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Note for Network Analysis - NA By Manjunatha Parameshwarappa

  • Network Analysis - NA
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0.1. SOURCE TRANSFORMATION TECHNIQUE The other method to find current in the above circuit is 3A 4Ω i1 = Currrent × Resistance in other baranch T otal Resistance 8 3.75 × = 1.142A 3 3.75 + 5 i1 = Q 2) In the circuit shown in Figure ?? determine current I by source transformation. 5Ω 5V +- Replace the current source and parallel resistance by voltage source in series with resistor V = I × R = 3 × 4 = 12V 2Ω 12V +- 3Ω b Figure 11 Now 4 Ω and 2 Ω are in series which are replaced by single resistance 6 Ω Figure 7 Solution: 6Ω First the parallel resistance is replaced by single resistor which is as shown in Figure 5Ω 5V 8Ω V0 +- 12V Figure 12 Replace the voltage sources by current source in parallel with resistance 6 Ω and 3 Ω a I 3Ω 12V +- 20 × 30 600 = = 12Ω 20 + 30 50 +- +- 12V V0 8Ω 30Ω 20Ω R= + 12V - Solution: 4Ω 0.1A V0 8Ω Figure 10 a I 3Ω 2Ω 0.1A 12Ω 2A 6Ω 8Ω V0 3Ω 4A b Figure 13 Figure 8 Replace two current sources by single current source Now replace the current source in parallel with resistor and two parallel resistors 6 Ω and 3 Ω with single by voltage source in series with resistor, which is as resistance. Current sources are in opposite directions. Again replace current source by voltage source in series shown in Figure with resistor 2 Ω 12Ω 5Ω 5V +- I +- 1.2V Figure 9 The current I in the circuit is I= 5 − 1.2 3.8 = = 0.224A 5 + 12 17 Q 3) In the circuit shown in Figure ?? determine the V0 using source transformation. R= 6×3 18 = = 2Ω 6×3 9 The current in the circuit is I= 4 4 = = 0.4A 8+2 10 The voltage drop across 8 Ω is V = 0.4A × 8 = 3.2V Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 2

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0.1. SOURCE TRANSFORMATION TECHNIQUE 3Ω 2Ω 1A 8Ω V0 2Ω + 4V - 2A V0 8Ω 5i1 5Ω - + i1 5Ω Figure 14 Q 4) In the circuit shown in Figure ?? determine the current I1 6/5A 3A 1Ω 2Ω - + i1 5i1 2Ω Figure 19 5Ω 3Ω Figure 15 Solution: Current source 1 A and 65 A are in opposite directions, replace by single current source and also replace parallel resistors 3 Ω and 5 Ω by a single resistor Ieq = For the given circuit there is a current source of 3A. Shift the current source between resistors 1 Ω 2 Ω. The modified circuit is as shown in Figure 16 1Ω 3A 2Ω - + 2Ω 5Ω i1 5i1 Req = 6 1 − 1 = = 0.2A 5 5 3×5 15 = = 1.875Ω 3+5 8 5i1 - + 3Ω 5Ω 1.875Ω 3A Figure 16 0.2A Convert current sources into voltage sources in series with resistor 1 Ω and 2 Ω. Figure 20 Replace the 0.2A current source and parallel resistor 1.875 Ω by voltage source in series with 1.875 Ω resistor 1Ω 2Ω 3V +- - + 6V +- 5Ω i1 5i1 3Ω 5i1 - + 5Ω i1 2Ω - + Figure 17 3V 1.875Ω 0.375 V +- 3Ω Figure 21 Apply KVL for the loop - + 5Ω 5i1 - + 6V 5Ω Figure 18 i1 0.375 − 6.875i1 − 5i1 = 0 0.375 − 11.875i1 = 0 11.875 i1 = = 31.67A 0.375 Now convert voltage sources into current sources in Q 5) In the circuit shown in Figure ?? determine the parallel with resistors as shown in Figure current I Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 3

