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1. DC CIRCUITS (Only Independent Sources) Module- 01

2. Contents- Kirchhoff’s Laws Ideal and Practical Voltage and Current Sources Source Transformation Mesh and Nodal Analysis (no super node and super mesh) Star-Delta / Delta-Star Transformations Superposition Theorem Thevenin’s Theorem Norton’s Theorem and Maximum Power Transfer Theorem.

3. INTRODUCTION- When electrical circuit is driven by DC source then it is called as DC Circuit. Essential part of the circuits are source of power (battery), conductor through which current flows and load (Lamp, Heater, Resistor and Motor). Source supplies electrical energy to load which convert into other form of energy. The direction of current is positive terminal of battery- bulb – Negative terminal of battery. If the load in DC circuit is resistance then it can be connected in following ways- Series circuit Parallel circuit Series-Parallel circuit

4. SERIES CIRCUIT- In an electric circuit, the different components are connected either in series or in parallel to produce different resistive networks. Resistors in Series- Two or more resistors are said to be connected in series when the same amount of current flows through all the resistors. In such circuits, the voltage across each resistor is different. For the above circuit, the total resistance is given as: R total  = R 1  + R 2  + ….. + R n The total resistance of the system is just the total sum of individual resistances.

5. SERIES CIRCUIT- Consider a circuit shown below. The resistances R 1 , R 2 and R 3 are connected in series across voltage source V, same current I will be flows through it. According to ohm’s law Voltage across R 1 is V R1 = IR 1 Voltage across R 2 is V R2 = IR 2 Voltage across R 3 is V R3 = IR 3 The total voltage (V) applied is sum of the voltages across R 1 , R 2 and R 3 . V = V R1  + V R2  + V R3 = IR 1 + IR 2 + IR 3 = I(R 1 + R 2 + R 3 ) = IR total R 1 + R 2 + R 3 = V/I

6. SERIES CIRCUIT- Voltage Division Rule- We know the current through circuit is- Voltage across R 1 is V R1 = IR 1 Similarly Voltage across R 2 is V R2 = IR 2 Voltage across R 3 is V R3 = IR 3 Note- The voltage drop across each resistor in series circuit is directly proportional to the ratio of its resistance to total series resistance of the circuit and applied voltage.

7. PARALLEL CIRCUIT- Consider a circuit in which three resistors R 1 , R 2 and R 3 are connected in parallel across supply voltage V. Due to voltage V current I flowing through circuit divided into I 1 through R 1 , I 2 through R 2 and I 3 through R 3 . Voltage across each resistor is same which is supply voltage V. For the above circuit, the total resistance is given as: By ohm’s law

8. PARALLEL CIRCUIT- From the given circuit the total current is sum of branch currents.

9. PARALLEL CIRCUIT- Current Division Rule- Consider a circuit in which two resistors R 1 and R 2 are connected in parallel across supply voltage V. Total current I is divided into two parts I 1 through R 1 , I 2 through R 2 . Total Resistance R total is Voltage V is given as

10. PARALLEL CIRCUIT- By ohm’s law In a parallel circuit of two resistors, the current through one resistor is the line current times the opposite resistance divided by sum of two resistances.

11. SERIES & PARALLEL CIRCUITS- Examples- Determine the current through & voltage across three resistances of ohmic values 5 Ω , 7 Ω and 8 Ω connected in series across 100 V source . Examples- Determine the current through & voltage across three resistances of ohmic values 5 Ω , 7 Ω and 8 Ω connected in parallel across 100 V source . Examples- Two resistors R 1 and R 2 are connected in parallel to a certain supply. The current taken from the supply is 5A. Calculate the value of R 1 if R 2 = 6 Ω and current through R 2 is 2 A. Also calculate total power absorbed by the circuit. Examples- Calculate effective resistance of the circuit and the current through 8 Ω resistance, when potential difference of 60 V is applied between point A & B. Examples- Calculate effective resistance between point A & B .

