Thevenin's Theorem: Solving Homework Statement on Load Current

[Gremlin]

Thanks.

 

[Gremlin]

Thevenin equivalent voltage is

Vth = 6j * i + v2
Vth = 299.26∠56.31

What formula have you used to calculate this?

 

[NascentOxygen]

What formula have you used to calculate this?

Are you asking how to go from this rectangular form: Vth = 6j * i + v2
to this polar form: Vth = 299.26∠56.31 ?

Or are you asking about the substituted values for i and v2?

 

[Gremlin]

No i can convert rectangular to polar, I'm looking to find out how you go about getting Vth when you've multiple sources of voltage.

 

[Gremlin]

It's ok, I worked it out. This site was useful for anyone struggling:

http://www.electronics-tutorials.ws/dccircuits/dcp_7.html

 

[Gremlin]

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:

(a) applying Thévenin’stheorem
(b) applying the superposition theorem
(c) by transforming the two voltage sources and their associated
reactances into current sources (and thus form a pair of Norton
generators)
View attachment 74253
[/B]

Would i be correct in saying that i = 5.71 + j0.892 A?

That's from Vth / (Rth + Rl) = 166 + j249 / (0 + j2.4 + 35 + j35.71).

 

[The Electrician]

That's correct; now all you have left is part (b) and (c).

 

[MrBondx]

Here is a link for anyone still struggling, it will cover the grey areas. I found it helpful, I hope you will too. Check out the 2nd/last example he gives -http://youtu.be/25axDabtoFk

 

[KatieMariie]

How did you guys get on with part c? I've got so far as a refined circuit with a current source in parallel with Rn (j2.4) and the 35+j35.707 Load, but can't seem to get past this as my numbers don't add up! I'm not getting the same as i did in a or b?

 

[The Electrician]

How did you guys get on with part c? I've got so far as a refined circuit with a current source in parallel with Rn (j2.4) and the 35+j35.707 Load, but can't seem to get past this as my numbers don't add up! I'm not getting the same as i did in a or b?

What is the value of the current source? How did you determine it? It''s easier to see where you've gone wrong when you show your work.

 

[KatieMariie]

Hi, apologies for the lack of detail in the first post.

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?

 

[KatieMariie]

Hi, apologies for the lack of detail in the first post.

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?

Sorry, not 'as above', that applies to another thread!

 

[The Electrician]

With the angle and the magnitude you have the load impedance in polar form. Thus
$$Z_L = 50 \Omega ~~\angle 45.57°$$
You can convert this to rectangular form if you need it that way.

By the way, before you dive into the calculations for the Thevenin equivalent, take a close look at the definitions of the voltage supplies. While they both have the same frequencies they won't have the same phase angle (why is that do you think?).

Hi, apologies for the lack of detail in the first post.

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?

You need to include the j2.4 ohms in series with the load impedance, all driven by the voltage of 166 + j249 volts.

 

[KatieMariie]

Thankyou, I've realized my mistake, I've had the correct method all along, just one of my values was out by 0.1. Typical.

Thanks again.

 

[Gremlin]

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:[/B]

(c) by transforming the two voltage sources and their associated
reactances into current sources (and thus form a pair of Norton
generators)
View attachment 74253

Straight question: would it be seen as incorrect here if i just converted the thevenin's equivalent voltage that i calculated in thevenin's equivalent circuit in part a) using the I_N = V_TH/R_TH formula?

 

[The Electrician]

Straight question: would it be seen as incorrect here if i just converted the thevenin's equivalent voltage that i calculated in thevenin's equivalent circuit in part a) using the I_N = V_TH/R_TH formula?

That would be incorrect, because you only had a single Thevenin equivalent for part a). You must convert both sources separately into their Norton equivalents.

 

[gneill]

Straight question: would it be seen as incorrect here if i just converted the thevenin's equivalent voltage that i calculated in thevenin's equivalent circuit in part a) using the I_N = V_TH/R_TH formula?

That would depend upon the objective of the exercise. If you just want an answer, then that's fine. If you are expected to demonstrate a different solution approach, then no, you should start from scratch and solve the circuit by this alternative method.

 

[Gremlin]

Ha. I didn't think they'd let me off that easily.

 

[Birchyuk]

Hello everyone,

I am really struggling with this question also, i have an answer for (a) and i have two methods of solving (b) but i keep getting different answers. I do not know if i have just made a simple mistake in my calculations or i am just completely wrong. I have been stuck on this for 5 days straight so any help would be greatly appreciated. I have also attached my workings out.

Many thanks in advance.

 

[The Electrician]

Hello everyone,

I am really struggling with this question also, i have an answer for (a) and i have two methods of solving (b) but i keep getting different answers. I do not know if i have just made a simple mistake in my calculations or i am just completely wrong. I have been stuck on this for 5 days straight so any help would be greatly appreciated. I have also attached my workings out.

Many thanks in advance.

