# On the superposition method

Discussion in 'General Electronics Chat' started by shubham161, Sep 11, 2012.

1. ### shubham161 Thread Starter Member

Jul 22, 2012
50
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i know that i didn't use the superposition theorem properly. It's my own method to analyze very simple circuits by inspection.

the current from current source (only) will divide into four parts i.e 2+2+2+2. The first three part will flow from 1 ohm resistor while one part will flow from 3 ohm resistor. and we also know that the current due to voltage source alone is -1 A in the direction of I4. when i can clearly see the current then there is no need to include that 3 ohm resistor in the solution. so i removed it along with voltage source.

2. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
That's completely different than what you said previously. You said the entire 8A would flow "from" the 1 ohm resistor. You said nothing about it spliting up or about any of it going through the 3 ohm resistor.

You keep using the phrase "flow from" some resistor. This is all but meaningless. Current flows "through" a resistor and you need to indicate what direction it is flowing..

3. ### shubham161 Thread Starter Member

Jul 22, 2012
50
3
oh sorry, english is not my native language and so it becomes very hard for me to explain just in text. if you can provide me with two example circuits with two mesh and all integral values, then i can show you how i compute current through 1 ohm resistor by neglecting everything that's on the right hand side of 1 ohm resistor. This is my own method so that i don't need to pick pen or paper.

4. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
Perhaps I am just misinterpretting what you wrote originally. So please just explain what you meant in Post #11 when you said:

I took this to mean that all 8A is flowing through the 1 ohm resistor and none of it is flowing through the 3 ohm resistor.

5. ### shubham161 Thread Starter Member

Jul 22, 2012
50
3
yes, i mean that 8A will flow through 1 ohm resistor if i replace voltage source by "infinite resistance". In superposition theorem we replace voltage source by its internal resistance which in ideally zero i.e we short it. but i do it differently.

Edit: Only for circuits like this. Just to make the calculations simple and fast.

6. ### shubham161 Thread Starter Member

Jul 22, 2012
50
3
okay, i think i should stick to the known and trusted methods. In our entrance exam for engineering, we are forced to solve problems under 2 minute without use of calculator and so i formulated this "crude" method. After entering engineering college, we are allowed to use calculators where i can solve equations. But i became so used to it that i forget that it was not the correct method and i wrote it here.

i waited for you to provide me with some circuits but you didn't. perhaps i have confused you badly.

7. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
I don't have time to craft any circuits right now. Even if I did, you need to allow at least a few days for someone to respond.

But this circuit will do.

Please explain how you go from saying that you have 8A flowing (downward, I assume) in the 1 ohm resistor if you ignore the voltage source and that you have 1A flowing (again, downward) if you ignore the current source and arrive at the correct answer.

Or describe your method, step by step, and what result you would get for the current in the center branch if we make the voltage source 20V, the 1 ohm resistor is replaced by a 4 ohm resistor, and the 3 ohm resistor is replaced by a 6 ohm resistor.

8. ### shubham161 Thread Starter Member

Jul 22, 2012
50
3
Its 8 AM here, and i am leaving for my college. When I'll return home i'll google search some circuit and explain the steps.

Nov 25, 2009
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