1 / 39

Circuit Analysis Techniques: Mesh and Nodal Analysis

Learn how to analyze circuits using mesh and nodal analysis techniques and understand concepts such as linearity, superposition, and Thevenin's theorem. This chapter provides step-by-step examples and explanations to help you master electrical and electronics engineering circuit analysis.

andes
Download Presentation

Circuit Analysis Techniques: Mesh and Nodal Analysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. EEE1012Introduction to Electrical &Electronics EngineeringChapter 2: Circuit Analysis Techniquesby 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 10th Ed. E.g. 8.11 - modified] R2 R1 4Ω R3 1Ω 2Ω a b I1 I2 I3 Ia Ib 6V 2V

  8. Mesh Analysis Example: (cont) Loop a: 2 = 2Ia+4(Ia−Ib) = 6Ia−4Ib Loop b: −6 = 4(Ib−Ia)+Ib = −4Ia+5Ib After solving: Ia = −1A, Ib = −2A Hence: I1 = 1A, I2 = −2A, I3 = 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 10th Ed. E.g. 8.21 - modified] a R2 b I2 12Ω I3 I1 R1 2A 6Ω 4A 2Ω R3

  12. Nodal Analysis Example: (cont) Node a: Node b: After solving: Va = 6V, Vb = − 6A Hence: I1 = 3A, I2 = 1A, I3 = −1A

  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. Superposition Theorem • 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.

  19. Superposition Theorem Example: [Boylestad 10th Ed. E.g. 9.5 - modified] I2 I1 4Ω R2 R1 3A 2Ω 6V 12V

  20. Superposition Theorem Example: [Boylestad 10th Ed. E.g. 9.5 - modified] Consider only the 12V source: I2a I1a 4Ω R2 R1 2Ω 12V

  21. Superposition Theorem Example: [Boylestad 10th Ed. E.g. 9.5 - modified] Consider only the 6V source: I2b I1b 4Ω R2 R1 2Ω 6V

  22. Superposition Theorem Example: [Boylestad 10th Ed. E.g. 9.5 - modified] Consider only the current source: I2c I1c 4Ω R2 R1 3A 2Ω

  23. Superposition Theorem Example: [Boylestad 10th Ed. E.g. 9.5 - modified] Hence: I1 = I1a + I1b + I1c = 1A I2 = I2a + I2b + I2c = 2A

  24. Thevenin’s Theorem

  25. Thevenin’s Theorem 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. Thevenin’s Theorem • Calculate the Thevenin’s resistance, RTh, 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, VTh, by switching back on all powers and calculate the open circuit voltage between the terminals.

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

  28. Thevenin’s Theorem Example: [Boylestad 10th Ed. E.g. 9.6 - modified] RTh calculation: 3Ω 6Ω

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

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

  31. Norton’s Theorem

  32. Norton’s Theorem 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. Norton’s Theorem • Calculate the Norton’s resistance, RN, 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, IN, by switching back on all powers and calculate the short circuit current between the terminals.

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

  35. Norton’s Theorem Example: [Boylestad 10th Ed. E.g. 9.6 - modified] RN calculation: 3Ω 6Ω

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

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

  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. 2Ω 2Ω 3A 6V V = IR

More Related