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Note for Network Analysis - NA By Manjunatha Parameshwarappa

  • Network Analysis - NA
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Manjunatha Parameshwarappa
Manjunatha Parameshwarappa
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0.1. NODE ANALYSIS 0.1 Node Analysis Q 2) In the circuit shown in Figure 3 determine all branch currents and the voltage across the 5 Ω resistor Steps to find a current flowing in a circuit using Node by node analysis. Analysis 3Ω 1. Identify nodes in a circuit and Label all the nodes. 2Ω +50V 2. Select one of node as reference node. +25V 5Ω 6Ω 3. Apply KCL to each node. 8Ω Figure 3 4. Solve the resulting simultaneous linear equations for the unknown node voltages using Cramer’s Solution: The circuit is labeled by nodes which is as shown in Rule. Figure 4 5. Branch currents can be calculated using node 3Ω V1 2Ω voltages +50V Q 1) In the circuit shown in Figure 1 determine all branch currents by node analysis. 6Ω +42V 4Ω +42V 4Ω 8Ω +25V 2 (Ref) 6Ω Figure 4 +- 8Ω 5Ω 10V Apply KCL to node V1   V1 − 50 V1 − 25 1 Figure 1 + + = 0 3+6 2+8 5 Solution: The circuit is labeled by nodes which is [0.111 + 0.1 + 0.2] V1 − 5.556 − 2.5 = 0 as shown in Figure 2 [0.4111] V1 = 8.056 8Ω V1 6Ω V1 = 19.59 +10V Current through 5 Ω resistor is 2 (Ref) I5 = Figure 2 Apply KCL to node V1   V1 − 42 1 V1 + 10 + + = 0 8 4 6 [0.125 + 0.25 + 0.167] V1 − 5.25 + 1.67 = 0 [0.5416] V1 = 3.58 V1 = 6.61 8.056 = 3.918A 5 Voltage across 5 Ω resistor is V = I5 × 5 = 3.918 × 5 = 19.59V Q 3) In the circuit shown in Figure 6 determine all branch currents and the voltage across the 5 Ω resistor by node analysis. 1Ω 2Ω Current through 8 Ω resistor is I8 = 10 V +- 42 − 6.61 = 4.42A 8 10 + 6.61 = 2.77A 6 Current through 4 Ω resistor is I4 = 2A Figure 5 Current through 6 Ω resistor is I6 = 10Ω 5Ω Solution: 1Ω V1 10 V +- 6.61 = 1.65A 4 Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 2Ω 5Ω V2 10Ω 2A 3 (Ref) 1

