Solve Superposition Theorem Homework: 50Ω Load

[Ebies]

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

Homework Equations

[/B]

The Attempt at a Solution

:[/B]

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...

 

[Ebies]

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

 

[gneill]

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...

 

[gneill]

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]

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 what's waaaay out. I used cos trig function in both thevenins and superposition theorem so can't really understand why the difference... let me quickly add the two currents again just to eliminate any possible error I could have made...

 

[Ebies]

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)

 

[Ebies]

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]

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.

 

[Ebies]

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?

 

[Ebies]

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

 

[Ebies]

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]

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.

 

[Ebies]

thanks gneill, you are a star

 

[KatieMariie]

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

 

[gneill]

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?

 

[KatieMariie]

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]

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?

 

[KatieMariie]

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

 

[gneill]

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?

 

[KatieMariie]

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 into two? if so then what can i do with it? I'm not sure i understand what I'm supposed to be doing.

 

[gneill]

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 into 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.

 

[KatieMariie]

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?

 

[KatieMariie]

You know what, scratch all of that. I'm barking up the wrong tree..

 

[gneill]

You should be converting the two voltage sources to current sources to begin with:

Once that's done a further simplification, thanks to parallel connections, should be obvious -- combining parallel components to reduce the circuit complexity.

Can you spell out the steps for converting a Thevenin voltage source to a Norton current source? Suppose you start with Vth = 415V and Zth = j6 Ω (Just like the voltage source on the right in the figure above).

 

[KatieMariie]

Converting the thevein voltage to nortons current involved using ohms law and the conjugate (complex number devision);
I = V/R = 415/j6 = (415/j6)*(-j6/-j6) etc etc?

To be fair i never saw the circuit as depicted above, thankyou.

Am i following the right tracks...
Steps I've taken so far;
Take the original equation, remove the load and replace with short, calculate current through short (In)
Remove all sources and replace with shorts, find Rn
Use In and Rn in the equivalent circuit, calculate current in the branch across the Load.

If this is so, I've made a mistake somewhere as the total currents flowing though the branches do not Sum up to In (which i imagine it should).

Do i repeat this again for the 2nd equivalent circuit? If so what do i do when I'm left with two figures?

I hope I'm grasping some basics.

 

[gneill]

Converting the thevein voltage to nortons current involved using ohms law and the conjugate (complex number devision);
I = V/R = 415/j6 = (415/j6)*(-j6/-j6) etc etc?

That's the idea. Do that for both of the voltage sources. What do you end up with for this example? ($I_N$ and $Z_N$)?

To be fair i never saw the circuit as depicted above, thankyou.

Am i following the right tracks...
Steps I've taken so far;
Take the original equation, remove the load and replace with short, calculate current through short (In)
Remove all sources and replace with shorts, find Rn
Use In and Rn in the equivalent circuit, calculate current in the branch across the Load.

If this is so, I've made a mistake somewhere as the total currents flowing though the branches do not Sum up to In (which i imagine it should).

Do i repeat this again for the 2nd equivalent circuit? If so what do i do when I'm left with two figures?

I hope I'm grasping some basics.

You don't need to remove loads or create shorts or any of that "Find the Thevenin Equivalent" algorithm stuff here. You are given a circuit where you can easily identify two Thevenin-type source models which you can convert directly to their Norton equivalents. The idea is to render the resulting remodeled circuit more amenable to further, easy simplification.

For the circuit in this problem that means first changing the voltage sources to current sources so that you end up with the circuit in the form of the bottom circuit in the image I gave above. Then you can simplify the circuit considerably thanks to the parallel connections, leaving a single Norton source (current source and its parallel impedance) and the separate load impedance.

At that point you can proceed in a couple of ways. You could apply the current division rule to determine the current through the load, or you might convert the Norton current source back to a Thevenin equivalent leaving you with a strictly series circuit comprising Vth, Zth and Z_load. I'd probably choose the latter since it is in keeping with the spirit of this part of the question: Convert the voltage sources; easily simplify down to a single current source; convert that source back to a voltage source and a trivial series circuit; solve for the current.

 

[KatieMariie]

Indevidually, -103.75 for I1, -j69.16 for I2, added together would be -103.75-j69.16(reccuring) -this would be my single norton current source?

So i have a current source in parallel with three impedances, "convert that source back to a voltage source and a trivial series circuit; solve for the current"

I'll go away and do some number crunching, then get back to you :)

 

[gneill]

Indevidually, -103.75 for I1, -j69.16 for I2, added together would be -103.75-j69.16(reccuring) -this would be my single norton current source?

The value magnitudes are good, but why did you end up with negative values? A voltage source will "want" to push current from it's + terminal, so the current source that replaces it should "want" to deliver current in the same direction.

So i have a current source in parallel with three impedances, "convert that source back to a voltage source and a trivial series circuit; solve for the current"

You can also combine the two Norton impedances into a single impedance...

 

[KatieMariie]

I've tried to follow these instructions but fell flat on my face. I don't know why they're negative, it's simply what the calculations told me.

I'm close to giving up with this.

 

[KatieMariie]

So I've now got a simplified circuit; current source In (103.75-j69.167) in parallel with Rn (j2.4) and the Load (35+j35.70714)

Now is it a case of converting this current source back into a voltage and a circuit re-drawn as a voltage source with Rn and L in series?

I want to confirm this before i spend another 2 days working on something that isn't going to work.

 

[gneill]

So I've now got a simplified circuit; current source In (103.75-j69.167) in parallel with Rn (j2.4) and the Load (35+j35.70714)

Yes!

Now is it a case of converting this current source back into a voltage and a circuit re-drawn as a voltage source with Rn and L in series?

Yes!

 

[KatieMariie]

waheey, progress made!

there is hope for me yet...

 

[KatieMariie]

Hi gneill,

I've been away on work for a while and have come back to this question.

I've tried the above, as in converted the current back into a voltage source ( via ohms law, so now V = 0.601355-j17.82945 ?!) and then used ohms law to calculate the current in the series circuit, but it doesn't add up. Were am i going wrong?

 

[gneill]

Hi gneill,

I've been away on work for a while and have come back to this question.

I've tried the above, as in converted the current back into a voltage source ( via ohms law, so now V = 0.601355-j17.82945 ?!) and then used ohms law to calculate the current in the series circuit, but it doesn't add up. Were am i going wrong?

Can you show some of your calculations and intermediate values?

 

[KatieMariie]

So i have 103.75-j69.16 for the current source, as stated above, and also have the circuit written out as the current source with Rn on one parallel branch, and the Load on the other.

To convert the current to voltage i used (103.73-j69.16)*(j2.4) = 165.98 + j248.95, so this becomes the voltage and the circuit is then rewritten in series. Using ohms law, the load resistance and this voltage, the current value comes out at 5.87 + j1.11, which is close, but not exactly what I'm looking for?

 

[gneill]

So i have 103.75-j69.16 for the current source, as stated above, and also have the circuit written out as the current source with Rn on one parallel branch, and the Load on the other.

To convert the current to voltage i used (103.73-j69.16)*(j2.4) = 165.98 + j248.95, so this becomes the voltage and the circuit is then rewritten in series. Using ohms law, the load resistance and this voltage, the current value comes out at 5.87 + j1.11, which is close, but not exactly what I'm looking for?

Your method seems to be fine, so perhaps you're losing significant figures by truncating / rounding intermediate values?

Don't round intermediate values and carry two or three extra decimal places ("guard digits") in intermediate values of a multistage calculation.

 

[KatieMariie]

Hi, thankyou, that seemed to be the case, I've now finised the question. Thanks for you help. :)

 

[js3]

I have completed the part b) successfully this morning after using the rms values like myself and gneill discussed in a separate thread. I am now on the final part c).
Is it really as simple as
IL1=j415/j4
IL1 = 103.75A

And thus the same calculation for IL2 and its series j6 impedance?
Then draw them in parallel with the 50ohm load?

 

[gneill]

I have completed the part b) successfully this morning after using the rms values like myself and gneill discussed in a separate thread. I am now on the final part c).
Is it really as simple as
IL1=j415/j4
IL1 = 103.75A

And thus the same calculation for IL2 and its series j6 impedance?
Then draw them in parallel with the 50ohm load?

