1 / 21

E E 2315

E E 2315. Lecture 06 - Thévenin and Norton Equivalent Circuits. Th é venin and Norton Equivalents. Combining Voltage Sources. Voltage sources are added algebraically. Combining Voltage Sources. Voltage sources are added algebraically. Combining Voltage Sources. Don’t do this.

gouldj
Download Presentation

E E 2315

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. E E 2315 Lecture 06 - 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 Simplify Circuits (1/5)

  10. Source Transformations Simplify Circuits (2/5)

  11. Source Transformations Simplify Circuits (3/5)

  12. Source Transformations Simplify Circuits (4/5)

  13. Source Transformations 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) 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

More Related