Week 9: Superposition

Superposition

What is Superposition?

- Superposition is another algorithmic circuit analysis method, with a prescribed procedure, like the node-voltage or loop-current methods.

It applies only in circuits with 2 or more independent sources.

It comes from the mathematical notion of superposition in which a final solution can be built as the sum of two or more partial solutions. It works well when the partial solutions can be obtained very easily. (Using voltage dividers or resistor reduction techniques.)

Superposition only applies to linear circuits, which are made up of linear circuit elements: resistors, voltage and current sources, capacitors, inductors, and linear amplifiers. It does not apply when the circuit has non-linear elements like diodes and transistors (electronics).

The Superposition Procedure:

Here are the steps: (Recall that the method only applies to circuits with two or more independent sources.)

1. Make a partial circuit by choosing one source to keep and turning of all of the other independent sources.

 -To “turn off” a voltage source, we turn it into a short circuit (zero volts and carries any amount of current.) To turn off a current source, we convert it to an open circuit (zero amps and any amount of voltage across it.)

2. Solve this “partial circuit” for the quantities of interest. Use whatever techniques that are applicable to the partial circuit: resistor reductions, voltage dividers, source transformations, or even node voltage.

3. Go back to the original circuit and create another partial circuit by choosing another source to keep (different than the first one) and turning off all of the other sources in the circuit.

4. Solve the second partial circuit for the quantities of interest.

5. Repeat the process until each independent source has been used in a partial circuit.

6. Sum up the all of the partial results to get the total results for the quantities of interest.

 -If there are n independent sources in a circuit, then when using superposition, you would expect to solve n partial circuits and find n partial results for whatever you are looking for.

Example:

4Ω and 2Ω are in series and also 3Ω and 1Ω :

Now 4Ω and 6Ω are parallel:

4Ω||6Ω=4×6/4+6=2.4Ω

So using Ohm's law:

Ix1=5V/2.4Ω=2.083A

For a moment forget Ix and concentrate on finding current of resistors. If we have the current of resistors, we can easily apply KCL and find Ix2 . So, 4Ω and 2Ω are parallel and also 3Ω and 1Ω are parallel:

4Ω||2Ω=4×2/4+2=43Ω

3Ω||1Ω=3×1/3+1=34Ω

Now, we can find their voltage drops:

V4Ω||2Ω=4/3×−3A=−4V

V3Ω||1Ω=3/4×−3A=−2.25V

Please note that the voltage drop on 4Ω||2Ω is the same as 4Ω and 2Ω voltage drops, because the circuits are equivalent and all are connected to the same nodes. The same statement is correct for 3Ω||1Ω voltage drop and 3Ω and 1Ω voltage drops. So

V4Ω=V2Ω=V4Ω||2Ω=−4V

V3Ω=V1Ω=V3Ω||1Ω=−2.25V

To find Ix2 all we need is to write KCL at one of the nodes:

−I2Ω+Ix2+I3Ω=0

→Ix2=I2Ω−I3Ω 

I2Ω and I3Ω can be found using Ohm's law:

I2Ω=V2Ω2Ω=−42=−2V

I3Ω=V3Ω3Ω=−2.253=−0.75V

Therefore,

Ix2=−1.25A

And

Ix=Ix1+Ix2=2.083−1.25=0.8333A

Ix=0.8333A

Reflection:

Superpositon overall is a topic with an easy-to grasp concept. To solve a circuit using the superposition theorem, We need to choose one source and then we need to turn off the other sources. The choosing of the sources is the most crucial part because choosing the wrong source to turn on might lead to a wrong output. After that, its only a matter of applying the basic circuit laws and equations to solve the problem.

Video:

Superposition

Superposition problem with dependent source

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