Adding two current sources give me the wrong answer....

[yosimba2000]

I have to find current Ix.

Equations: None

My approach:
1) Turn 60<0 and 2+4j impedances (top left of circuit) into current source parallel to impedances. 2+4j = 4.47<63.43
Current source = 60<0 / 4.47<63.43 = 13.423 < 296.57

2) The two impedances 2 +4j are now in parallel with the 6 ohms, so I comebine those to get impedance Z1 = 3<36.9

3) Then I add the current source on the left to the current source on the right. So net current source is
13.423 < 296.57 + 5<90 = 9.23<310.56

4) Then I transform net current source and Z1 into voltage source in series with Z1, so voltage source is 27.69<347.46

I think that adding the current sources should not give me the wrong answer for Ix, but after solving it this way, I get Ix = 3.87<14.33. Apparantly this is wrong.

The book did NOT add the current sources. After getting impedance Z1, it turned it back into voltage source in series with Z1. Then it calculated current Ix, and got 5.25<17.4.

I know how to get the answer the book did, but I don't understand why my method (adding both current sources, then turning back into voltage source) doesn't give me the right answer!

 

[gneill]

Yosimba2000, please remember to use the homework formatting template when you post a problem in the homework areas. It is a forum rule.

What happened to the -j2 Ω capacitor at the top of the circuit? I don't see where you've taken that into consideration in your simplification.

My suggestion would be to first find the Thevenin equivalent including that capacitor. There's a single voltage divider to deal with and the Thevenin impedance is pretty straightforward too. Then convert the result to its Norton equivalent.

 

[Hesch]

1) Turn 60<0 and 2+4j impedances (top left of circuit) into current source parallel to impedances. 2+4j = 4.47<63.43
Current source = 60<0 / 4.47<63.43 = 13.423 < 296.57

So you have made a Norton equivalent to the left. But you have forgotten the 6Ω resistor, which should have been included.

Instead I will suggest that you use KCL ( Kirchhoff's Current Law ) to solve the problem. Place two nodes:

Va above the 6Ω resistor.
Vb above the 4Ω resistor.

and define a reference voltage below the 4Ω resistor to be 0V.

Now, an equation regarding Va could look like:

( I1 ) - ( I2 ) - ( I3 ) = 0 →
( ( 60V - Va ) / ( 2+j4 Ω ) ) - ( ( Va - 0V ) / 6Ω ) - ( ( Va - Vb ) / -2j Ω ) = 0

Make another equation as for Vb.

Now you have two equation, by which you can determine Vb ( and Va ).

Ix = Vb / ( 4-j3 Ω )

 

[yosimba2000]

Yosimba2000, please remember to use the homework formatting template when you post a problem in the homework areas. It is a forum rule.

What happened to the -j2 Ω capacitor at the top of the circuit? I don't see where you've taken that into consideration in your simplification.

My suggestion would be to first find the Thevenin equivalent including that capacitor. There's a single voltage divider to deal with and the Thevenin impedance is pretty straightforward too. Then convert the result to its Norton equivalent.

Yes, I'll remember next time. Sorry.

I cannot use any transformations besides Source Transformations since that's what the problem wants.

The -2j capacitor was taken care of when I added both current sources together, and treating the Z1, -2j, 4 ohm, and -3j as inpedances in series.

But do you guys think that by adding both current sources together, Ix would still be the same?

I think it would be easier if I showed my HW. It is problem 10.52 in My Work.pdf

 

[gneill]

Yes, I'll remember next time. Sorry.

I cannot use any transformations besides Source Transformations since that's what the problem wants.

A Thevenin equivalent is a source transformation. So it is fair game here. In fact, you can find the Thevenin equivalent by repeated individual source transformations, accumulating series and parallel components into the model as opportunities arise.

Your first source transformation would be a first step, taking in the 6 Ohm resistor that ends up in parallel with the current source and its parallel resistance. Then if you were to change the result back to a voltage source in series with an impedance, you could "consume" the -j2 capacitor into the total series impedance. The result is the Thevenin equivalent of all those components. THEN you can convert that into a current source (Norton equivalent) which places your current sources in parallel.

The -2j capacitor was taken care of when I added both current sources together, and treating the Z1, -2j, 4 ohm, and -3j as inpedances in series.

Draw the circuit. How can the current sources be in parallel AND those impedances be in series?

When you've absorbed the 6 Ohm resistor into the current source's parallel impedance you end up with something like this:

As you can see those components are not in series, nor are the current sources in parallel (yet). That -j2 Ohm capacitor is in the way.

But do you guys think that by adding both current sources together, Ix would still be the same?

Yes, so long as you leave the Ix branch alone and all your simplifications are correct.

 

[yosimba2000]

Oh man... I totally didn't see the -2j capacitor blocking the current sources from being in parallel!

Thanks!