1 / 19

E E 2315

Learn how the Maximum Power Transfer Theorem and Superposition Method can be used to solve circuits and find maximum power transfer. Understand the rules and steps to apply these theorems effectively.

jeanneg
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 07 - Network Theorems,

  2. Maximum Power Transfer Theorem • Given a Thévenin equivalent circuit of a network, the maximum power that can be transferred to an external load resistor is obtained if the load resistor equals the Thévenin resistance of the network. Any load resistor smaller or larger than the Thévenin resistance will receive less power.

  3. Maximum Power Demonstrated (1/2) PL is max if and

  4. Maximum Power Demonstrated (2/2)  • Maximum Power Transfer Theorem applied when matching loads to output resistances of amplifiers • Efficiency is 50% at maximum power transfer and

  5. Superposition Theorem (1/2) • In a linear system, the linear responses of linear independent sources can be combined in a linear manner. • This allows us to solve circuits with one independent source at a time and then combine the solutions. • If an independent voltage source is not present it is replaced by a short circuit. • If an independent current source is not present it is replaced by an open circuit.

  6. Superposition Theorem (2/2) • If dependent sources exist, they must remain in the circuit for each solution. • Nonlinear responses such as power cannot be found directly by superposition • Only voltages and currents can be found by superposition

  7. Superposition Example 1 (1/4) Find I1, I2 and Vab by superposition

  8. Superposition Example 1 (2/4) Step 1: Omit current source. By Ohm’s law and the voltage divider rule:

  9. Superposition Example 1 (3/4) Step 2: Omit voltage source. By the current divider rule and Ohm’s law :

  10. Superposition Example 1 (4/4) Combining steps 1 & 2, we get:

  11. Superposition Example 2 (1/5) Find Ix by superposition

  12. Superposition Example 2 (2/5) Activate only the 16 A Current source at the left. Then use Current Divider Rule:

  13. Superposition Example 2 (3/5) Activate only the 16 A Current source at the right. Then use Current Divider Rule:

  14. Superposition Example 2 (4/5) Activate only the 64 V voltage source at the bottom. Then use Ohm’s Law:

  15. Superposition Example 2 (5/5) Sum the partial currents due to each of the sources:

  16. Superposition Example 3 (1/4) • Solve for Vab by Superposition method • Dependent source must remain in circuit for both steps

  17. Superposition Example 3 (2/4)

  18. Superposition Example 3 (3/4)

  19. Superposition Example 3 (4/4) Combining the solutions:

More Related