On the superposition method

Thread Starter

shubham161

Joined Jul 22, 2012
47
He's describing the concept of superposition, but not doing so correctly or validly. You don't "neglect" the sources, you turn them off (i.e., set them to zero output). For a current source, this means effectively removing it from the circuit. But for a voltage source it means shorting it so that it have 0V across it but no restriction on the current rhrough it.

That his method is invalid is evidenced by the fact that it produces the wrong answer for this problem. He gets 1A flowing downward in the 1R resistor when neglecting the current source and 8A flowing downward in it when neglecting the voltage source for a total of 9V. But, used properly, superposition gives 1A flowing downward with the current source turned off and 6A flowing downward when the voltage source is turned off.
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.
 

WBahn

Joined Mar 31, 2012
29,978
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.
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..
 

Thread Starter

shubham161

Joined Jul 22, 2012
47
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..
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.
 

WBahn

Joined Mar 31, 2012
29,978
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.
Perhaps I am just misinterpretting what you wrote originally. So please just explain what you meant in Post #11 when you said:

now current leaving/entering the current source is given as 8A. This 8 ampere current will flow from 1 ohm resistor (this time i have neglected voltage source).
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.
 

Thread Starter

shubham161

Joined Jul 22, 2012
47
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.
 

Thread Starter

shubham161

Joined Jul 22, 2012
47
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.
 

WBahn

Joined Mar 31, 2012
29,978
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.
 

Thread Starter

shubham161

Joined Jul 22, 2012
47
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.
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.
 
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