12. SHORT & OPEN CIRCUITS- Sort circuits- When two terminals of a circuit are connected by a wire, they are said to be short circuited. A short circuit has following features: It has zero resistance. Current through it is very large. There is no voltage across it. Open circuits- When two terminals of a circuit have no direct connection between them, they are said to be open circuited. An open circuit has the following features: It has infinite resistance. Current through it is zero. The entire voltage appears across it.

13. Open Circuits and Short Circuits in a Series Circuit - When an open circuit appears in a series circuit, the equivalent resistance becomes infinite and no current flows through the circuit. Current through circuit is Voltage across open circuit is When a short circuit appears in a series circuit, as shown, the resistance R2 becomes zero. Current through circuit is Voltage across open circuit is

14. Open Circuits and Short Circuits in a Parallel Circuit - When an open circuit appears in a parallel circuit, no current flows through that branch. The other branch currents are not affected by the open circuit. Current through circuit is Voltage across open circuit is When a short circuit appears in a parallel circuit, the equivalent resistance becomes zero. Current through circuit is Voltage across open circuit is

15. Open Circuits and Short Circuits- Examples- Find an equivalent resistance between terminal A & B. Examples- Determine the current delivered by the source in the circuit shown below- Examples- Determine the current delivered by the source in the circuit shown below-

16. KIRCHHOFF’S LAWS- The entire electric circuit analysis is based mainly on Kirchhoff’s laws. Some of the term related to Kirchhoff’s laws are as follows - Kirchhoff’s law are classified as- Kirchhoff’s Current law/Kirchhoff’s First Law (KCL) Kirchhoff’s Voltage law/Kirchhoff’s Second Law (KVL)

17. KIRCHHOFF’S LAWS- Kirchhoff’s Current law/Kirchhoff’s First Law (KCL)- It is related to current at a junction point. It states that- “ The algebraic sum of currents meeting at a point/junction/node is zero. ” Consider a conductor, carrying currents I 1 , I 2 , I 3 , I 4 and I 5 at point o. According to Kirchhoff’s law- Ʃ I = 0 Assuming incoming current positive and outgoing current negative. Above equation becomes- + I 1 + I 2 + I 3 +( - I 4 ) +( - I 5 ) = 0 I 1 +I 2 +I 3 -I 4 -I 5 = 0 It also state that, “ Sum of the currents flowing towards any junction in an electric circuits equal to the sum of the currents flowing away from that junction .”

18. KIRCHHOFF’S LAWS- Example- Find the currents in all branches Example- Find V XY as shown in figure

19. KIRCHHOFF’S LAWS- Kirchhoff’s Voltage law/Kirchhoff’s Second Law (KVL)- It is related to electromotive force and voltage drop in a circuit and stated that- “ The algebraic sum of all the voltages in a closed circuit/mesh/loop is zero. ” If we start from any point in a closed circuit and go back to that point, after going around the circuit, there is no increase or decrease in potential at that point. This means that sum of emf’s and sum of voltage drops or rise meet on the way is zero. According to Kirchhoff’s law- Ʃ V = 0 In a given diagram V is supply voltage and V 1 , V 2 and V 3 are voltage drop across R 1 , R 2 and R 3 . V = V 1 +V 2 +V 3 = IR 1 +IR 2 +IR 3

20. KIRCHHOFF’S LAWS- Sign Convention- While applying KVL to closed loop circuit, algebraic sum is considered. It is important to assign proper sign to emf and voltage drops in a closed loop circuits. Assumptions – Consider rise in voltage as positive and fall in voltage as negative. If we go from positive terminal of battery/source to the negative terminal there is fall in potential and so the emf should be assigned a negative sign, While if we move negative to positive terminal of battery then there is rise if potential so emf should be assigned positive sign.

21. KIRCHHOFF’S LAWS- When current flows through the resistor there is voltage drop across it. If we go from resistor in the same direction as current there is fall in potential and sign of voltage drop is negative. If we go to opposite to the direction of direction flow, there is rise in potential and voltage drop is trated as positive.

22. KIRCHHOFF’S LAWS- While Solving problems on Kirchhoff’s Law- Assume unknown current and show directions. Write KVL for all loops . In a solution if the value of any unknown current come out to be negative, it means that actual direction of current is opposite to that of assumed direction.