Yes, that's what's depicted in post #27

 

[js3]

So To confirm my result, i have added the two current sources, and multiplied them by Zt.

103.75-j69.16 x j2.4 = 165.984+j249V

Then dividing the answer by the load impedance 35+35.71 I get 5.88+j1.115A
Am I correct in going on to draw this current value in parallel with the j2.4ohm and again in parallel with the 35+35.71ohm impedances respectively?

 

[gneill]

So To confirm my result, i have added the two current sources, and multiplied them by Zt.

103.75-j69.16 x j2.4 = 165.984+j249V

Okay. If you'd kept more significant figures during intermediate steps you'd have obtained 166 + j249 V. But close enough! You have now reduced the sources to a single Thevenin source:

Then dividing the answer by the load impedance 35+35.71 I get 5.88+j1.115A

No. Note that Zth and ZL are in series with the source.

Am I correct in going on to draw this current value in parallel with the j2.4ohm and again in parallel with the 35+35.71ohm impedances respectively?

No. You're done when you find the current in the load (paying attention to what the circuit looks like as in the figure above).

However, if you wish to keep to the spirit of part (c), you'll just combine the current sources into a single current source and combine their parallel impedances into a single impedance yielding a single Norton source driving the load impedance:

Then use the current division rule to find the load current.

 

[js3]

i used the latter as you suggested
In x (Zn/(Zn+Zl)).

Came out with IL= 5.7136+j0.8921A
Some rounding errors have occurred but will iron those out when i write up in neat. Thanks for your help gneill.

 

[GeorgeSparks]

Hi Guys,

I am not going to lie I am stuck! I have followed numerous different methods now and still coming up with the closest answer of 5.617919153+j1.37425425A.
I have also been using the method from http://www.electronics-tutorials.ws for both Thevenin and Norton equivalents. Also Superposition I am getting a completely different answer its driving me nuts. I am following these step and step and coming up with completely different answers to my first Current from the Thevenin circuit of part a which seems correct according to previous posts.
For this answer in regards to Current in the Norton circuit I have used the formula for current division in the last comment posted In x (Zn/(Zn+Zl)) giving me 5.617919153+j1.37425425A.
All of my calculations have been done in complex rectangular form I have In = 103.75-j69.16666667, Zn = j2.4 and Zl = 35+j35.71.

Cheers
George

 

[gneill]

Hi Guys,

I am not going to lie I am stuck! I have followed numerous different methods now and still coming up with the closest answer of 5.617919153+j1.37425425A.
I have also been using the method from http://www.electronics-tutorials.ws for both Thevenin and Norton equivalents. Also Superposition I am getting a completely different answer its driving me nuts. I am following these step and step and coming up with completely different answers to my first Current from the Thevenin circuit of part a which seems correct according to previous posts.
For this answer in regards to Current in the Norton circuit I have used the formula for current division in the last comment posted In x (Zn/(Zn+Zl)) giving me 5.617919153+j1.37425425A.
All of my calculations have been done in complex rectangular form I have In = 103.75-j69.16666667, Zn = j2.4 and Zl = 35+j35.71.

Cheers
George

Hi George, Welcome to PF.

In all cases you need to show your solution attempt work in detail so that helpers can see what you've done right and perhaps spot where you've gone wrong. We (usually) can't tell what happened from a numerical answer alone.

Since your $Z_n$ and $I_n$ values look good as does your $Z_L$ value, it's odd that your current division result went awry. You might want to review your calculation there.

 

[GeorgeSparks]

Hi gneill,

Please find attached my working for superposition where I end up with an answer of 5.726421015 + j1.12310464 A. I must be doing something incorrect in my calculations as I end up with a different answer for Thevenin, Superposition and Norton haha :/ any help is much appreciated!

 

[gneill]

Somehow your determination of the parallel combination of R2 and R3 has ended up being equal to R3. Can't be true unless R2 is an open circuit (infinite real impedance). This has thrown off your value for VL1.

 

[GeorgeSparks]

Somehow your determination of the parallel combination of R2 and R3 has ended up being equal to R3. Can't be true unless R2 is an open circuit (infinite real impedance). This has thrown off your value for VL1.

Thank you for the guidance gneill I must have finger bashed my calculator on the first part. All looks as it should now