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Thevenin and Norton Equivalence in Circuit Analysis

Learn about the Thevenin and Norton equivalent circuits, combining voltage and current sources, and source transformations that simplify circuits. Discover how to obtain Thevenin circuits with dependent sources and solve for open circuit voltages.

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Thevenin and Norton Equivalence in Circuit Analysis

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  1. E E2415 Lecture 03 - Thévenin and Norton Equivalent Circuits

  2. Thévenin and Norton Equivalents

  3. Combining Voltage Sources Voltage sources are added algebraically

  4. Combining Voltage Sources Voltage sources are added algebraically

  5. Combining Voltage Sources Don’t do this. Why is this illogical? Whose fundamental circuit law is violated by this?

  6. Combining Current Sources Current sources are added algebraically

  7. Combining Current Sources Current sources are added algebraically

  8. Combining Current Sources Don’t do this. Why is this illogical? Whose fundamental circuit law is violated by this?

  9. Source Transformations can Simplify Circuits (1/5)

  10. Source Transformations Simplify Circuits (2/5)

  11. Source Transformations can Simplify Circuits (3/5)

  12. Source Transformations can Simplify Circuits (4/5)

  13. Source Transformations can Simplify Circuits (5/5)

  14. Obtaining Thévenin Circuit with Dependent Sources • Replace all independent voltage sources with short circuits (0 resistance). • Replace all independent current sources with open circuits ( resistance). • Apply a 1.0 amp current source to the terminal pair. • Resulting terminal voltage numerically equal to Thévenin resistance

  15. Another Thévenin Circuit (1/4) Find open circuit voltage Vab:

  16. Another Thévenin Circuit (2/4) Solve mesh equations for I2 Then Vab can be found:

  17. Another Thévenin Circuit (3/4) Inject One Amp: Now get Thévenin Resistance by node voltage solution:

  18. Another Thévenin Circuit (4/4)

  19. The Result

  20. Check on Previous Example (1/2): • VTh = 12 V and RTh = 3   IN = 4 A • We will calculate IN directly. KVL 1: 20 = 15I1 - 10I2 KVL 2: 0 = 10ix + 8ix ix = 0 I1 = I2

  21. Check on Previous Example (2/2): • VTh = 12 V and RTh = 3   IN = 4 A • We will calculate IN directly. KVL 1: 20 = 15I1 - 10I2 KVL 2: 0 = 10ix + 8ix ix = 0  I1 = I2

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