Thevenin's Theorem: Solving Homework Statement on Load Current

[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.

 

[David J]

Ok thanks for the reply. I looked at an earlier post (81) where this was discussed.

If I change $V1$ to sine
$V1= \sqrt2 (415) \sin (100\pi + 90)$ I add $90$ because it leads $V2$ by $90^0$
$V2$ remains the same
$V2 = \sqrt2 (415) \sin (100 \pi)$

If I then remove the $\sqrt2$ from each equation this also removes the trig function and arguments. In this case the $\sin$ function and $100 \pi$ will be removed from each equation leaving

$V1=415 \angle 90$
$V2=415 \angle 0$

Is this correct?

Do I now have to change the above to rectangular form ??

 

[gneill]

That looks good. Whether or not you'll need their rectangular form will depend upon the mathematical operations you want to perform. It's the usual complex number shuffle...

 

[David J]

I need to find the $Vth$ of the circuit. I have read some of the past posts and I see in some cases they have converted:

$V1=415V \angle 90^0$ to $0 +j415$
$V2=415V \angle 0^0$ to $415 +j0$

so $V1 =0 +j415$ and $V2= 415 +j0$

So can I just make these voltages as follows:-

$V1 = j415V$
$V2 = 415V$

What I am trying to understand now is how to get the $Vth$ from these values. What do I have to do to proceed? I have already worked out the $Zt$ to be $j2.4$ but do I go back now and use the individual values $j4$ and $j6$ in some current calculations? I feel confident I can do the math, its just knowing exactly what math I need to do that is confusing me.

Thanks again

 

[gneill]

You proceed as you would for any circuit: To find the Thevenin voltage you remove the load and find the potential across the open terminals. How you accomplish that is up to you; there are several ways to analyze the circuit. Nodal analysis appears to be a straightforward choice...

 

[David J]

Thanks again, I have no experience of Nodal analysis but I will try to find some info on this

 

[David J]

Good Morning, I started looking at Nodal analysis last night. I have no experience of this and I was wondering if you could take a look at the attachment and advise if I am on the right track or not. I think its correct but I am unsure how to proceed and was wondering if you could show me how this is done. I am not looking for the answer, I am just trying to learn how to correctly carry this out using Nodal analysis.

Thanks

 

[gneill]

Well, you've identified the sole essential node, which is good. Only one essential node means there's only one node equation to write.

For this problem you're looking for $V_{th}$, which means removing the 50 Ω load and finding the potential across the open terminals.

In this circuit the node voltage happens to coincide with the voltage you're looking for, so that's very handy!

Your next step is to write the node equation. There are only two current carrying paths to the node, so it shouldn't hard to sum the currents...

 

[David J]

Thanks, I have been studying online tutorials for this and started trying to write the equation but everything I have seen has been shown with (in the case of this circuit) the $50 \Omega$ resistor "in circuit".

$I1 +I2 + I3 = 0$

$I1 = \frac {415-Vth} {4}$ $I2 = \frac {-Vth} {50}$ $I3 = \frac {415-Vth} {6}$$I1 = \frac {415-Vth} {4}$ $+$ $I2 = \frac {-Vth} {50}$ $+$ $I3 = \frac {415-Vth} {6}$ $=0$

If I remove the $50 \Omega$ then this equation above is of no use

 

[gneill]

First, you MUST remove the load to find the Thevenin voltage. It's the open-circuit voltage after all.

Second, your node equation will not be useless! It will give you Vth. There will be two terms, one for each "branch" that can carry current leading to the node.

Note that you must use the complex values for the component quantities when you write your node equation. Your V1, for example, is 415j. And the two inductor impedances are 4j and 6j respectively.

 

[David J]

Ok I have attempted it as shown below

$\frac {V1-VL} {j4}$ $=$ $\frac {VL-V2} {j6}$

so $\frac {j415-VL} {j4} = \frac {VL-415} {j6}$

$(j415-VL)(j6) = (VL-415)(j4)$

$j2490-j6VL = j4VL-1660$

$j2490 + 1660 = j10VL$

$VL = \frac {j2490} {10} + \frac {1660}{10}$

so $VL = j249 + 166$

I think this is just about correct

I need to find the current through the load so I used $iL = \frac {Vth} {Zth + ZL}$

I added $j249$ and $166$ and converted to polar and this gave me $299.26 \angle56.310$

I added $Zth = j2.4$ and $ZL = 35 +j35.7$ and this gave me $35 + j38.1$ Converted to Polar this is $51.74 \angle47.43$

I then did the calculation $\frac{299.26\angle56.310} {51.74\angle47.43}$ and this resulted in $5.78\angle8.88$

Convert this back to rectangular i get $5.71 + j0.892$

I feel this is the correct value of load current

 

[gneill]

That looks good. Nicely done.

 

[David J]

Thanks for your advice with this, much appreciated

 

[Davey345]

I wonder if someone could point me in the right direction i have completed the first part and obtained
IL = (15296.9 + j2390.4) / 26761 = 5.71503 + j0.89037
I have being doing the superposition part and after doing the calculation about 6 times i have come to the conclusion i have the original equation wrong
I have used from the 2 voltage sources (V1=j415 and V2 =415) where ZL = 35+35.7j from part a
ILH = j415/ (j4 + ((j6)(ZL))
IRH = 415 / j6 + ((j4) + ZL)
the answers i obtained were
ILH = 43.61 +1.95j
IRH = 0.868 = 42.43j

I am not sure where i have gone wrong, i am not looking for the answer just a point in the right direction please

 

[gneill]

Hi Davey345, Welcome to Physics Forums!

