Finding complex voltage in AC circuit.

The Electrician

Joined Oct 9, 2007
2,971
I haven't ignored it, i tried it and i still had Ig2 in final result, so i thought that i could try different approach, anyway, what is wrong with approach in my previous post?
Plain old arithmetic errors:

You should have: which means that the voltage of the branch with E2 and R2 (which is the voltage that i am looking for) is Et=E2-R2(Ig4-Ig3) = 1 - 2*10^3(0.5-j)10^(-3) = 1-1+2j =2 jV

And: So now I=Et/(Rt+R6) = j/6 10^(-3) A which means that the voltage across the R6 is U=I*R6 = 4j/3 V
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
Plain old arithmetic errors:

You should have: which means that the voltage of the branch with E2 and R2 (which is the voltage that i am looking for) is Et=E2-R2(Ig4-Ig3) = 1 - 2*10^3(0.5-j)10^(-3) = 1-1+2j =2 jV

And: So now I=Et/(Rt+R6) = j/6 10^(-3) A which means that the voltage across the R6 is U=I*R6 = 4j/3 V

Thank you a lot, i was looking for the simplest way to solve this since the node analysis includes relatively lot of arithmetic and it's very easy to make a mistake somewhere, you see, even in this shorter approach i made arithmetic mistake, anyway, thanks again for help!
 

The Electrician

Joined Oct 9, 2007
2,971
The solution to your two equations should have been:

U20 = (1/2 - j) + 1000 Ig2
U30 = (1/2 - (7 j)/3) + 1000 Ig2

and U20 - U30 = 4j/3

You see how Ig2 vanishes from the final result.
 

The Electrician

Joined Oct 9, 2007
2,971
One final comment about this problem. If you had been required to find the two voltages that you've called U20 and U30, the fact that there are some unknowns doesn't preclude finding a solution. Just leave the unknowns in symbolic form--the solutions would be just this:

U20 = (1/2 - j) + 1000 Ig2
U30 = (1/2 - (7 j)/3) + 1000 Ig2

Your might get another problem that leads to just such a mixed solution.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
One final comment about this problem. If you had been required to find the two voltages that you've called U20 and U30, the fact that there are some unknowns doesn't preclude finding a solution. Just leave the unknowns in symbolic form--the solutions would be just this:

U20 = (1/2 - j) + 1000 Ig2
U30 = (1/2 - (7 j)/3) + 1000 Ig2

Your might get another problem that leads to just such a mixed solution.
Well, i tried to, but i probably made a mistake in calculations so i couldn't get rid of Ig2.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
Do it and show the resulting circuit and, if you don't see how it leaves the solution staring you in the face, we can pick up the discussion from that point.

Here is the picture:

josjednokolo.png

Now i don't see the solution here, but i see that the voltage across R6 is the same as the voltage across R2 and across the E2/R2.
 

MrAl

Joined Jun 17, 2014
11,389
When you say that we cannot solve that one node, you mean the voltage of that one node not the current, since we can't find correct voltage of that node since we changed the value of impedance, but everything else will remain the same as in the original circuit.


Ok, i have another idea now, as i stated in my first post, if i am looking for thevenin generator, i need to know it's voltage and it's inner resistance, Thevenin resistance is R2 and now i have an idea for Thevenin voltage, and here it is(note that i removed the branch with R6 since i am looking for Thevenin generator):

View attachment 103863

which means that the voltage of the branch with E2 and R2 (which is the voltage that i am looking for) is Et=E2-R2(Ig4-Ig3) = 1 - 2*10^3(0.5-j)10^(-3) = 1-1+j = jV

so now i have the following circuit:

View attachment 103865

So now I=Et/(Rt+R6) = j/6 10^(-3) A which means that the voltage across the R6 is U=I*R6 = 2j/3 V

Is this the correct way to solve this and is this the correct solution?
Hello again,

Yes, that one node voltage is not solvable once the capacitor is eliminated, and this becomes obvious if you short out the cap.

But my question now is, why did you not short out the cap yet???

You are looking to solve this without nodal equations, and you can do that, but when you seek to do things this way you need to find every possible simplification first, especially the simple ones like the cap elimination.
You may also need to use a little more complicated sub circuit elimination theory to eliminate another component, but if you havent done the cap yet it's pointless to talk about the more complicated ideas.
Once you do the simplifications, the circuit shouts out it's own solution :)
 

The Electrician

Joined Oct 9, 2007
2,971
Hello again,

Yes, that one node voltage is not solvable once the capacitor is eliminated, and this becomes obvious if you short out the cap.

But my question now is, why did you not short out the cap yet???

