Solve Superposition Theorem Homework: 50Ω Load

[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

 

[gneill]

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

Excellent

 

[Student12345]

Hi I was hoping someone can help me with part c)
I seem to be getting negative current and I'm not sure if this is correct
So far I have
I1= v1/j4
415/j4
(415)(-j4)/(j4)(-j4)
-j1660/16
I1= -j103.75

I2=v2/j6
-j415/j6
(-j415)(-j6)/(j6)(-j6)
-2490/36
I2=-69.167

Not sure where to go after this can I add the 2 currents together and add the 2 parallel reactances (1/x1+1/x2=1/xtotal)?

 

[gneill]

Not sure where to go after this can I add the 2 currents together and add the 2 parallel reactances (1/x1+1/x2=1/xtotal)?

Sure.

 

[Student12345]

Sure.

I have got an answer but it doesn't match my answers from a) and b) can you please guide me in the direction of where I am going wrong?

My calculations give I1 =-j103.75 and I2=-69.167 as shown in the post above. I then formed Zn=-j2.4 which I believe is correct.

I then used IL= In * Zn/Zn+Zl and I think this is where I went wrong.
IL= (-69.167-j103.75)* (-j2.4/(-j2.4*(35+J35.71)))

IL= (-69.167-j103.75)*(-j2.4/(85.704-j84)

IL= (-69.167-j103.75)*(0.01399-j0.01428)

IL=-0.51-j0.46

Obviously I have calculations inbetween.

 

[gneill]

I have got an answer but it doesn't match my answers from a) and b) can you please guide me in the direction of where I am going wrong?

My calculations give I1 =-j103.75 and I2=-69.167 as shown in the post above. I then formed Zn=-j2.4 which I believe is correct.

The currents are good but the sign on your combined impedance is not.

I then used IL= In * Zn/Zn+Zl and I think this is where I went wrong.
IL= (-69.167-j103.75)* (-j2.4/(-j2.4*(35+J35.71)))

IL= (-69.167-j103.75)*(-j2.4/(85.704-j84)

You have multiplied the impedance terms in the denominator rather than adding them!

 

[Student12345]

The currents are good but the sign on your combined impedance is not.

You have multiplied the impedance terms in the denominator rather than adding them!

Haha so I did. Thanks for that!

Zn I got from
1/Zn = 1/z1+1/z2 = 1/j4+1/j6
=(-j4/(j4)(-j4))+(-j6/(j6)(-j6))
=(-j4/-j²16)+(-j6/-j²36)
=(-j4/16)+(-j6/36)
1/Zn=-j0.417 Zn=-j2.4
Where have I gone wrong?

 

[gneill]

Haha so I did. Thanks for that!

Zn I got from
1/Zn = 1/z1+1/z2 = 1/j4+1/j6
=(-j4/(j4)(-j4))+(-j6/(j6)(-j6))
=(-j4/-j²16)+(-j6/-j²36)
=(-j4/16)+(-j6/36)
1/Zn=-j0.417 Zn=-j2.4
Where have I gone wrong?

Your last step where you take the reciprocal does not handle the j properly.

 

[Student12345]

Your last step where you take the reciprocal does not handle the j properly.

Ok thanks I'm a little confused because normally if I have a figure say -j²35.71² I will change the sign and drop the j² to give +35.71²but it doesn't seem to be right for this one

 

[gneill]

$Zn = \frac{1}{-j 0.417} = \frac{2.4}{-j} \cdot \frac{j}{j} = \frac{j 2.4}{+1} = j 2.4$

The bottom line being that when you "move" a j from denominator to numerator (or vice versa) you change its sign

 

[Student12345]

$Zn = \frac{1}{-j 0.417} = \frac{2.4}{-j} \cdot \frac{j}{j} = \frac{j 2.4}{+1} = j 2.4$

The bottom line being that when you "move" a j from denominator to numerator (or vice versa) you change its sign

Thanks so much I think I've been staring at this question so long it's driven me crazy!