Can you show more detail for how you arrived at your expressions for ILH and IRH?

Remember, when you suppress a voltage source you replace it with a short circuit (wire), but anything in series with that source is still part of the circuit.

 

[Davey345]

Sorry the images ore not fantastic, i am still in the rough working out stages. I have put a short circuit across the Voltage source V1 and taken the currents from that source and repeated the exercise from V2 i believed.

 

[Davey345]

The equation comes from the Current division in a circuit, i have said that the current through the circuit from the source is
J6 and ZL are is series and in parallel to j4 so with current divider rule resistors and impedances in parallel are the product of the two resistors over the sum.
and then i have used ohms law to obtain an equation for the current

 

[gneill]

I'm not so much interested in the mechanics of the complex number handling as the methods/steps taken.

The equation comes from the Current division in a circuit, i have said that the current through the circuit from the source is
J6 and ZL are is series and in parallel to j4 so with current divider rule resistors and impedances in parallel are the product of the two resistors over the sum.
and then i have used ohms law to obtain an equation for the current

I'm not following your description. It looks like something got lost in the editing.

Lets deal with the active V1 source first. If V2 is suppressed then ZL and j6 are in parallel, and j4 is in series with that parallel pair. You can find the total current as an intermediate step, then apply current division for the parallel pair to find the load current. Is that more or less what you've done?

 

[gneill]

The currents that you gave in post #163 correspond to the source currents. I can confirm that they are good (or close enough) values for those currents.

All that would remain is to apply current division to each to pull out the load branch currents and then sum the results.

 

[Davey345]

Oh Thank you, I forgot that last step.

 

[Davey345]

I have a query as to why I have subtle different answers, I understand rounding errors but I have a different j component to my last answer, albeit it only marginal.
I used Vth as 166+249j, and RL as 35+35.7j giving Zth of the load as j2.4ohms.
I calculated the I short-circuit as 103.75-69.166j, which enabled me to calculate Zth of the norton circuit of 0.07532+2.31801j,
then using I = Vth/ (Zth+RL) I get I= 5.71413+0.90549j
my confusion is for parts a i get I = 5.71503 + 0.89037j
part b I= get 5.7152+0.89295j
is this acceptable or have I made a mistake ?

 

[gneill]

I have a query as to why I have subtle different answers, I understand rounding errors but I have a different j component to my last answer, albeit it only marginal.
I used Vth as 166+249j, and RL as 35+35.7j giving Zth of the load as j2.4ohms.
I calculated the I short-circuit as 103.75-69.166j, which enabled me to calculate Zth of the norton circuit of 0.07532+2.31801j,

Not sure what "Zth of norton circuit" is. The Norton impedance should be identical to the Thevenin impedance for a given circuit.

then using I = Vth/ (Zth+RL) I get I= 5.71413+0.90549j
my confusion is for parts a i get I = 5.71503 + 0.89037j
part b I= get 5.7152+0.89295j
is this acceptable or have I made a mistake ?

This is a tough call. I know that when I solved by the individual methods I kept all intermediate values to full precision (no truncation or rounding other than by the limitations of the calculator/software) and all my results agreed to three decimal places. I didn't bother comparing more digits than that at the time.

Pulling up my spreadsheet I can confirm that all my results are identical to at least six decimal places. For the load current, to six places, I have:

$I_L = 5.714530 + 0.892446j~A$

 

[The Electrician]

This particular problem is quite popular. People have been asking for help for at least a couple of years.
Just for reference, here's the exact result, and its 20 digit approximation:

 

[js3]

Hi, i have been following the forum and have a query. I have calculated my answers in peak i.e. keeping the √2 on my voltage sources V1 and V2.
Did i need to change that?
My answer for part a) IL is 5.78<9.42°

My answer for part b) is very similar, just wondered whether i have the correct method?

 

[gneill]

Hi, i have been following the forum and have a query. I have calculated my answers in peak i.e. keeping the √2 on my voltage sources V1 and V2.
Did i need to change that?
My answer for part a) IL is 5.78<9.42°

If that's a peak value, then it doesn't look right. Should be closer to 8 amps if the rms value is about 5.8 amps (as found by others). Also, the angle should match what others find regardless of whether the result is peak or rms.

You can use peak values to do the calculations. The only place you might run into issues is if you were asked to calculate power values. Then you'd want to use rms values for the voltages and currents. You can always convert sinusoidal voltages and currents to their rms values later. It's just a scaling factor.

My answer for part b) is very similar, just wondered whether i have the correct method?

You'd need to post the details of your work.

 

[js3]

Thankyou for your response.
When i say i have left them as peak, i mean that for V1 i have used j415x√2. And for V2 i have used 415x√2.
So using nodal analysis part a) leaves me with Vt = 166√2+j249√2Did i just need to leave the √2 out is my question? Because if that's the problem, my mistake carries over to part b.