1. @ McGraw-Hill Education 1 Kulshreshtha, D. C. Basic Electrical Engineering McGraw-Hill Education © 2010 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it without permission. BASIC ELECTRICAL ENGINEERING PowerPoint Slides D. C. KULSHRESHTHA, Next

2. @ McGraw-Hill Education 2 Kulshreshtha, D. C. Basic Electrical Engineering McGraw-Hill Education © 2010 Chapter 3 Network Analysis–Part I  D D.C. Kulshreshtha Next

3. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 3 Thought of The DAY Whatever THE MIND OF MAN can CONCIEVE and BELIEVE, it can ACHIEVE. Next

4. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 4 Topics to be Discussed Electric Circuit, The Resistance Parameter, The Capacitance Parameter, The Inductance Parameter, Energy Sources-- Classification I Ideal Voltage Source. deal Current Source. Series and Parallel Combinations. Practical Voltage Source. Practical Current Source. Source Transformation. Kirchhoff’s Laws. Next

5. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 5 Electric Circuit It is a closed path, composed of active and passive elements. Active Element : It supplies energy to the circuit. Passive Element : It receives energy and then 1) either converts it to heat, as in a Resistance (R). 2) or stores it in (a) Electric Field, as in a Capacitor (C). (b) Magnetic Field, as in an Inductor (L). Next

6. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 6 Energy Sources Classification Independent Source Or Dependent Source Voltage Source Or Current Source DC Source Or AC Source Ideal Source Or Practical Source Next

7. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 7 Independent Ideal Voltage Source The source has zero internal resistance. Note that the source determines the voltage, but the current is determined by the load. Next Load Source

8. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 8 The voltage source is said to be idle if the output terminals are open such that i = 0. When turned off (killed or made inactive), so that v = 0, it is equivalent to a short circuit. Reference Marks : One terminal is marked plus and the other minus. (Oversimplification; one mark can be omitted.) When actual polarity is opposite to the reference marks, the voltage is a negative number. Next

9. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 9 Independent Ideal Current Source The Source has infinite internal resistance (R i ). Next Source Load Note that the source determines the current, but the voltage is determined by the load.

10. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 10 The current source is said to be idle if the output terminals are shorted together, such that v = 0. When turned off (killed or made inactive), so that i = 0, it is equivalent to an open circuit. Reference Marks : An arrow is put. When actual direction of current is opposite to the reference (arrow) direction, the current is a negative number. Next

11. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 11 Do you observe duality ? The roles for the current and voltage are interchanged in the two sources. Next

12. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 12 Practical Voltage Source It is represented by an ideal voltage source in series with an internal resistance (R SV ). Next

13. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 13 Practical Current Source It is modelled as an ideal current source in parallel with an internal resistance (R SI ). Next

14. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 14 Source Transformation A practical current source can be converted into its equivalent practical voltage source, and vice versa. This conversion is valid only for the external load connected across the terminals of the source. Next

15. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 15 Next

16. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 16 Equivalence between Voltage Source and Current Source Two sources would be equivalent if they produce identical values of V L and I L, when they are connected across the same load. Next

17. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 17 Series and Parallel Combinations What would be the net emf of the combination if two ideal voltage sources of 2 V and 4 V are connected in series so as to aid each other? Ans. 6 V What would be the net emf of the combination if two ideal voltage sources of 4 V and 4 V are connected in parallel ? 4 V or 8 V ? Ans. Obviously, it should be 4 V Next Click

18. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 18 What would be the net emf of the combination if two ideal voltage sources of 2 V and 4 V are connected in parallel ? 2 V or 4 V or 3 V ? The question seems to be quite tricky! Ans. The question is wrong. The question contradicts itself. Ideal Voltage Sources in parallel are permissible only when each has the same terminal voltage at every instant of time. What is its dual ? Next Click

19. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 19 Ideal Voltage Sources Connected in Series Next

20. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 20 Ideal Current Sources Connected in Parallel Next

21. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 21 Practical Current Sources Connected in Series Next

22. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 22 Practical Voltage Sources Connected in Parallel Next

23. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 23 Example 1 : R educe the network shown in figure to its simplest possible form by using source transformation. Next

24. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 24 Solution Next

25. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 25 Next

26. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 26 Example 2 In the given figure, (a) If R L = 80 Ω, find current i L. (b) Transform the practical current source into a practical voltage source and find i L if R L = 80 Ω again. (c) Find the power drawn from the ideal source in each case. Next

27. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 27 Solution : Next Click

28. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 28 Benchmark Example 3 Take the benchmark example of the circuit given in figure. Using source transformation, determine the voltage v across 3-Ω resistor. Next

29. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 29 Solution : Transforming the 4-A current source into a voltage source, Next

30. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 30 Combining the two voltage sources, Next

31. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 31 Again transforming the voltage source into current source, Next

32. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 32 Combining the two current sources we get Fig. (e). Transforming this current source into voltage source (Fig. f ) Next

33. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 33 Combining the two resistances, we get Fig. (g). Finally, using voltage divider, we get Next

34. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 34 Kirchhoff’s Laws ( 1) KCL : Algebraic sum of currents meeting at a junction of conductors in a circuit is zero. IIt is simply a restatement of the principle of conservation of charge. Next

35. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 35 (2) KVL : The algebraic sum of voltages around a closed circuit or a loop is zero. IIt is simply a restatement of the principle of conservation of energy. Next

36. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 36 Polarity of Voltages Note Note that polarity of the voltage (emf) across a battery does not depend upon the assumed direction of current. Next

37. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 37 Applying KVL 1. Select a closed loop. 2. Mark the voltage polarity (+ and -) across each element in the closed loop. 3. Go round the selected loop, and add up all the voltages with + or – signs. 4. Any one of the following two rules can be followed : Next

38. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 38 (i) Rule 1 : While travelling, if you meet a voltage rise, write the voltage with positive sign ; if you meet a voltage drop, write the voltage with negative sign. (ii) Rule 2 : While travelling, write the voltage with positive sign if + is encountered first; write the voltage with negative sign if – is encountered first. We shall be following Rule 1, as it has a strong analogy with the physical height (altitude) of a place. Next

39. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 39 Example 5 : Use KVL to find v R2 and v x. Next

40. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 40 For finding v R2, we write KVL eqn. going around loop abgha clockwise : If you choose to go around the loop anticlockwise, you get Giving the same result. Next Click

41. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 41 There are two ways to determine v x 1) We can consider this voltage as the voltage across the gap from d to f. Writing KVL (habcdfgh) : 2) Knowing v R2, apply a short-cut (bcdfgb) : Next Click

42. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 42 Important Note about KVL The assumed direction of current through a resistor and the polarity of voltage across it are always in conformity. The end into which the current enters is marked positive. Passive-element sign convention. Next

43. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 43 Example 6 : Find the current supplied by the 60-V source in the network. Next

44. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 44 Solution : We need not find the currents I 1, I 2 and I 3. Instead, we reduce the network. Next

45. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 45 Example 7 : D etermine the value of current I. 2 – 3 – I – 4 = 0 2 – 3 – I – 4 = 0 or I = -5 A Next

46. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 46 Example 8 Using KCL and KVL, determine the currents i x and i y in the network shown. Next

47. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 47 Solution : Using KCL, the currents in other branches are marked as shown. Writing KVL equations for the loops 1, 2 and 3, Next

48. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 48 Next Writing the above equations in matrix form, Click Using Calculator, we solve for I x and I y,

49. Wednesday, November 18, 2015 Ch. 3 Network Analysis- Part I 49 Review Electric Circuit, The Resistance Parameter, The Capacitance Parameter, The Inductance Parameter, Energy Sources-- Classification I Ideal Voltage Source. deal Current Source. Series and Parallel Combinations. Practical Voltage Source. Practical Current Source. Source Transformation. Dependent Sources. Kirchhoff’s Laws. Next