 3 j5 j4 j2 -j4 1 -j3 V a V b V c V d V e V f I 1 I 2 I 3 I 4 I S Z 1 i 1 i 2 i 3 + Find: I S , I 1 through I 4 , V a through V f , and Z T , the total impedance presented by the circuit to the 25V voltage source. (Note: all voltages in reference to ground, 0V.) Approach 1: Combine impedances and determine I x and V x using Ohm's Law for AC circuits ( V=IZ) . Need to find Z T to start: For the right legs (call it Z 1 ) we have (1 - j3) // (j2 -j4) = {(1 - j3)(-j2) / (1 - j3) + (-j2)} Z 1 = 1.24 -82.9 0 = (.154 -j1.23) Now add this in series with the -j6 capacitor, (.154 -j1.23) + (-j6) = (.154 -j7.23), and again add this in parallel with the left side of the centre loop (call it Z 2 ) Z 2 = (.154 -j7.23) // (3 + j5) = {(.154 -j7.23)(3 + j5) / (.154 -j7.23) + (3 + j5)] Z 2 = 10.92 5.5 0 = (10.87 +j1.05) Now Z T = Z 2 + (2 + j4) = (12.87 +j5.05) = 13.82 21.4 0 Ω ( Note: mainly inductive in nature) We can now calculate I s = V S /Z T = (25 0 0 ) / 13.82 21.4 0 = 1.809 -21.4 0 A V a = V S - I S (2 Ω ) = (25 0 0 ) - (1.809 -21.4 0 A)(2 Ω ) = 21.67 3.5 0 V V c = 0 + I S (j4 Ω ) = (0V) + (1.809 -21.4 0 A)(j4 Ω ) = 7.23 68.6 0 V To find V b we can proceed a couple of different ways: 1 st , we can find I 1 by using (V a - V c ) / (3 + j5) = (21.67 3.5 0 V - 7.23 68.6 0 V) / (3 + j5) I 1 = 3.386 -74.9A

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EE201.3 Mesh/Nodal Analysis Denard Lynch Page 2 of 10 Nov., 2008 Similarly, I 2 = (V a - V c ) / (Z 1 + cap. in series) = (21.67 3.5 0 V - 7.23 68.6 0 V) / (7.23 - 88.8) I 2 = 2.730 72.9 0 A 2 nd , we can use Current Division to divide I S into I 1 and I 2 . (be careful!) If we call Z 1 ' = Z 1 + (-j6), then I 1 = I S {Z 1 ' / (Z 1 ' + (3 + j5))}, and I 2 = I S {(3 + j5) / (Z 1 ' + (3 + j5))}. These calculations result in I 1 and I 2 as above. Alternatively, once we have either I 1 or I 2 , we can subtract is from I S to get the other. We can now find V b = V c + I 1 (j5) = 7.23 68.6 + (3.386 -74.9)(5 90) = 22.02 30.4 0 V (A good check at this point is that V b also = V a - (3.386 -74.9)(3 0) ) We can continue to find V d = V a - (I 2 )(-j6) = 21.67 3.5 0 V - (2.730 72.9 0 )(6 -90) = 8.57 45.8V Again, we can use different methods to find I 3 . (Feel free to test the other alternative as a check!) I 3 = (V d - V c ) / (j2 -j4) = {(8.57 45.8) - (7.23 68.6)} / (2 -90) = 1.693 80.0A Continuing with these techniques, we can find the remaining values: I 4 = 1.071 61.6A V e = 10.85 30.8V V f = 7.55 43.6V V eb = V e - V b = 11.17 -150.1V Approach 2: Use Mesh analysis to find the three loop currents first, then calculate the other values: First write the mesh equations: Loop 1: (2 + 3 + j5 + j4)i 1 - (3 + j5)i 2 + (0)i 3 = 25 0; rearrange and combine. .. (5 + j9)i 1 + (-3 - j5)i 2 + (0)i 3 = 25 0 (Eq. 1) Loop 2: (3 + j5)i 1 - (3 + j5 -j6 +j2 -j4)i 2 + (j2 - j4)i 3 = 0; rearrange and combine. .. (3 + j5)i 1 + (-3 +j3)i 2 + (0 - j2)i 3 = 0 (Eq. 2)
EE201.3 Mesh/Nodal Analysis Denard Lynch Page 3 of 10 Nov., 2008 Loop 3: (0)i 1 - (j2 -j4)i 2 + (1 -j3 -j4 +j2)i 3 = 0; again rearranging. ... (0)i 1 + (0 + j2)i 2 + (1 -j5)i 3 = 0 (Eq. 3) Equations 1, 2, and 3 are simultaneous equations in 3 unknowns (i 1 - i 3 ), and can be solved using an advanced scientific calculator or computer program capable of doing matrix math. As matrices, these equations can be arranged in the general form Ai = V (Eq. 4), where A = (5 + j 9) ( 3 j 5) 0 (3 + j 5) ( 3 + j 3) (0 j 2) 0 (0 + j 2) (1 j 5) , i = i 1 i 2 i 3 , and V = 25 0 0 0

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This note was uploaded on 09/25/2010 for the course EE 201 taught by Professor Linch during the Spring '10 term at University of Saskatchewan- Management Area.

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Mesh_Nodal_Analysis - EE201.3 Consider the circuit shown...

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