Lecture 9 Superposition Superposition Theorem (1/2) - PowerPoint Presentation

Slide1

Lecture 9

SuperpositionSlide2

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.Slide3

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 superpositionSlide4

Superposition Example 1 (1/4)

Find I

1

, I

2

and V

ab

by superpositionSlide5

Superposition Example 1 (2/4)

Step 1: Omit current source.

By Ohm’s law and the

voltage divider rule:Slide6

Superposition Example 1 (3/4)

Step 2: Omit voltage source.

By the current divider rule and Ohm’s law :Slide7

Superposition Example 1 (4/4)

Combining steps 1 & 2, we get:Slide8

Superposition Example 2 (1/5)

Find I

x

by superpositionSlide9

Superposition Example 2 (2/5)

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

use Current Divider Rule:Slide10

Superposition Example 2 (3/5)

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

use Current Divider Rule:Slide11

Superposition Example 2 (4/5)

Activate only the 64 V voltage source at the bottom. Then use Ohm’s Law:Slide12

Superposition Example 2 (5/5)

Sum the partial currents due to each of the sources: