Applying the superposition theorem

[David J]

I am wondering is someone could comment on a question I have recently answered. I have attached the question and my answer. Apologies for not following the standard procedure of Latex but there are drawings associated with this question. I answered section A and my results are written on the sheet marked as "Question 1b"
$V1=j415v$, $V2= 415v$ and i calculated the current through the load to be $5.71+j0.892$

When i attempted question 1b i did not use the polar or rectangular values. I followed the hand out notes literally and my answers are posted on the attached Q1b answer PDF.

I am just wondering if someone could take a look and advise have I done this correctly or not. I think its correct as I have followed my notes but what I cannot understand is the value I got for $Ic$ is 8.71A using the superposition theorem. Using the the Thevenins theorem i got a different value, $5.71+j0.892$

2 different value of which I am struggling to understand. Any help would be appreciated

 

[gneill]

Your result for superposition is incorrect because you summed only the magnitudes of the individual current phasors, not the phasors themselves. That is, the load currents you calculated will have both magnitude and phase angle (complex values) and must be summed as such.

You also need to use the complex impedance for the load in your calculations (the load has a power factor specification).

 

[David J]

Hello gneill, thanks for the heads up. I am still a little confused by this system of math. I had a feeling my method was incorrect. So in order to get the correct answer the calculations should be done using the complex values as you stated

As shown below I think I have converted these correctly:-

$V1= j415$ or $415 \angle 90^0$
$V2=415$ or $415 \angle 0^0$
$50\Omega$ @ $0.7pf$ is actually written as $50\Omega \angle 45.6^0$ or $35+j35.71$ in rectangular form
$j4$ is written as $4\Omega \angle 90^0$ or simply $0 + j4$
$j6$ is written as $6\Omega \angle 90^0$ or simply $0+j6$

So assuming the values above are converted correctly (please advise) and my initial equation steps were correct then it would be the normal complex numbers arithmetic

 

[gneill]

Yes. Your complex values look good, and yes, the same steps you would use for DC voltages and resistors apply.

 

[David J]

Thanks for that, I started again and came up with this. Apologies for the attachment but I have included drawings and I cannot put the drawings in Latex. However the attachment is clear. I think I am correct. There is a slight difference in the value achieved using superposition and the value achieved using Thevenins theorem but on the whole I think it both values make sense. Does this make sense to you?

Using Thevenins theorem i calculated a current through the $50\Omega$ resistor of $5.78a$ and $\angle 8.9^0$

Using the superposition theorem i got $6.08a$ and $\angle 8.1^0$

Appreciated

 

[gneill]

Your value for $I_3$ is off a bit. The method is okay, so perhaps a typo or rounding error slipped in along the way. You should have reached a value of 3.544 + 3.255j A.

I'm not sure why you bothered to calculate all the extraneous currents. After you found the current from the source, you are presented with a current divider situation which you used to find $I_3$, but then when you had the same situation for $I_6$ you calculated the other currents too, and then employed KCL to find $I_6$. Too much extra work, and too many chances for rounding errors to slip in! You only need $I_3$ and $I_6$ to find the current through the load.

Anyways, enough ranting . The value you found for $I_6$ is close enough. I'm seeing 2.170 - 2.363j A, so nothing to worry about there.

Fix up your $I_3$ and you should be able to duplicate your Thevenin approach's results with your superposition results. You really should be able to achieve identical answers.

 

[David J]

Hello gneill, thanks for the update. I will review the issue of $I3$. The approach was based purely on the hand out information but I did wonder why I needed to calculate the extraneous currents when I only required 2 out of 6 so to speak. I am understanding it a little bit better now. Thanks for your help once again

 

[David J]

Hello again, I reworked the $I3$ and i have no idea where i got the first value from. it must have been an arithmetic mistake. I have the correct, identical values now using superposition and Thevenin approach. Thanks for your help.

Regarding part C of this question (looking back at the attachment in post 1) I can't even understand what this question is asking "forming a pair of norton generators" can anyone advise on this? What is the question asking me to do ? I think I need to confirm the correct value of current using this method as was done with the first 2 methods but where to begin ??

 

[gneill]

For part (c) they want you to convert both of the voltage sources into current sources (think: Thevenin to Norton). Once that is done, a further simplification should become obvious.