23. KIRCHHOFF’S LAWS- Example- By using Kirchhoff’s law calculate the branch current shown in figure, I 1 = 5 A( ) I 2 = 5A( ) Example- By using Kirchhoff’s law calculate the branch current shown in figure, Example- Find the current through 2 Ω resistor, I = 0.3 A( ) I 2 Ω = 1.66 A

24. KIRCHHOFF’S LAWS- Example- Determine the current through 20 Ω resistor shown in figure, Example- For the network shown in figure, Determine a) I 1 , I 2 and I 3 b) resistance R c) value of emf E. Example- Determine potential difference V AB for given network.

25. IDEAL & PRACTICAL VOLTAGE AND CURRENT SOURCE- There are two types of sources available- Voltage source Current source Ideal & Practical Voltage source- Ideal voltage source is the type of source whose output remains constant whatever changes in load current. It has zero internal resistance. In practice ideal voltage source is not possible because it has some internal resistance. Smaller is the value of internal resistance, more it will approaches to ideal voltage source.

26. IDEAL & PRACTICAL VOLTAGE AND CURRENT SOURCE- Ideal & Practical Current source- Ideal current source is the type of source whose output remains constant whatever changes in load resistance. It’s internal resistance is infinity. At any load resistance it supplies constant current. In practice ideal current source has very large internal resistance. Higher is the value of internal resistance, more it will approaches to ideal current source.

27. SOURCE TRANSFORMATION- As we know, we have two sources i.e. current source and voltage source. As per the requirement for circuit simplification of complex network one type of source can be converted to other type of source. This process is called as source transformation. There are different types of source transformation can be used which are- Voltage to current transformation Current to voltage transformation Voltage series source Current parallel source Voltage to Current (V-I) Transformation- A voltage source in series with resistance can be transfer/replace/converted to current source in parallel with resistance. Example-

28. SOURCE TRANSFORMATION- Current to Voltage (I-V) Transformation- A current source in parallel with resistance can be transfer/replace/converted to voltage source in series with resistance. Example- Series Voltage Source- Voltage sources are connected in series with additive polarity, the resultant voltage source will be the addition of voltage source. For subtraction property it will be subtracted. Example-

29. SOURCE TRANSFORMATION- Parallel Current Source- Current sources connected in parallel can be added or clubbed together as per direction of currents. Example-

30. SOURCE TRANSFORMATION- Example- Using source transformation find the current through 4 Ω resistor. I 4 Ω = 1.33 A (↓) Example- Using source transformation find the current through 3 Ω resistor. I 10 Ω = 0.0516 A (↓) Example- Using source transformation find the current through 10 Ω resistor. Example- Find the voltage across 4 Ω resistor. V 4 Ω = 7.12 V I 3 Ω = 2 A (↓)

31. MESH & NODAL ANALYSIS- Mesh Analysis(Maxwell’s Mesh current Method)- In this method, Kirchhoff’s voltage law is applied to each mesh in terms of mesh current instead of branch current. Each mesh assigned a mesh current. The mesh current is assumed to be flow in clockwise direction. Once mesh current is known we can determine branch current. Steps to follow in Mesh Analysis- Consider the network as shown in figure. Identify the mesh and assign the unknown current in each mesh in clockwise direction only.

32. MESH & NODAL ANALYSIS- Assign polarity of voltage across the branches. Apply KVL to each mesh around the mesh and use ohm’s law to express in terms of unknown current & resistance. Apply KVL to Mesh 1 ………………………… (1) Apply KVL to Mesh 2 ………………………… (2) Apply KVL to Mesh 3 ………………………… (3) Solve the above simultaneous equations to find branch currents.