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0.1. SOURCE TRANSFORMATION TECHNIQUE 2A 16 + 10 − 7 × I = 0 3 16 + 30 − 7I = 0 3 46 7I = 3 46 I = = 2.19A 3×7 3Ω 8Ω I +10V 4Ω 6Ω Figure 22 Solution: KVL cannot be applied directly due to the presence of current source. Replace the current source into voltage Q 6) In the circuit shown in Figure ?? determine the voltage v0 across 100 Ω resistor source in series with 8 Ω and current source as 100Ω 8V v0 100Ω - + V = 2 × 8 = 16V olts Now the 8 Ω and 4 Ω are in series which are placed in series with voltage source. Solution: 4Ω 6Ω Replace the current source by voltage source in series with 100 Ω resistor which is as shown in Figure. 100Ω 8V v0 100Ω + Figure 23 Replace the voltage source into current source in parallel with 12 Ω 10 V +- +- 3 V Figure 27 3Ω +10V 100Ω - - + 8Ω I 30 mA 100Ω Figure 26 16 V 3Ω +10V 10 V +- I 6Ω 12Ω 4/3 A Voltage sources of 8 and 3 are in series which are replaced by single voltage source. 100Ω 100Ω Figure 24 12 Ω and 6 Ω are in parallel. Current source can be replaced by voltage source the details are as shown in Figure 6 × 12 R= = 4Ω 6 + 12 10 V 100Ω +- 11 V Figure 28 100 mA 4 16 V = I × R = × 4 = V olts 3 3 v0 +- v0 100Ω 100Ω 100Ω 110 mA Figure 29 3Ω +10V I 4Ω +- 16/3 V Replace the voltage source of 11 volts with current source in parallel with 100 Ω resistor. I= Figure 25 Current I is 16 + 10 − 7 × I = 0 3 11 = 110mA 100 100 mA current source with 100 Ω resistor and 110 mA current source with 100 Ω resistor are in parallel which are replaced by single current source and single resistor as Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 4

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0.1. SOURCE TRANSFORMATION TECHNIQUE I = 100 + 110 = 210mA 100 × 110 R= = 50Ω 100 + 110 Finally replace 210 mA in parallel 50 Ω resistor by voltage source of 10.5 V in series with 50 Ω R = 2 + 4 = 6Ω 24Ω 12Ω 240 V +- V = I × R = 210 × 10−3 × 50 = 10.5V 50Ω 210 mA 50Ω v0 100Ω v0 +10.5 V 100Ω + 180 V - 48 A 6Ω Figure 33 Replace 240 V voltage source in series resistor 24 Ω by a current source and 180 V voltage source in series resistor 6 Ω by a current source I= V 240 = = 10A R 24 I= 180 V = = 30A R 6 Figure 30 The current flowing in the circuit is I= V 10.5 = = 0.07A R 50 + 100 Voltage across the 100 Ω resistor is 24Ω 12Ω 10 A 6Ω 48 A 30 A V = I × R = 0.07 × 100 = 7V +- Q 7) In the circuit shown in Figure 31 determine Figure 34 the current in the 12 Ω resistor using source Current sources 10 A, 48A, and 30 A are in parallel. transformation method Replace these by single current source. Also replace 60 V 2Ω parallel resistor by a single resistor. 24Ω 12Ω 48 A 4Ω I = 10 + 48 − 30A = 28A 30 A 240 V +- Figure 31 R= Solution: Replace the current source and parallel resistor 4 Ω by voltage source in series with resistor 4 Ω 24 × 6 144 = = 4.8Ω 24 + 6 30 4.8Ω 12Ω 4.8Ω 28 A 12Ω +- 134.4 V V = I × R = 30 × 4 = 120V 2Ω Figure 35 +- 60 V 12Ω 24Ω 48 A 240 V +- Figure 32 +- 120 V 4Ω The current in 12 Ω resistor is sources 10 A, 48A, and 30 A are in parallel. Replace by single current source. Replace parallel resistors by a single resistor. I= 134.4 = 8A 12 + 4.8 Replace the voltage sources and source resistors by single voltage source in series with single resistor 4 Ω Q 8) In the circuit shown in Figure 39 determine the current in the 3 Ω resistor using source transformation V = 60 + 120 = 180V method Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 5

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