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0.1. NODE ANALYSIS Figure 6 Apply KCL at node V1   V1 − 10 V1 V1 − V2 = 0 + + 1 5 2 V1 [1 + 0.2 + 0.5] − 10 − 0.5V2 = 0 V1 [1 + 0.2 + 0.5] − 0.5V2 = 10 1.7V1 − 0.5V2 = 10 Apply KCL at node V2   V2 V2 − V1 + −2 = 0 10 2 −0.5V1 + V2 [0.1 + 0.5] = 2 Apply KCL at node V1   V1 − V2 V1 − V3 +3+8 = 0 + 3 4 V1 [0.25 + 0.333] − 0.25V2 − 0.0.25V3 + 11 = 0 0.583V1 − 0.333V2 − 0.25V3 = −11 Apply KCL at node V2   V2 − V1 V2 − V3 V2 = 0 + + 3 2 1 −0.333V1 + V2 [0.5 + 0.333 + 1] − 0.5V3 + 11 = 0 −0.333V1 + 1.833V2 − 0.5V3 = 0 Apply KCL at node V3   V3 − V1 V3 − V2 V3 + (−25) = 00 + + 4 2 5 −0.25V1 − 0.5V2 + V3 [0.25 + 0.5 + 0.2] = 25 −0.5V1 + 0.6V2 = 2 The simultaneous equations are −0.25V1 − 0.5V2 + V2 + 0.95V3 = 25 1.7V1 − 0.5V2 = 10 The simultaneous equations are −0.5V1 + 0.6V2 = 2 1.7 −0.5 = 1.02 − .25 = 0.25 = 0.77 ∆= −0.5 0.6 10 −0.5 2 0.6 6+1 V1 = = = 9.09 ∆ 0.77 1.7 10 −0.5 2 3.4 + 5 V2 = = = 10.91 ∆ 0.77 V1 is the voltage across 5 Ω resistor which is 9.09 V 0.583V1 − 0.333V2 − 0.25V3 = −11 −0.333V1 + 1.833V2 − 0.5V3 = 0 −0.25V1 − 0.5V2 + V2 + 0.95V3 = 0.583 −0.333 −0.25 −0.5 ∆ = −0.33 1.833 −0.25 −0.5 0.95 = 0.583[1.741 − .25] + 0.333[−0.313 − .125] −0.25[.165 + .458] = 0.583[1.491] + 0.333[−0.438] − 0.25[0.623] = 0.87 − 0.1458 − 0.155 = 0.5692 −11 −0.333 −0.25 0 1.833 −0.5 25 −0.5 0.95 V1 = ∆ Q 4) In the circuit shown in Figure 8 determine number of nodes and its voltages. 4Ω -3A −11[1.74 − 0.25] + 0.333[12.5] − 0.25[45.82] 0.5692 −16.39 + 4.1625 − 11.45 −23.677 = = = −41.43 0.5692 0.5692 Q 5) In the circuit shown in Figure 9 determine the current Ix 2Ω = 3Ω -8A -25A 5Ω 1Ω Figure 7 4Ω 8A -3A V1 -8A 25 V2 2Ω V3 b 3Ω 1Ω 5Ω 8Ω c I x 2Ω e +100V Solution: Figure 8 10Ω -25A a Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 3Ω 4Ω d g f 5Ω h 2

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0.1. NODE ANALYSIS Figure 9 Figure 11 Solution: Solution: V1 For the given circuit there is current source, to apply KCL current source has to be converted into voltage source, the modified circuit is as shown in Figure 10 +- +100V 3Ω 5Ω 3i1 2Ω 3Ω V2 15 A 4Ω i1 1Ω 8Ω V1 I x 2Ω V2 10Ω 80V + Vx - Figure 10 3 (Ref)   V1 − 100 V1 V1 − V2 = 0 + + 8 4 2 [0.125 + 0.25 + 0.5] V1 − 12.5 − 0.5V2 = 0 0.875V1 − 0.5V2 = 12.5  1V2 V2 − 80 V2 − V1 + + = 0 3 15 2 −0.5V1 + [0.33 + 0.067 + 0.5] V2 − 5.33 = 0  Figure 12 Applying KCL to mesh V1 V1 − V2 V1 + − 15 = 0 1 2 (1 + 0.5)V1 − V2 = 15 1.5V1 − V2 = 15 −0.5V1 + 0.897V2 = 5.33 The simultaneous equations are i1 = 0.875V1 − 0.5V2 = 12.5 −0.5V1 + 0.897V2 = 5.33 0.875 −0.5 = 0.7848 − .25 = 0.5348 ∆= −0.5 0.897 12.5 −0.5 5.33 0.897 11.212 + 2.665 = V1 = ∆ 0.5348 13.877 = = 25.94V 0.5348 0.875 12.5 −0.5 5.33 4.66 + 6.25 10.91 V2 = = = = 20.4V ∆ 0.5348 0.5348 V1 − V2 25.94 − 20.4 = = 2.77A 2 2 Q 6) Find the loop currents i1 , i2 , i3 in the circuit shown in Figure 11 Ix = i1 1Ω 3Ω 15 A 3i1 + Vx - 2Ω V1 2 Applying KCL to mesh V2 V2 − V1 V2 + − 3i1 = 0 1 3 −V1 + [1 + 0.333]V2 − 3i1 = 0 −V1 + 1.333V2 − 3i1 = 0 −V1 + 1.333V2 − 3(0.5V1 ) = 0 −2.5V1 + 1.333V2 = 0 1.5V1 − V2 = 15 −2.5V1 + 1.333V2 = 0 1.5 −1 = 2 − 2.5 = −0.5 ∆ = −2.5 1.333 15 1 0 1.333 20 = = −40V V1 = ∆ −0.5 1.5 15 −2.5 0 37.5 V2 = = = −75V ∆ −0.5 Power delivered by dependent current source is i1 = V1 −40 = = −20 2 2 Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 3

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0.1. NODE ANALYSIS 2.786 − 50.6 + 1.6626 − 56.3 0.2826 − 62.79 1.764 − j2.148 + 0.922 − j1.38 = 0.2826 − 62.79 4.436 − 52.7 2.686 − j3.528 = = 0.2826 − 62.79 0.2826 − 62.79 = 15.876 10.01 V2 × 3i1 = −75 × 3 × −20 = 4.5kW = Q 7) In the network shown in Figure ?? find the current I by node voltage method. 5Ω 30∠0o V A 2+j3 Ω B 4Ω j5 Ω 6Ω 20∠0o V I 15.876 10.01 VB = = 2.646 10.01A 6 6 Q 8) In the network shown in Figure 14 find the value of E2 such that current through the 8+j8Ω is zero. I= Figure 13 Solution: Applying KCL to node V1 10 Ω VA − 306 0 VA − VB VA + + 5 2 + j3 j5     1 30 1 1 1 − VB − + + VA 5 j5 2 + j3 2 + j3 5   6 VB 1 30 0 VA − .2 − j0.2 + − 3.66 56.3 3.66 56.3 5 [.2 − j.2 + .2776 − 56.3] VA − .2776 − 56.3VB B 15 Ω = 0 j10 Ω 50∠0o V 10 Ω = 0 E2 = 0 = 6 [.2 − j.2 + .153 − j.23] VA − .2776 − 56.3VB = 6 [.353 − j.43] VA − .2776 − 56.3VB = 6 [.5566 − 50.6] VA − .2776 − 56.3VB = 6 Figure 14 Solution: It is given that current through the 8+j8Ω is zero this is possible only when VA = VB i., VA − VB = 0 Applying KCL to node VA Applying KCL to node V2 VB − 206 0 VB − VA VB + + 4 2 + j3 6     1 1 206 1 1 + + − VA − VB 6 4 2 + j3 2 + j3 4   1 VA 206 0 .16 + .25 + VA − − 3.66 56.3 3.66 56.3 4 6 6 [.16 + .25 + .277 − 56.3] VB − .277 − 56.3VA VA − 50 VA + = 0 10 j10   1 1 50 + VA − = 0 10 j10 10 [0.1 − j0.1] VA = 5 = 0 = 0 0.1416 − 45VA = 5 = 0 VA = 35.466 45 = 5 −.2776 − 56.3VA + [.16 + .25 + .153 − j.23] VB = 6 −0.2776 − 56.3VA + [0.563 − j0.23] VB = 5 Applying KCL to node VB −0.2776 − 56.3VA + 0.66 − 22.3VB = 5 0.5566 − 50.6VA − 0.2776 − 56.3V2 = 6 −0.2776 − 56.3VA + 0.66 − 22.3VB = 5 0.5566 − 50.6 −0.2776 − 56.3 ∆ = −0.2776 − 56.3 0.66 − 22.3 8+j8 Ω A VB − E2 VB + 15 10  1 1 E2 + VB − 10 15 15 E2 [0.1 + 0.066] VB − 15 E2 15 E2 = 0 = 0 = 0 = 0.166VB = 2.5VB = 0.3366 − 72.9 − 0.0766 − 112.6 0.1−j0.321+0.029+j0.07 = 0.129−j0.251 = 0.2826 −62.79 0.5566 − 50.6 66 0 −0.2776 − 56.3 56 0 VB = = ∆ VA = VB E2 = 2.5 × 35.466 45 E2 = 88.656 45 Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 4

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