33. MESH & NODAL ANALYSIS- Example- Find the value of current flowing through 1 Ω resistor. I 1 Ω = 0.5 A Example- Find the value of current flowing through 5 Ω resistor. I 5 Ω = 0.78 A Example- Find the value of current flowing through 8 Ω resistor. I 2 Ω = 5.71 A

34. MESH & NODAL ANALYSIS- Example- Find the current supplied by battery . I = 0.66 A Example- Find the value of current flowing through 5 Ω resistor. I 10 Ω = 0.78 A Example- Find the value of current flowing through 5 Ω resistor. I 5 Ω = 2 A

35. MESH AND NODAL ANALYSIS- Super Mesh- When a current source is present in branch and are common to two meshes then it is known as super mesh. In this cases we need some manipulation and concept of super mesh to solve the problem of super mesh. Example- By using Mesh analysis find the current through 10 Ω resistor.

36. MESH AND NODAL ANALYSIS- Nodal Analysis (Node Voltage Method)- This method is based on Kirchhoff’s current law to determine voltage at different nodes with respect to reference node. After determination of node voltage, current in all branches can be determine. Step to solve the Node Analysis problem- Mark all nodes. Assign any one node as reference node. Assign unknown potential to all nodes with respect to reference node. Assume unknown current and directions. Apply KCL for simultaneous node voltage and solve it. Using these voltage find any branch current which is required.

37. MESH AND NODAL ANALYSIS- Example- Find the current through 2 Ω resistor. V A = 11 V, V B = 12 V I 2 Ω = 6A Example- Find the current through 4 Ω resistor. V A = 5.5 V, V B = 8.5 V I 10 Ω = 0.75A Example- Find the current through 100 Ω resistor. V A = 67.25 V, V B = 48 V I 100 Ω = 0.48A

38. MESH AND NODAL ANALYSIS- Example- Find the current through 15 Ω resistor. V A = 15 V, V B = -26.25 V I 15 Ω = 2.75A Example- Find the voltage at node A and B. V A = 8 V, V B = 8 V Example- Find the current through 4 Ω & 3 Ω resistor. V 1 = 10 V, V 2 = 14.09 V, V 3 = 11.16 V I 4 Ω = 0.29A, I 3 Ω = 0.98A

39. MESH AND NODAL ANALYSIS- Super Node- When a voltage source is present between two nodes make them to be super node. In this cases separate equation is obtained between two nodes and these two nodes are combines during nodal analysis. Example- Determine the current in 5 Ω resistor for the network shown.

40. STAR ( Υ ) TO DELTA ( ∆ ) TRANSFORMATION- When a circuit cannot be simplified by normal series–parallel reduction technique, the star-delta transformation can be used. Consider delta and star network as shown. Delta Connection Star Connection Resistors R A , R B and R C forms delta network and resistors R 1 , R 2 and R 3 forms star network. These two networks will be electrically equivalent if the resistance as measured between any pair of terminals is the same in both the arrangements.

41. STAR ( Υ ) TO DELTA ( ∆ ) TRANSFORMATION- In figure (a) of delta network, Resistors between point 1 and 2 is …………….. (1) In figure (b) of star network, Resistors between point 1 and 2 is …………….. (2) Since the two networks are electrically equivalent, if equation (1) and (2) are equal. …………….. (3) Similarly, resistors between point 2 and 3 is Delta to Star Transformation- …………….. (4)

42. STAR DELTA TRANSFORMATION- Also resistors between point 3 and 1 is Subtracting equation (4) from (3) Adding equation (5) and (6), we get Similarly …………….. (5) …………….. (6) …………….. (7) …………….. (8) …………….. (9)

43. STAR DELTA TRANSFORMATION- Thus, star resistor connected to a terminal is equal to the product of the two delta resistors connected to the same terminal divided by the sum of the delta resistors. Star to Delta Transformation- Multiply equation (7) & (8), (8) & (9) and (9)& (7) …………….. (10) …………….. (11) …………….. (12)

44. STAR DELTA TRANSFORMATION- Adding equations (10), (11) & (12) Hence

45. STAR DELTA TRANSFORMATION- Thus, delta resistor between the two terminals is the sum of two star resistors connected to the same terminals plus the product of the two resistors divided by the remaining third star resistor. Note- If all resistances are of equal values-

