Superposition method and proof

In summary, the conversation involves using the superposition method to find the magnitude and direction of current in a circuit with three resistors. The person asking the question is struggling with understanding the concept and is unsure about how to approach the problem. After some attempts, they eventually figure out the solution.
  • #1
schmiggy
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Homework Statement


Use the superposition method to find the magnitude and direction of the current flowing through the 10 ohm resistor. Then use any technique other than that used previously to check the validity of your answer


Homework Equations





The Attempt at a Solution


First I'd like to point out that I have a very basic understanding of the content being covered in class, I understand the theory but have a hard time actually putting it into practice.

First of all I simplified the circuit down to three resistors by finding the resistance total of the 5 ohm and 15 ohm in parallel then adding the 16 ohm in series.

A number of things confused me about this question. Firstly, every worked example in class has never included 2 of the same type sources; the circuits included either a single voltage source, a single current source, or one of each, never two of the same. So my first question is, does the same concept apply? Do I remove one source, calculate the current using V = IR by finding resistance total, and then do the same with the other source? Do I combine the sources, find Rtotal and use V=IR?

My attempt at superposition by taking out one of the sources and calculating I through the entire circuit yielded a different value than that found using mesh analysis - it was very close, however still different. I'm not confident at all in my calculations and as such am a little uncomfortable posting calculations (embarrassment, shame etc).

My mesh analysis calculations yielded 6.25Amps flowing through the 10 ohm resistor.
I designated two loops using the simplified diagram attached and using KVL I found..
Loop 1: 200 - (10+20)I1 + 20I2 = 0
Loop 2: -100 - (20+19.75)I2 + 20I1 = 0
I rearranged in terms of I1 and using simultaneous equations found I1 which is equal to the current flowing through the 10ohm resistor.

One more questions regarding superposition.. if using V = IR for the two different sources, is Rtotal different depending on the current flow? Ie will the 200V source experience resistance equal to (10||20)+19.75 and for the 100V (19.75||20)+10 for Rtotal?
As you can see I'm not very good at this!

Is anyone able to verify my calculation of I through the 10ohm, or point out where I'm going wrong.. any help is greatly appreciated!
 

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  • #2
Managed to figure it out after a lot of smashing my head against a brick wall, thanks anyway! :)
 

1. What is the Superposition Method?

The Superposition Method is a mathematical technique used to solve linear differential equations by breaking them down into simpler equations that can be solved individually. It involves finding the sum of the individual solutions to the simpler equations to obtain the general solution to the original equation.

2. How does the Superposition Method work?

The Superposition Method works by assuming that the general solution to a linear differential equation can be expressed as a linear combination of the individual solutions to simpler equations. This assumption is based on the principle of superposition, which states that the total response of a linear system is equal to the sum of the individual responses to each input.

3. What is the proof behind the Superposition Method?

The proof behind the Superposition Method involves using the linearity property of differential equations, which states that the sum of two solutions to a linear differential equation is also a solution. By breaking down the original equation into simpler equations, we can use this property to show that the sum of the individual solutions is indeed the general solution to the original equation.

4. When is the Superposition Method most useful?

The Superposition Method is most useful when solving a linear differential equation with multiple inputs or boundary conditions. By breaking down the equation into simpler equations, we can solve each one separately and then combine the solutions using the Superposition Method to obtain the general solution.

5. Are there any limitations to the Superposition Method?

Yes, there are limitations to the Superposition Method. It can only be used to solve linear differential equations, and it assumes that the individual solutions are independent of each other. It also cannot be applied to non-homogeneous equations or equations with discontinuous inputs. In these cases, other methods such as variation of parameters may be necessary.

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