I have trouble reading your work because your images are low contrast and somewhat blurry.

Part (b) of this problem asks you to solve for the current in the load using the superposition theorem. This means you should solve the problem twice, once for each source.

I think you would make better progress to start over and show only one part of your solution at a time. Why don't you replace the right hand source with a short circuit and solve for the current in the load with only the left hand source active. You are allowed to use any method you choose, but tell us what method you're using and show your work.

 

[Birchyuk]

Hello and thanks for your reply.

I apologise for the images, i hope the attached are easier to read.

Ok so i calculated Thevenins voltage ( i hope is correct) and it is my understanding that i should have a similar answer for question b.

So for method 1 my thinking was:

Remove v2 and calculate the impedance of j6 and Z-load
Divide the voltage between j4 and (j6+Z-load) which will give me the voltage on the load
Then calculate the current by V-load/Z-load

Same process but for V2 and then add the currents.

For method 2

Remove v2 and calculate current a which in turn will allow me to calculate current b from a current divide.
Then,
Remove v1 and calculate current d which in turn will allow me to calculate current f from a current divide.

Add these together to calculate current on the load.

I hope this explains a little better and i hope you can ready my working out now.

 

[The Electrician]

Hello and thanks for your reply.

I apologise for the images, i hope the attached are easier to read.

Ok so i calculated Thevenins voltage ( i hope is correct) and it is my understanding that i should have a similar answer for question b.

So for method 1 my thinking was:

Remove v2 and calculate the impedance of j6 and Z-load
Divide the voltage between j4 and (j6+Z-load) which will give me the voltage on the load
Then calculate the current by V-load/Z-load

Same process but for V2 and then add the currents.

For method 2

Remove v2 and calculate current a which in turn will allow me to calculate current b from a current divide.
Then,
Remove v1 and calculate current d which in turn will allow me to calculate current f from a current divide.

Add these together to calculate current on the load.

I hope this explains a little better and i hope you can ready my working out now.

I see a couple of numerical errors in your complex arithmetic where you are working out the Thevenin method--part (a).

Where you are getting ready to calculate the Thevenin voltage, at one point you have:

$\frac{415 + j415}{j10}\frac{-j10}{-j10}$

But then you get $\frac{-j4150+4150}{-100}$

The sign of the denominator is wrong. This error is propagated and gives wrong results later on.

On the next page you have this calculation:

-j415 + (41.5-j41.5) * (0+j6)

which then becomes:

j415 + (0+j249+0+249) What happened to the minus sign in front of j415?

and then:

249 + j166 Which should be 249 - j166

This error also propagates and gives wrong results later.

Because of these errors, you have a wrong result for the current in ZL. You then compare this to the result from the superposition method and conclude that the result from the superposition method is wrong.

I looked over the calculations for your first method using superposition, and it looks like you have a good result, but you aren't carrying enough digits in your complex arithmetic, so your results are a little off.

I didn't check your results for the 2nd method using superposition; fix these errors first.

 

[NascentOxygen]

I apologise for the images, i hope the attached are easier to read.

If you wish to ask for help with further homework questions, you must avoid attaching images of working. You need to learn Latex and present only the essential working and include it within your post. It is unfair to expect helpers to plough through screensful of difficult-to-read handwriting. Make no mistake, ALL images of handwriting are difficult to read on small screens in typical lighting.

Click on INFO to find a link to an introduction to Latex. There is a host of online tutorials, too.

 

[Birchyuk]

Firstly NascentOxygen, apologies and hopefully my effort below has come out ok!

The Electrician- Thanks for your help and a little embarrassed that I missed that.

I have re-done calculations and I have the below, does this seem better? I will need to go through my superposition theory again

$(415+j415)/j10*(-j10/-j10)$
$=(-j4150+4150)/100$
$=41.5-j41.5$

To find Thevenins Voltage
$j6*I+V2$
$-j415+(41.5-j41.5)*(0+j6)$
$-j415+(0+j249+249)$
$=249-j166$

To find current through the load

$I=Vth/(Zth+ZL)$
$249-j166/j2.4+(35+j35.7)$
$(249-j166/35+j38.1) * (35-j38.1/35-j38.1)$
$(8715-j5810-j9486.9+j^2 6324.6)/(1225-j1333.5+j1333.5-j^2 1451.61)$
$(2391-j15296)/2676$
$=0.89-j5.72$

 

[The Electrician]

Firstly NascentOxygen, apologies and hopefully my effort below has come out ok!

The Electrician- Thanks for your help and a little embarrassed that I missed that.

