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.

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Lecture 9

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Lecture 9

Lecture 9

Superposition


Superposition theorem 1 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.


Superposition theorem 2 2

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

Superposition Example 1 (1/4)

Find I1, I2 and Vab by superposition


Superposition example 1 2 4

Superposition Example 1 (2/4)

Step 1: Omit current source.

By Ohm’s law and the

voltage divider rule:


Superposition example 1 3 4

Superposition Example 1 (3/4)

Step 2: Omit voltage source.

By the current divider rule and Ohm’s law :


Superposition example 1 4 4

Superposition Example 1 (4/4)

Combining steps 1 & 2, we get:


Superposition example 2 1 5

Superposition Example 2 (1/5)

Find Ix by superposition


Superposition example 2 2 5

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

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

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

Superposition Example 2 (5/5)

Sum the partial currents due to each of the sources:


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