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# Note for Network Analysis - NA By Manjunatha Parameshwarappa

• Network Analysis - NA
• Note
• Javaharlalnehru national college of engineering - JNNCE
• Electronics and Communication Engineering
• 1853 Views
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#### Note for Network Analysis - NA By Manjunatha Parameshwarappa

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0.1. MESH ANALYSIS 0.1 Mesh Analysis Hence loop2 equation is. Steps to find a current flowing in a circuit using Mesh Analysis 2i2 + 8i2 + 5(i2 − i1 ) + 25 = 0 −5i1 + 15i2 = −25 1. Identify loops or meshes in a circuit and Label a The two simultaneous equations are: mesh current to N meshes 14i1 − 5i2 = 50 (1) −5i1 + 15i2 = −25 2. Apply KVL to each mesh with the corresponding mesh current to generate N equations. Multiply eqn 1 by 3 and adding with equation 2 42i1 − 15i2 = 150 3. Solve the resulting simultaneous linear equations for the unknown mesh currents using Cramer’s Rule. Q 1) In the circuit shown in Figure 1 determine all branch currents and the voltage across the 5 Ω resistor by loop current analysis. 3Ω 2Ω +50V +25V 5Ω 6Ω 8Ω (2) −5i1 + 15i2 = −25 −−−−− = −− 37i1 = 125 i1 = 3.3784A i2 = −0.541A iab = 3.3784A ieb = −i2 = 0.541A ibc = i1 − i2 = 3.3784 − (−0.541) = 3.9194A voltage across the 5 Ω resistor is 5ibc = 19.597V Q 2) In the circuit shown in Figure 3 determine the mesh currents i1 , i2 , i3 b f c Figure 1 3Ω b a +50V d i1 2Ω 7V +- i1 d 3Ω 6V +- 2Ω a e i2 5Ω i2 1Ω Solution: In the given circuit there are two meshes and named as i1 and i2 as shown in Figure 2. e +25V i3 2Ω g 1Ω h Figure 3 Solution: 6Ω c 8Ω f Figure 2 Applying the KVL for the loop abcda. The loop is passing through 3, 5 and 6 Ω resistors and it touches negative terminal of the battery of 50 volts. The 5 Ω resistor is common to loop currents i1 and i2 . For loop1 the current i2 in 5 Ω resistor is opposite to the i1 current. Hence loop1 equation is. Applying the KVL for the loop abcdea 1(i1 − i2 ) + 2(i1 − i3 ) + 6 − 7 = 0 3i1 − i2 − 2i3 = 1 For the loop cfgdc 2i2 + 3(i2 − i3 ) + 1(i2 − i1 ) = 0 −i1 + 6i2 − 3i3 = 0 For the loop dghed 3i1 + 5(i1 − i2 ) + 6i1 − 50 = 0 14i1 − 5i2 = 50 3(i3 − i2 ) + 2(i3 − i1 ) + i3 − 6 = 0 −2i1 − 3i2 + 6i3 = 6 Similarly for the loop befcb , The loop2 is passing The three mesh equations are, through 2, 8 and 5 Ω resistors and it touches positive 3i1 − i2 − 2i3 = 1 terminal of the battery of 25 volts. The 5 Ω resistor −i1 + 6i2 − 3i3 = 0 is common to loop currents i1 and i2 . For loop2 the current i1 in 5 Ω resistor is opposite to the i2 current. −2i1 − 3i2 + 6i3 = 6 Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 1