46. STAR DELTA TRANSFORMATION-

47. STAR DELTA TRANSFORMATION- Example- Find the equivalent resistance between point X and Y R XY = 8.67 Ω Example- Find the equivalent resistance between point A and B R AB = 7.45 Ω R AB = 32.36 Ω

48. STAR DELTA TRANSFORMATION- Example- Find the equivalent resistance between terminal A and B R AB = 10 Ω Example- Calculate R AB for the circuit shown in figure R AB = 1.64 Ω Example- Find the equivalent resistance between terminal A and B R AB = 23 Ω

49. STAR DELTA TRANSFORMATION- Example- Find the equivalent resistance between terminal A and B R AB = 23.52 Ω Example- Calculate R AB for the circuit shown in figure R AB = 4.23 Ω Example- Find the current flowing through 8 Ω resistor I 8 Ω = 1.18 A

50. SUPERPOSITION THEOREM- This theorem is used when source transformation is not possible and the network have number of sources acting on it. Statement- It states that ‘In a linear network containing more than one independent sources, the resultant current in any element is the algebraic sum of the currents that would be produced by each independent source acting alone, all the other independent sources being represented meanwhile by their respective internal resistances.’ The independent voltage sources are represented by their internal resistances if given or simply with zero resistances, i.e., short circuits if internal resistances are not mentioned. The independent current sources are represented by infinite resistances, i.e., open circuits. A linear network is one whose parameters are constant, i.e., they do not change with voltage and current.

51. SUPERPOSITION THEOREM- Steps for superposition theorem- Consider the network as shown in figure- Calculate current flowing through terminal A and B. Case ( i )- According to superposition theorem, each source act independently. Consider only source V 1 is acting independently while replacing V 2 by its internal resistance if not short it. Apply KVL to each mesh and find current through AB (I AB ’) Case (ii)- Consider only voltage source V 2 is act independently while replacing V 1 by its internal resistance, if not then short it. Apply KVL to each mesh and find current through AB (I AB ’’)

52. SUPERPOSITION THEOREM- Case (iii)- Total current through terminal A & B is- I AB = Current through AB because of only V 1 + Current through AB because of only V 1 I AB = I AB ’ + I AB ’’ Example- Using superposition theorem find the value of current flowing through 1 Ω resistor. I 3Ω = 5 A, I 3Ω = 10 A , I 4 Ω = 5 A Example- Using superposition theorem find the value of current flowing through 1 Ω resistor. I 1Ω = 4 A

53. SUPERPOSITION THEOREM- Example- Using superposition theorem find the value of current flowing through 10 Ω resistor. I 8Ω = 0.316 A Example- Using superposition theorem find the value of current flowing through 8 Ω resistor. I 10Ω = 0.316 A Example- Find the value of current flowing through 4 Ω resistor. I 4Ω = 4.29 A

54. THEVENIN’S THEOREM- This theorem is used to find the current through complex network. Statement- It states that ‘Any two terminals complex network can be replaced by an equivalent voltage source (EMF/ Thevenin;s voltage/V TH ) in series with equivalent resistance (Thevenin’s resistance). The voltage source is the voltage across the two terminals with load, if any, removed. The series resistance is the resistance of the network measured between two terminals with load removed. Constant voltage source being replaced by its internal resistance (or if it is not given with zero resistance, i.e., short circuit) and constant current source replaced by infinite resistance, i.e., open circuit.

55. THEVENIN’S THEOREM- Steps for Thevenin’s Theorem- Remove the load resistance R L . Find open circuited voltage V TH across the point A & B. (remove the resistance across which we have to calculate current through it). Find resistance R TH as seen from point A and B with voltage source and current source replace by it’s internal resistance. Replace the circuit with Thevenin’s circuits i.e Voltage source V TH in series with equivalent resistance R TH . Find the current through R L using Ohm’s law.