You are looking to solve this without nodal equations, and you can do that, but when you seek to do things this way you need to find every possible simplification first, especially the simple ones like the cap elimination.
You may also need to use a little more complicated sub circuit elimination theory to eliminate another component, but if you havent done the cap yet it's pointless to talk about the more complicated ideas.
Once you do the simplifications, the circuit shouts out it's own solution :)
Did you read any posts after #38? His solution in #38 was correct except for a simple arithmetic error, which was pointed out in subsequent posts.
 

MrAl

Joined Jun 17, 2014
11,389
Did you read any posts after #38? His solution in #38 was correct except for a simple arithmetic error, which was pointed out in subsequent posts.
Hello again,

Yes but he also mentioned that he was looking for a simpler solution which might come from doing the easier simplifications first. He redrew the schematic and still showed the cap, so i assumed he was still dealing with it. Although that might not be the case it seems better to draw the schematic without the cap if taking the time to draw it at all.

No big deal, if he is satisfied with the complexity of arriving at the solution then we might be done here :)
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
What is the combined current flowing in R2 and R6?

In R2 there's I2=E2/R2 +Ig3 - Ig4 + I6 (I6 current through R6)


So by KCL sum of all currents going through the node is zero so E2/R2 +Ig3 - Ig4 + I6 - I2= 0 but i don't know I6 and I2. I solved this another way, but i'd really like to see how can it be done this way.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
Hello again,

Yes, that one node voltage is not solvable once the capacitor is eliminated, and this becomes obvious if you short out the cap.

But my question now is, why did you not short out the cap yet???

You are looking to solve this without nodal equations, and you can do that, but when you seek to do things this way you need to find every possible simplification first, especially the simple ones like the cap elimination.
You may also need to use a little more complicated sub circuit elimination theory to eliminate another component, but if you havent done the cap yet it's pointless to talk about the more complicated ideas.
Once you do the simplifications, the circuit shouts out it's own solution :)
To be honest, i've never done much of simplifications in the circuits (the ones that include masked generators and these things with impedances in series with current generators), so it's something new for me, so i was looking for the fastest way to solve so i thought that the way i solved this was faster, anyway i am really thankful to you guys for pointing out those things to me because they could be useful in some other problems where i wouldn't be able to determine unknown values the way i could here.
 

WBahn

Joined Mar 31, 2012
29,976
In R2 there's I2=E2/R2 +Ig3 - Ig4 + I6 (I6 current through R6)


So by KCL sum of all currents going through the node is zero so E2/R2 +Ig3 - Ig4 + I6 - I2= 0 but i don't know I6 and I2. I solved this another way, but i'd really like to see how can it be done this way.
Okay, this has dragged on long enough.

Look at that bottom node.

You have three current sources attached to the bottom node taking a total net current off that node of:

Inet = Ig3 + (E2/R2) - Ig4

That current has to come into the node via the parallel combination of R2 and R6 and the voltage you are looking for is the voltage across R6 which IS the voltage across the parallel combination of R2 and R6, therefore:

U = Inet · (R2 || R6)

U = (Ig3 + (E2/R2) - Ig4) · (R2 · R6) / (R2 + R6)

Using the values given in the first post:

U = (j1 mA + (1 V / 2 kΩ) - 0.5 mA) * (2 kΩ · 4 kΩ)/(2 kΩ + 4 kΩ)

U = (j1 mA + 0.5 mA - 0.5 mA) * (8 kΩ²)/(6 kΩ)

U = (j1 mA) * (4 kΩ)/(3)

U = j (4/3) V

This same equation can be obtained directly by simply applying KCL to the bottom node in the original diagram. Summing the current leaving the node, we have

[Ig3] + [-Ig4] + [(0V - (U - E2))/R2] + [- U/R6] = 0
Ig3 - Ig4 - (U - E2)/R2 - U/R6 = 0
Ig3 - Ig4 + E2/R2 = U( 1/R2 + 1/R6 ) = U (R2 + R6)/(R2 · R6)
Ig3 - Ig4 + E2/R2 = U (R2 + R6)/(R2 · R6)

U6 = (Ig3 - Ig4 + E2/R2) · (R2 · R6) / (R2 + R6)

To check this result, we can merely look at the sum of the currents flowing in the resistors R2 and R6 and see if this is consistent with the sum of the currents in the sources Ig3 and Ig4.

The current in R6 is simply

I6 = U / R6 = j(4/3) V / 4 kΩ = j(1/3) mA

The current in R2 is

I2 = (U - 1 V) / R2 = (-1 + j(4/3)) V / 2 kΩ = (-0.5 + j(2/3)) mA

I2 + I6 = (-0.5 + j1) mA

The sum of Ig3 and Ig4 is

Ig3 + Ig4 = (-0.5 + j1) mA

So the solution is most likely correct.
 
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