1. EEE1012 Introduction to Electrical & Electronics Engineering Chapter 2: Circuit Analysis Techniques by Muhazam Mustapha, July 2010

2. Learning Outcome Understand and perform calculation on circuits with mesh and nodal analysis techniques and superposition Be able to transform circuits based on Thevenin’s or Norton’s Theorem as necessary By the end of this chapter students are expected to:

3. Chapter Content Mesh Analysis Nodal Analysis Linearity and Superposition Source Conversion Thevenin’s Theorem Norton’s Theorem

4. Mash Analysis Mesh

5. Mesh Analysis Assign a distinct current in clockwise direction to each independent closed loop of network. Indicate the polarities of the resistors depending on individual loop. [*] If there is any current source in the loop path, replace it with open circuit – apply KVL in the next step to the resulting bigger loop. Use back the current source when solving for current. Steps:

6. Mesh Analysis Apply KVL on each loop: –Current will be the total of all direction –Polarity of the sources will maintained Solve the simultaneous equations. Steps: (cont)

7. Mesh Analysis Example: [Boylestad 10 th Ed. E.g. 8.11 - modified] R1R1 IaIa IbIb 2V 2Ω2Ω R2R2 1Ω1Ω 6V R3R3 4Ω4Ω a b I1I1 I3I3 I2I2

8. Mesh Analysis Example: (cont) Loop a : 2 = 2I a +4(I a −I b ) = 6I a −4I b Loop b : −6 = 4(I b −I a )+I b = −4I a +5I b After solving: I a = −1A, I b = −2A Hence: I 1 = 1A, I 2 = −2A, I 3 = 1A

9. Noodle Analysis Nodal

10. Nodal Analysis Determine the number of nodes. Pick a reference node then label the rest with subscripts. [*] If there is any voltage source in the branch, replace it with short circuit – apply KCL in the next step to the resulting bigger node. Apply KCL on each node except the reference. Solve the simultaneous equations.

11. Nodal Analysis Example: [Boylestad 10 th Ed. E.g. 8.21 - modified] R1R1 4A 2Ω2Ω R2R2 6Ω6Ω 2A R3R3 12Ω I1I1 I2I2 I3I3 a b

12. Nodal Analysis Node a : Node b : After solving: V a = 6V, V b = − 6A Hence: I 1 = 3A, I 2 = 1A, I 3 = −1A Example: (cont)

13. Mesh vs Nodal Analysis Mesh: Start with KVL, get a system of simultaneous equations in term of current. Nodal: Start with KCL, get a system of simultaneous equations on term of voltage. Mesh: KVL is applied based on a fixed loop current. Nodal: KCL is applied based on a fixed node voltage.

14. Mesh vs Nodal Analysis Mesh: Current source is an open circuit and it merges loops. Nodal: Voltage source is a short circuit and it merges nodes. Mesh: More popular as voltage sources do exist physically. Nodal: Less popular as current sources do not exist physically except in models of electronics circuits.

15. Linearity and Superposition

16. Linearity Concept of Circuit Elements Due to Ohm’s Law, the effect of voltage across a circuit element is linear. –Can be added linearly depending on how much potential is applied to each of them. This is true for the effect of current too.

17. Superposition Theorem Statement: The current through, or voltage across, an element is equal to the algebraic sum of the currents or the voltages produced independently by each source

18. Choose one power source to consider, then switch off other sources: –Voltage source: remove it and replace with short circuit –Current source: remove it and replace with open circuit Calculate the voltages and currents in the elements of concern based on the resulting circuit. Do the above for all sources, then sum the respective voltages or currents by considering the polarities. Superposition Theorem

19. Example: [Boylestad 10 th Ed. E.g. 9.5 - modified] R1R1 3A 2Ω2Ω 4Ω4Ω R2R2 I1I1 I2I2 6V 12V

20. Superposition Theorem Example: [Boylestad 10 th Ed. E.g. 9.5 - modified] R1R1 2Ω2Ω 4Ω4Ω R2R2 I 1a I 2a 12V Consider only the 12V source:

21. Superposition Theorem Example: [Boylestad 10 th Ed. E.g. 9.5 - modified] Consider only the 6V source: R1R1 2Ω2Ω 4Ω4Ω R2R2 I 1b I 2b 6V

22. Superposition Theorem Example: [Boylestad 10 th Ed. E.g. 9.5 - modified] Consider only the current source: R1R1 3A 2Ω2Ω 4Ω4Ω R2R2 I 1c I 2c

23. Superposition Theorem Example: [Boylestad 10 th Ed. E.g. 9.5 - modified] Hence: I 1 = I 1a + I 1b + I 1c = 1A I 2 = I 2a + I 2b + I 2c = 2A

24. Thevenin’s Theorem

25. Statement: Network behind any two terminals of linear DC circuit can be replaced by an equivalent voltage source and an equivalent series resistor Can be used to reduce a complicated network to a combination of voltage source and a series resistor

26. Calculate the Thevenin’s resistance, R Th, by switching off all power sources and finding the resulting resistance through the two terminals: –Voltage source: remove it and replace with short circuit –Current source: remove it and replace with open circuit Calculate the Thevenin’s voltage, V Th, by switching back on all powers and calculate the open circuit voltage between the terminals. Thevenin’s Theorem

27. Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 3Ω3Ω 6Ω6Ω 9V Convert the following network into its Thevenin’s equivalent:

28. Thevenin’s Theorem Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 3Ω3Ω 6Ω6Ω R Th calculation:

29. Thevenin’s Theorem Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 3Ω3Ω 6Ω6Ω 9V V Th calculation:

30. Thevenin’s Theorem Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 2Ω2Ω 6V Thevenin’s equivalence:

31. Norton’s Theorem

32. Statement: Network behind any two terminals of linear DC circuit can be replaced by an equivalent current source and an equivalent parallel resistor Can be used to reduce a complicated network to a combination of current source and a parallel resistor

33. Calculate the Norton’s resistance, R N, by switching off all power sources and finding the resulting resistance through the two terminals: –Voltage source: remove it and replace with short circuit –Current source: remove it and replace with open circuit Calculate the Norton’s voltage, I N, by switching back on all powers and calculate the short circuit current between the terminals. Norton’s Theorem

34. Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 3Ω3Ω 6Ω6Ω 9V Convert the following network into its Norton’s equivalent:

35. Norton’s Theorem Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 3Ω3Ω 6Ω6Ω R N calculation:

36. Norton’s Theorem Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 3Ω3Ω 6Ω6Ω 9V I N calculation:

37. Norton’s Theorem Example: [Boylestad 10 th Ed. E.g. 9.6 - modified] 2Ω2Ω 3A Norton’s equivalence: OR, We can just take the Thevenin’s equivalent and calculate the short circuit current.

38. Maximum Power Consumption An element is consuming the maximum power out of a network if its resistance is equal to the Thevenin’s or Norton’s resistance.

39. Source Conversion Use the relationship between Thevenin’s and Norton’s source to convert between voltage and current sources. 2Ω2Ω 3A 2Ω2Ω 6V V = IR