# Superposition Theorem

## Homework Statement

FIGURE 1 shows a 50 Ω load being fed from two voltage sources via their associated reactances. Determine the current i flowing in the load by:

Superposition Theorem

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## The Attempt at a Solution

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see attached files as I can not write in itex and its too complicated to try and write it using ()()().

Problem is that once I have calculated both Vl1 and Vl2 and I calculate he current I for each and add the currents together my answer for current is different to what it should be by quite a lot! am I doing something wrong here? I have also tried a different method where by I remove each voltage source the work out I1, I2, I3 for Voltage source V1 and I4, I5 and I6 for voltage source V2 in turn adding or deducting them as required and my answer still isn't what its supposed to be. The reason I know what the answer should be is due to the fact that the first part of this question asks me to calculate it using Thevenins Theorem...

#### Attachments

• Figure 1.jpg
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## Answers and Replies

these are the equations used. As for Vl1 it should be j4+ the contents in the bracket for the overall division numerator but its written as 4 and the software I used to create the equations in is a ball ache to change it to j4

#### Attachments

gneill
Mentor
Your expressions look okay, but it's hard to tell if computational errors have occurred without seeing some steps and results.

Why not type out some of your work and show some intermediate results? For example, in both cases you have a repeated parallel impedance calculation (load impedance in parallel with the impedance associated with the suppressed source). Can you show the values that you obtained for those sub-expressions?

• Ebies
yes I can supply the figures... this is what I worked out and from both IL1+IL2 you can see it does not match up with the Thevenins figure...

#### Attachments

• Working out.pdf
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gneill
Mentor
Okay, I don't see anything wrong with those results. The magnitude of the sum of the currents looks fine to me. Where do you see a problem?

• Ebies
well should they not equal the current previously calculated in part a) of the original question when added together? (the current calculated when I done the thevenins part)

gneill
Mentor
well should they not equal the current previously calculated in part a) of the original question when added together? (the current calculated when I done the thevenins part)
Yes. Keep in mind that if you choose a different basis function (cos versus sin) that you can expect a 90° phase shift in the result. Did you compare the magnitudes of the results?

• Ebies
magnitudes were fine but the phase angle is whats waaaay out. I used cos trig function in both thevenins and superposition theorem so cant really understand why the difference... let me quickly add the two currents again just to eliminate any possible error I could have made...

ok problem solved... made a mistake in my addition on the two currents... thanks for the assistance gneill... yet again very reliable and helpful... enjoy xmas eve and stay safe... :0)

for this particular problem it has an extension as part 1c which states the following:

(c) by transforming the two voltage sources and their associated reactances into current sources (and thus form a pair of Norton generators)....

Now am I correct in saying the way I answered 1B I kind of covered 1c as well in the sense that I already calculated IL1 and IL2 and as such have both current generators and thus my redrawn circuit will be two current generators either side of the load with their respective values?

gneill
Mentor
Now am I correct in saying the way I answered 1B I kind of covered 1c as well in the sense that I already calculated IL1 and IL2 and as such have both current generators and thus my redrawn circuit will be two current generators either side of the load with their respective values?
You found those currents by suppressing one source or the other but leaving the impedances intact. As such that was applying the superposition theorem. This time they want you to apply source transformations beforehand. The transformations are applied to just a voltage source and its series impedance without regard to the rest of the circuit.

so use ohms law with the voltage source and its series impedance to find i for each respective loop independently and i will be my current source respectively?

in fact I will calculate what I think it is and post it here for confirmation

this just seems too simple, thus casting doubt in my mind... I have IL1=V1/j4 and IL2=V2/j6 and hat would be the two current generators IL1 and IL2 respectively

gneill
Mentor
this just seems too simple, thus casting doubt in my mind... I have IL1=V1/j4 and IL2=V2/j6 and hat would be the two current generators IL1 and IL2 respectively
Yes. The result is two current generators and three impedances all in parallel. Simple indeed.

thanks gneill, you are a star

I've attempted the calculations as above, but still having no luck. Can someone help me out of this rutt?? Please? :)

gneill
Mentor
I've attempted the calculations as above, but still having no luck. Can someone help me out of this rutt?? Please? :)
Can you show some details of your attempt?

I've calculated the currents I1 and I2 (correlating to voltage sources V1 and V2) as follows;

Please excuse the formatting, i'm new to all of this.

I1 = V1/J4 = 415/j4 = -103.75
I2 = V2/j6 = j415/j6 = -j69.16(reccurring)

Am i along the right tracks or falling at the first hurdle?

I'm not sure on how to proceed.

gneill
Mentor
That looks fine so far.

Draw the transformed circuit and note the configuration. What type of connections do you see (series/parallel/other)? How might you simplify the circuit?

I see a pair of loops? and as stated above there are two generators and three impedances in parallel?

gneill
Mentor
I see a pair of loops? and as stated above there are two generators and three impedances in parallel?
Okay, how might you simplify things? What can you do with parallel components?

I'm sure it will all seem so simple when it's complete but at the moment i just don't know.

You mention "The transformations are applied to just a voltage source and its series impedance without regard to the rest of the circuit." so do i split it in to two? if so then what can i do with it? i'm not sure i understand what i'm supposed to be doing.

gneill
Mentor
I'm sure it will all seem so simple when it's complete but at the moment i just don't know.

You mention "The transformations are applied to just a voltage source and its series impedance without regard to the rest of the circuit." so do i split it in to two? if so then what can i do with it? i'm not sure i understand what i'm supposed to be doing.

Going back to the initial circuit diagram you can see two voltage sources. Each has a series-connected impedance (j4 and j6 Ohms respectively) associated with it. Essentially each pair forms a Thevenin model, right?

You do a source conversion (Thevenin to Norton) to turn each of the voltage source / series impedances into their Norton equivalents, thus replacing the voltage sources with current sources in the circuit. This renders the circuit amenable to further simplification.

Yes, i see how we've now got two current sources and three impedances, but i just don't know what to do next.

Like i've said before i don't understand, please help me to understand because at the moment all i'm seeing are riddles!

I've resorted to watching a video on youtube - it's the most help i can get at this rate! - and i've calculated V1s Norton equivalent current as -145.71874-j162.61423 using the formula; (where j4 = R1, j6 = R2 and 35+j35.70714 = R3, V1=415)
In = Ij6 = (R3/(R1+R2)) * V1/(R1+((R2xR3)/(R2+R3)))

I don't know if i'm barking up the wrong tree?