Thevenin's Theorem: Solving Homework Statement on Load Current

[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.

 

[brabbit87]

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.

Thanks for the response. My figures are improved.

 

[Hndstudent]

Hi can someone give me a little bit of advice on question 1c please. The questions ask to convert the circuit in post #1 to be converted into a pair of current generators and then to determine the current flowing in the load. I should obtain the same value as that is question 1a and 1b correct? however... I have ended with a totally different answer.

I began by converting the two voltage sources into current sources by calculation v1/j4 and v2/j6. I then redrew my circuit so that the current sources were in parallel with both sources of inductance.

my course notes then implied that i added the current sources together and added the inductance together to obtain a norton equivalent circuit. which now has a current source of 103.75 - j69.167 and is in parallel with and inductance of j10

finally Il=I*(j10/j10+Rl)

where have i gone so wrong?

Thanks

 

[gneill]

How do inductors in parallel add?

Also, your numbers are hard to interpret unless you tell us your definitions for V1 and V2. They depend upon which supply you chose for providing the reference angle.

 

[Hndstudent]

How do inductors in parallel add?

Also, your numbers are hard to interpret unless you tell us your definitions for V1 and V2. They depend upon which supply you chose for providing the reference angle.

Hi Gneill, I am not too sure but it was just how tho learning material showed it. Even when if I use j4xj6/j4+j6=J2.4 I am still getting the wrong answer.

for voltage v1= SQRT2*415sin(100.PI.t+90) or 0+j415 v2= SQRT2*415sin(100.PI.t) or 415 +j0

Thanks

Nick

 

[gneill]

Well, your current from post #142 looks fine, and your new Norton impedance looks good. So what remains is your calculation of the load current. Perhaps post details of your calculation?

 

[Hndstudent]

Well, your current from post #142 looks fine, and your new Norton impedance looks good. So what remains is your calculation of the load current. Perhaps post details of your calculation?

i am calculation Il by using the equation #142, Il = I*(j2.4/j2.4+RL)

Il=(103.75 - j69.167)*[j2.4/j2.4+(35+j35.71)]

Thanks

 

[gneill]

That all looks fine. What was your result?

 

[Hndstudent]

annoyingly when calculating my brackets I wrote the answer down wrong! But thanks for making me go back and recalculating. (Y)

 

[David J]

Good Morning, I have started working on this question, question "a" to be precise and I was wondering if I could get some guidance. To date I have got to the following stage.
I had to convert the load to a complex impedance. I used $z=\sqrt {R^2+X^2}$
I know the magnitude of the load is $50\Omega$ and its angle is $45.57^0$ I got the angle by using $\cosh^{-1}$ of 0.7
In polar form it looks like $Zl=50\Omega\angle45.57$
I converted to rectangular using a calculator and got this: $Z=35\Omega + j35.71$
I then removed the load and voltage sources from the circuit and calculated the total impedance which I got to be $j2.4$
I would request someone please look at the attachment and advise if I am correct so far. I know its sometimes frowned upon to upload working out instead of using Latex but I don't know how to make diagrams on here. I have tried to make the drawing as clear as possible. I think I am on the right track having read through these posts on here and working through the learning materials I have been provided.
I know my next step is to work out the voltages
I am given
$V1=\sqrt2 (415) \cos (100 \pi t)$
and
$V2=\sqrt2 (415) \sin (100 \pi t)$
I understand $V1$ is leading $V2$ by $90^0$

My first question here is "what is or where does the $t$ come from? I have removed it from the equations below for the time being

So my second question is as follows.
$V1=\sqrt2 (415) \cos (100 \pi)$ so I get $586.89 \cos (100 \pi)$ which gives me an answer of $408.86 V$
$V2=\sqrt2 (415) \sin (100 \pi)$ so I get $586.89 \sin (100 \pi -90)$ which gives me $-408.86 V$

I don't think my values for V1 and V2 are correct and I think it maybe something to do with the missing $t$ and I was wondering if someone could explain and advise on the next step please or identify where I have gone wrong so far.

Appreciated

 

[gneill]

Your load impedance looks fine as does the work in your attachment for the Thevenin Impedance.

I know my next step is to work out the voltages
I am given
$V1=\sqrt2 (415) \cos (100 \pi t)$
and
$V2=\sqrt2 (415) \sin (100 \pi t)$
I understand $V1$ is leading $V2$ by $90^0$

My first question here is "what is or where does the $t$ come from? I have removed it from the equations below for the time being

The above are the time domain expressions for the voltages: they are AC voltage sources with an angular frequency of $100 \pi$ radians per second. The expressions give the voltage as a function of time.

You want to do your work in the frequency domain where reactive components have complex impedances and the sources can be represented by fixed phasor values.

So my second question is as follows.
$V1=\sqrt2 (415) \cos (100 \pi)$ so I get $586.89 \cos (100 \pi)$ which gives me an answer of $408.86 V$
$V2=\sqrt2 (415) \sin (100 \pi)$ so I get $586.89 \sin (100 \pi -90)$ which gives me $-408.86 V$

I don't think my values for V1 and V2 are correct and I think it maybe something to do with the missing $t$ and I was wondering if someone could explain and advise on the next step please or identify where I have gone wrong so far.

Appreciated

The given voltage source descriptions are time domain, peak voltages. You want to represent these as RMS phasors in order to work the problem (RMS is preferable for two reasons: 1: you'll be calculating power later on and RMS is the thing to use for that, and 2: it gets rid of the $\sqrt2$ terms on the voltages so the math is prettier!). The RMS part is easy to deal with, just drop the $\sqrt2$ from each of the voltage magnitudes. The trig functions and their arguments disappear too: phasors have implied angular velocity that you don't have to carry around in your math. All you need to worry about is that the relative phase of the two sources is respected. So your voltage phasors are both 415 V, but one will have a phase angle other than zero (as you said: $V1$ is leading $V2$ by $90°$).

You have a choice of changing the second voltage to a cosine, or the first voltage to a sine before turning them into phasors. As long as they are both using the same trig function then the all the subsequent math will be satisfied by their phasors.