Thevenin's Theorem: Solving Homework Statement on Load Current

[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.

 

[gneill]

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.

Okay, so that would be a correct value for the Thevenin voltage expressed as a peak value. Just drop the root 2's to make it an rms value. Presumably either result (peak or rms) should be acceptable since the original problem did not specify one or the other. You can leave the √2's in or out for all calculations. It's just a scaling constant. You can always compare a peak value to an rms result that was calculated by others by dividing your peak value by √2.

Peak vs rms will never affect the phase angle. So if your phase angle doesn't match an accepted result then you're probably in trouble somewhere in your calculations. If your peak value calculations yield a result that isn't the same as what others have found when you convert them to the same scale (peak or rms) then you need to look at your workings.

 

[Triopas]

Hello,

I am currently working through the same problem, question 1.(a).

Can anyone explain how we get this equation from nodal analysis?

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

Thanks.

 

[gneill]

Hello,

I am currently working through the same problem, question 1.(a).

Can anyone explain how we get this equation from nodal analysis?
Thanks.

Sure. First, can you describe what VL represents in your equation, and the precise circuit circumstances that you are analyzing?

 

[Triopas]

Okay, here goes;

I'm trying to work out the Thevenin's equivalent voltage across the load, which I am referring to as V_L.

Based on the notes that I have, the nodal analysis equation I end up with is:

$\frac {V_1 - V_L} {j4} + \frac {V_2-V_L}{j6} = \frac {V_L} {Z_L}$

Which seems correct as the voltage across the load is surely the sum of both sources?

But I see that others have used the equation I mentioned previously and got the correct answer, so I'm confused!

Thanks.

 

[The Electrician]

Okay, here goes;

I'm trying to work out the Thevenin's equivalent voltage across the load, which I am referring to as V_L.

Based on the notes that I have, the nodal analysis equation I end up with is:

$\frac {V_1 - V_L} {j4} + \frac {V_2-V_L}{j6} = \frac {V_L} {Z_L}$

Which seems correct as the voltage across the load is surely the sum of both sources?

But I see that others have used the equation I mentioned previously and got the correct answer, so I'm confused!

Thanks.

To solve using Thevenin equivalents, you must find the Thevenin voltage and Thevenin impedance. To find the Thevenin voltage, you remove the load ZL and find the voltage across the terminals where the load was. To do that, you solve:

$\frac { VTh-V_1} {j4} + \frac {VTh-V_2}{j6} =0$

where Vth is the voltage at the node where ZL was connected.

 

[Triopas]

Ahhh! Thank you! I was able to transpose that equation for V_Th and ended up with the answer I was expecting.

Thanks again.

 

[Spongecake]

I am currently on part c) and have converted each voltage source and series impedance into a current source and parallel impedance using Ohms Law, I = V/Z which i get 103.75 for v1 and -J69.17 for v2.
Then combining them to get 103.75-j69.17 as my voltage source.
Then to combine J4 and J6 which are in parallel which i make j2.4.

The next bit I'm slightly confused on which is to use current divider rule to calculate the load current which i think is Ix=[Rt/(Rt+Rx)]*It
It= the current source. Please can advise me if I'm on the right track.

 

[gneill]

I am currently on part c) and have converted each voltage source and series impedance into a current source and parallel impedance using Ohms Law, I = V/Z which i get 103.75 for v1 and -J69.17 for v2.
Then combining them to get 103.75-j69.17 as my voltage source.

I think you mean current source. You added the two Norton sources.

Then to combine J4 and J6 which are in parallel which i make j2.4.

The next bit I'm slightly confused on which is to use current divider rule to calculate the load current which i think is Ix=[Rt/(Rt+Rx)]*It
It= the current source. Please can advise me if I'm on the right track.

Yes, you're on the right track.

 

[bobmcgod]

Hello

I've been struggling with this question for a good three months now. I have read through this whole thread, been into see my tutor (distance learning so contact isn't regular) and got books out from the university library.

And I still don't have the slightest clue with where to even start. Not even a little bit. I have tried to go to other questions on the assignment but there's no luck there either. The course material I'm assuming is similar to that of the others posting on here as it's almost useless.

Attached is all I have so far.

Adam