# Lecture 9 - PowerPoint PPT Presentation

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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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## 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.

• 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.

### 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

### Superposition Example 1 (1/4)

Find I1, I2 and Vab by superposition

### Superposition Example 1 (2/4)

Step 1: Omit current source.

By Ohm’s law and the

voltage divider rule:

### Superposition Example 1 (3/4)

Step 2: Omit voltage source.

By the current divider rule and Ohm’s law :

### Superposition Example 1 (4/4)

Combining steps 1 & 2, we get:

### Superposition Example 2 (1/5)

Find Ix by superposition

### Superposition Example 2 (2/5)

Activate only the 16 A Current source at the left. Then

use Current Divider Rule:

### Superposition Example 2 (3/5)

Activate only the 16 A Current source at the right. Then

use Current Divider Rule:

### Superposition Example 2 (4/5)

Activate only the 64 V voltage source at the bottom. Then use Ohm’s Law:

### Superposition Example 2 (5/5)

Sum the partial currents due to each of the sources: