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Lecture 9

Lecture 9. Superposition. 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.

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Lecture 9

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  1. Lecture 9 Superposition

  2. 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.

  3. 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

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

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

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

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

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

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

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

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

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

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