I have re-done calculations and I have the below, does this seem better? I will need to go through my superposition theory again

$(415+j415)/j10*(-j10/-j10)$
$=(-j4150+4150)/100$
$=41.5-j41.5$

To find Thevenins Voltage
$j6*I+V2$
$-j415+(41.5-j41.5)*(0+j6)$
$-j415+(0+j249+249)$
$=249-j166$

To find current through the load

$I=Vth/(Zth+ZL)$
$249-j166/j2.4+(35+j35.7)$
$(249-j166/35+j38.1) * (35-j38.1/35-j38.1)$
$(8715-j5810-j9486.9+j^2 6324.6)/(1225-j1333.5+j1333.5-j^2 1451.61)$
$(2391-j15296)/2676$
$=0.89-j5.72$

Your result for the current through the load is correct, but you have not got very many significant digits. When you solve by another method, your final result may be different even in the 2 digits after the decimal point, and you may then think you have made a mistake.

I do these kind of numerical calculations on a HP50 calculator which does 12 significant digit complex arithmetic. The result I get for the load current is .892446 - j5.71453

You should carry a few more digits in the imaginary part of the load impedance; Z_L = 35 + j35.707142

Your work for the first method using superposition appears correct, but again, you don't have many digits in your final result.

 

[vela]

Your result for the current through the load is correct, but you have not got very many significant digits. When you solve by another method, your final result may be different even in the 2 digits after the decimal point, and you may then think you have made a mistake.

This is another good reason to plug numbers in only at the end. If you want an accurate final answer, you don't want to round off after each step, which means you have to keep extra digits around, which means a lot more writing and an increased chance of making a mistake.

In any case, you should ideally keep all the digits on intermediate calculations and round off the final answer to the correct number of sig figs.

 

[Birchyuk]

Hello,

Apologies for the delay in replying. Thank you for your help, I will carry out the calculations again with your recommendations and hope all works out (fingers crossed).

Many thanks, really appreciated.

 

[brabbit87]

hi, alittle help please.
im currently trying to calculate the current flow via v1 - v2 / z1 + z2 ... but what i am getting after the complex arithmetic is I = -41.5 + j41.5. which is different to others.

Is this purely because i have calculating the current to flow to be clockwise ie the polarity of the the circuit?

 

[gneill]

hi, alittle help please.
im currently trying to calculate the current flow via v1 - v2 / z1 + z2 ... but what i am getting after the complex arithmetic is I = -41.5 + j41.5. which is different to others.

Is this purely because i have calculating the current to flow to be clockwise ie the polarity of the the circuit?

You need to show more details of your work. What are the values of your parameters?

Also, be sure to employ parentheses (brackets) to disambiguate equations that you type out, thus clarifying the order of operations. You probably meant to write:

(v1 - v2)/(z1 + z2)

 

[The Electrician]

hi, alittle help please.
im currently trying to calculate the current flow via v1 - v2 / z1 + z2 ... but what i am getting after the complex arithmetic is I = -41.5 + j41.5. which is different to others.

Is this purely because i have calculating the current to flow to be clockwise ie the polarity of the the circuit?

Have a look at post #104 and 105.

 

[brabbit87]

My apologies.

i did essentially a kvl walk clock wise around the circuit with the load removed, ( to calculate the circuit current) then transformed for the value of the current

V1 - j4*i - j6*i - V2 = 0

V1 - V2 = j4*i + j6*

j415 - 415 = i(j4 + j6)

(-415 + j415) / j10 = i

multiplying by conjugate - j10

(-415 + j415)(-j10) / (j10)(-j10) = j4150 - 4150j^2 / -100j^2

Since j^2 is = to -1

(j4150 + 4150) / 100 so current = 41.5 + j41.5

however i noticed that people have a negative value for the imaginary component of the complex number.

so I am not sure if i am going wrong with the calculation,or the initial circuit equation i made up.

 

[brabbit87]

Have a look at post #104 and 105.

i made a typo. the complex value for the current i have is (415 + j415).. which is somewhat different to others values people have got.

i have also just put the values of the equation into the wolfram calculator and have the same values that i have just shown. i think i possibly did something wrong with my circuit equation.

 

[The Electrician]

i made a typo. the complex value for the current i have is (415 + j415).. which is somewhat different to others values people have got.

i have also just put the values of the equation into the wolfram calculator and have the same values that i have just shown. i think i possibly did something wrong with my circuit equation.

See post #66.

You can choose either V1 or V2 for the phase reference. Either will work, but it will change results.

 

[brabbit87]

See post #66.

You can choose either V1 or V2 for the phase reference. Either will work, but it will change results.

sorry I am not sure i follow. is there a reason i should pick V1 over V2 as the phase reference.

 

[The Electrician]

sorry I am not sure i follow. is there a reason i should pick V1 over V2 as the phase reference.

There's no particular reason. V1 is a cosine wave and V2 is a sine wave. Either will do as the reference. Some people might like sine waves rather than cosine waves, or vice versa. But, the phase you get for the various currents and voltages will depend on your reference. The sign of real and imaginary parts will be different depending on which reference you choose.

If you use the sine waveform as a reference, then V1 = 415 j volts and V2 = 415 volts. If the cosine waveform is your reference then V1 = 415 volts and V2 = -415 j volts. Either will work; this is what's mentioned in post #66 and 67.