56. THEVENIN’S THEOREM- Example- Find the value of current flowing through 2 Ω resistor. V TH = 23.33 V R TH = 3.33 Ω I L = 4.38 A Example- Find the value of current flowing through 8 Ω resistor. V TH = 50 V R TH = 12.5 Ω I L = 2.44 A Example- Find the value of current flowing through 2 Ω resistor connected between terminal A & B. V TH = 2.8 V R TH = 1.43 Ω I L = 0.82 A

57. THEVENIN’S THEOREM- Example- Find the value of current flowing through 10 Ω resistor. V TH = 10 V R TH = 17 Ω I L = 0.37 A Example- Find the value of current flowing through 40 Ω resistor. V TH = 4.2 V R TH = 8.33 Ω I L = 0.09 A Example- Find the value of current flowing through 30 Ω resistor. V TH = 70 V R TH = 26.09 Ω I L = 1.25 A

58. NORTON’S THEOREM- Norton’s theorem is converse of Thevenin’s theorem. It uses current source instead of voltage source and a resistance R N (which is same as R TH ) in parallel with the source. Statements- It states that, Any two terminal network can be replaced by a single current source with magnitude I N (called Norton’s current/Short circuit current) in parallel with single resistance R N (called Norton’s resistance).

59. NORTON’S THEOREM- Steps for Norton’s Theorem- Remove the load resistance R L and put a short circuit across the terminals. Find the short-circuit current I SC or I N across the terminals. Find the resistance R N as seen from points A and B by replacing the voltage sources and current sources by internal resistances. Replace the network by a current source I N in parallel with resistance R N . Find current through R N by current–division rule,

60. NORTON’S THEOREM- Example- Find the value of current flowing through 10 Ω resistor. I N = 5 A R TH = 0.95 Ω I L = 0.43 A Example- Find the value of current flowing through 15 Ω resistor in given ckt . Example- Find the value of current flowing through 10 Ω resistor. I N = 1.4 A R TH = 10 Ω I L = 0.56 A I N = -0.67 A R TH = 9.43 Ω I L = 0.33 A

61. NORTON’S THEOREM- Example- Find the value of current flowing through 10 Ω resistor. I N = 0.81 A R TH = 12.3 Ω I L = 0.45 A Example- Find the value of current flowing through 08 Ω resistor in given ckt . Example- Find the value of current flowing through 10 Ω resistor. I N = 6.58 A R TH = 3 Ω I L = 1.79 A I N = -13.17 A R TH = 1.46 Ω I L = 1.68 A

62. MAXIMUM POWER THEOREM- This theorem deals with transfer of maximum power from source to load. It states that ‘the maximum power is delivered from a source to a load when the load resistance is equal to the source resistance.’ Consider a complex circuit supplying power to load R L . The complex circuit is replaced by Thevenin’s equivalent circuit which consist of single source of emf V TH (Thevenin’s voltage) in series with single resistance R TH (Thevenin’s resistance), R L is load resistance. The load current I L through R L is given as-

63. MAXIMUM POWER THEOREM- Proof- Power delivered to the load R L ………….. (1) For a given circuit R TH and V TH are constants, therefore power delivered to load depends on R L . In order to find value of R L for which the value of P is maximum, it is necessary to differentiate equation (1) with respect to R L and set the result equal to zero. For P max , Differentiate equation (1) w. r. to R L , we get Hence Maximum power Transfer theorem proved. R L = R TH Maximum power transfer from source to load when load resistance equal to source resistance

64. MAXIMUM POWER THEOREM- Power delivered to load R L is given as- When R TH = R L and P = P max Steps for Maximum Power Transfer Function- Remove the variable load resistor R L . Find the open circuit voltage V TH across points A and B. Find the resistance R TH as seen from points A and B with voltage sources and current sources replaced by internal resistances. Find the resistance R L for maximum power transfer. R L = R TH Find the maximum power.

65. MAXIMUM POWER THEOREM- Example- Find the value of R L for maximum power transfer. Calculate maximum power. Example- Find the value of R L for maximum power transfer. Calculate maximum power. Example- Find the value of R L for maximum power transfer. Calculate maximum power.

66. MAXIMUM POWER THEOREM- Example- Find the value of R L for maximum power transfer. Calculate maximum power. Example- Find the value of R L for maximum power transfer. Calculate maximum power. Example- Find the value of R L for maximum power transfer. Calculate maximum power.