Thevenin's Theorem: Solving Homework Statement on Load Current

[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