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0.1. MESH ANALYSIS Cramer’s   i1 i2  =  i3 rule  1 0  6 ZI = V   i1 I =  i2  i3   v1 V =  v2  v3   Z11 Z12 Z13 Z =  Z21 Z22 Z23  Z31 Z32 Z33   v1 −1 −2  v2 6 −3  v3 −3 6 i1 = ∆ where ∆ is 3 −1 −2 ∆ = −1 6 −3 −2 −3 6 3[6 × 6 − (−3 × −3)] + 1[−1 × 6 − (−2 × −3)] −2[−1 × −3 − (−2 × 6)] =3(36-9)+1(-6-6)-2(3+12)=81-12-30=39 1 −1 −2 0 6 −3 6 −3 6 1(36 − 9) + 1(18) − 2(−36) i1 = = ∆ 39 27 + 18 + 72 = 3A 39 3 1 −2 −1 0 −3 −2 6 6 3(18) − 1(−6 − 6) − 2(−6) i2 = = ∆ 39 54 + 12 + 12 = 2A 39 3 −1 −1 6 −2 −3 i3 = ∆ 1 0 6 108 − 6 + 15 = 3A 39 i1 = 3A i2 = 2A i3 = 3A Q 3) In the circuit shown in Figure 4 determine all branch currents by mesh current analysis. 8Ω b a +42V i1 6Ω e i2 4Ω c d +- Solving these equations Using   3 −1 −2  −1 6 −3   −2 −3 6 10V f Figure 4 Solution: Applying the KVL for the loop abcda 8i1 + 4(i1 − i2 ) − 42 = 0 12i1 − 4i2 = 42 Similarly for the loop befcb 6i2 + 4(i2 − i1 ) − 10 = 0 −4i1 + 10i2 = 10 Simultaneous equations are 12i1 − 4i2 = 42 −4i1 + 10i2 = 10 12 −4 = 12×10−(−4×−4) = 120−16 = 104 ∆ = −4 10 42 −4 10 10 420 + 40 460 = = = 4.42A i1 = ∆ 104 104 12 42 −4 10 120 + 168 288 i2 = = = = 2.769A ∆ 104 104 Q 4) In the circuit shown in Figure 5 determine the current Ix 8A b 8Ω c I x 2Ω e +100V a 10Ω 3Ω 4Ω d g f 5Ω h Figure 5 Solution: For the given circuit there is current 3(36) + 1(−6) + 1(3 + 12) = 39 source, to apply KVL current source has to be Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 2

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0.1. MESH ANALYSIS converted into voltage source, the modified circuit is as shown in Figure 6 8Ω c I x 2Ω e +100V a i1 10Ω 80V 4Ω i2 d 3Ω +- b g i3 5Ω f = h Figure 6 12(0 − 240) − 100(−72 − 0) + 0(0 + 720) 1665 2880 + 7200 + 0 10080 = = 6.05A 1665 1665 12 −4 100 −4 9 0 0 −3 80 i3 = ∆ Applying the KVL for the loop abcda 8i1 + 4(i1 − i2 ) − 100 = 0 12i1 − 4i2 = 100 For the loop cefdc 2i2 + 3(i2 − i3 ) + 4(i2 − i1 ) = 0 −4i1 + 9i2 − 3i3 = 0 12(720 − 0) + 4(−320 − 0) + 100(12 + 0) 1665 8640 − 1280 + 1200 8560 = = 5.14A 1665 1665 i1 = 3A i2 = 2A i3 = 3A Q 5) Find the loop currents i1 , i2 , i3 in the circuit shown in Figure 7 = For the loop eghfe 10i3 + 5i3 + 3(i3 − i2 ) + 80 = 0 0i1 − 3i2 + 18i3 = −80 10 A + Vx i3 12i1 − 4i2 + 0i3 = 100 2Ω 6Ω −4i1 + 9i2 − 3i3 = 0 0i1 − 3i2 + 18i3 = −80 12 −4 0 ∆ = −4 9 −3 0 −3 18 4Ω 3Ω i1 Vx 8 The three mesh equations are, 12[162 + 9] + 4(−72 − 0)] + 0[12 − 0] = 2052 − 288 = 1764 100 −4 0 0 9 −3 −80 −3 18 i1 = ∆ i2 5Ω Figure 7 Solution: Vx = 3(i3 − i2 ) i1 = 10A Vx i3 − i1 = = 0.125Vx 8 i3 − 10 = 0.125 × 3(i3 − i2 ) i3 − 10 = 0.375i3 − 0.375i2 0.375i2 + 0.625i3 = 10 Applying KVL to mesh i2 = 100(162 − 9) + 4(0 − 240) + 0(0 + 720) 1665 15300 − 960 + 0 14340 = = 8.61A 1665 1665 12 100 0 −4 0 −3 0 −80 18 i2 = ∆ 4i2 + 3(i2 − i3 ) + 5(i2 − i1 ) = 0 −5i1 + 12i2 − 3i3 = 0 −5 × 10 + 9i2 − 4i3 = 0 12i2 − 3i3 = 50 The simultaneous equations are 0.375i2 + 0.625i3 = 10 12i2 − 3i3 = 50 Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 3

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0.1. MESH ANALYSIS 0.375 0.625 = −1.125 − 7.5 = −8.6255 ∆ = 12 −3 10 0.625 50 −3 −30 − 31.25 −61.25 i2 = = = = 7.1A ∆ −8.6255 −8.6255 0.375 10 12 50 i3 = ∆ = 18.75 − 120 −101.25 = = 11.74A −8.6255 −8.6255 The three loop currents i1 = 10A, i2 = 7.1A, i3 = 11.74A Dr. Manjunatha P Professor Dept of ECE, JNN College of Engineering, Shivamogga 4