Note for Network Analysis - NA By Manjunatha Parameshwarappa
- Network Analysis - NA
- Note
- Javaharlalnehru national college of engineering - JNNCE
- Electronics and Communication Engineering
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0.1. SOURCE TRANSFORMATION TECHNIQUE 0.1 Source Transformation Technique R a a Vs +- Is R b b Voltage Source Current Source Figure 1 Q 1) In the circuit shown in Figure ?? determine the current i1 through 5 Ω resistor by source transformation. second source is and each resistors are in parallel with 2 respective current sources I = VR = 10 15 = 3 A c 2A b +10V 5Ω i1 a c e 10Ω 5Ω 5Ω d i1 5Ω 1A 5Ω 15Ω 2 A 3 d Figure 5 Now two current sources are in parallel they can be added and total current is f Figure 2 2 8 I =2+ = A Solution: KVL cannot be applied due to the 3 3 presence of current source. Transform the current The parallel resistances are added source to voltage source. 5 × 15 15 R= = Ω 5 + 15 4 V = I × R = 1 × 10 = 10V The equivalent circuit is as shown in Figure. The current source is replaced by voltage source which is 5Ω 10V 10Ω as shown in Figure. The new voltage source is b c +- e 10V + - 5Ω 5Ω a 8 15 × = 10V 3 4 The total resistance in the circuit is 15 35 R= +5= = 3.75 + 5 = 8.75Ω 4 4 Current i1 through 5 Ω resistor is V =I ×R= i1 d f Figure 3 Now the 10 Ω and 5 Ω are in series 10V +a 5Ω c 10V +- b I= e V 10 = = 1.142A R 35/4 i1 15Ω 5Ω d f Figure 4 Replace the voltage source by current source. First current source is I = VR = 10 = 2A and the 5 c 8 A 3 15 Ω 4 c i1 10V 5Ω d +- 15 Ω 4 i1 5Ω d Figure 6 Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 1
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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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