Superposition currents?

[malmoit]

Please view the attached file, where you will find a simple electric circuit with resistances, V-sources and a A-source. I wonder how one calculates "I" in the example!? The answer is supposed to be 1,3 Ampere..

/Malmoit

 

[BeeBop]

Please view the attached file, where you will find a simple electric circuit with resistances, V-sources and a A-source. I wonder how one calculates "I" in the example!? The answer is supposed to be 1,3 Ampere..

/Malmoit

An ideal voltage source has zero internal resistance; an ideal current source has infinite internal resistance. Replace two of the three sources with their ideal resistance, and calculate the currents. Do this for each, then find the algebraic sum. (if the currents flow in the same direction add them, if opposites, subtract them.)

 

[asaini88]

use super position or thevnin theorem

 

[asaini88]

can u tell me how to start new threads

 

[silvarblade]

can u tell me how to start new threads

as for the problem in the post it would be easier to use superposition theorem!

 

[BeeBop]

as for the problem in the post it would be easier to use superposition theorem!

So silverblade, what is the superposition theorem? why don't you tell us?

 

[3iMaJ]

Writing loop equations would be far easier than super position.

 

[MrAl]

Hi there,


BeeBop's reply was basically the same as superposition.

Superposition states that the response at any point in a linear circuit
is equal to the sum of the responses from each source acting alone.

To get a circuit with each source acting alone we have to do what
BeeBop said, or a little simpler, 'kill' all sources and then 'unkill' them
one at a time and take the response for the node or branch we are
after, then after we have done this for each source we add the results.
To 'kill' a voltage source we short it out, and to 'kill' a current source
we open circuit it. To 'unkill' either we simply put it back the way it was.
Note that after we 'unkill' a source we have to 'kill' it again before unkilling
the next source. There should never be more than one source acting
in the circuit at a time as we go through all the sources one by one.

I agree for this circuit using Thev and Norton might be a little simpler
than superposition, but one has to be very careful to get all the
algebraic signs right for each step or the result gets badly skewed.

 

[dbx232629]

Try to find the current flowing through the two resistors using superposition, then use KCL in the node.

 

[silvarblade]

So silverblade, what is the superposition theorem? why don't you tell us?

I never said that your definition was wrong!

 

[steveB]

I wonder how one calculates "I" in the example!?

/Malmoit

In the old days we used signal flowgraphs and Mason's gain formula. It's a 30 second problem using that method. It seems that they don't teach this any more.

 

[dbx232629]

considering the effect of each source, find the current flowing in 20 and 30 ohm then, use KCL in the junction of the V source, 20 and 30 ohm res or since the source will be shorted if the current source is considered, you can find the opposing currents in that branch. The first method is easier though.