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supernode and finding Vth

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PG1995

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Hi :smile:

Please have a look on the following link:
https://img163.imageshack.us/img163/4862/img0006rw.jpg

We want to find Vth for the given circuit using nodal analysis. We need to create a supernode. I have drawn the directions the currents. How do I proceed? Please help me out. Thank you

Cheers
PG
 

jony130

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Simply you wrote one KCL equation, and you treat V1 and Vo as a one node.


10A = V1/6Ω + Vo/2Ω

And additional equation

V1 = Vo + 2*Vo

So we have

10A = (Vo + 2Vo)/6Ω + Vo/2Ω

Vo = Vth = 10V
 
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PG1995

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Thank you, Jony.

1: I was just wondering that whether you followed the whole procedure explained in the following link to deal with supernode:
https://img840.imageshack.us/img840/1959/img0002qc.jpg

You see the author also applied KVL.

2: I think you have taken Vo=Vx? Is this correct?

Thanks a lot for all your help. You have helped me with quite a few problems. Many thanks for that.

Regards
Pg
 

jony130

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1: I was just wondering that whether you followed the whole procedure explained in the following link to deal with supernode:
https://img840.imageshack.us/img840/1959/img0002qc.jpg
You see the author also applied KVL.
I think the author overcomplicated the subject too much.
I also apply KVL in this equation
V1 = Vo + 2*Vo
So as you can see we don't need to know the polarities of V3 and V2

All we need is this simply KVL equation

V2 = V3 + 5V or V2 - V3 = 5V or V3 = V2 - 5V

And if we change the polarity of a 5V voltage source
We will write

4_1309689857.jpg


V3 = V2 + 5V ; V3 - V2 = 5V; V2 = V3 - 5V

This is all we need form KVL when we dealing withe supernode.


2: I think you have taken Vo=Vx? Is this correct?
Yes
Vo = Vx = Vth
 
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PG1995

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All we need is this simply KVL equation

V2 = V3 + 5V or V2 - V3 = 5V or V3 = V2 - 5V

1: The quoted KVL equation above refers to the attached circuit diagram.

We start traversing the loop clockwise starting at V2. The V2 sees potential rise. It could be written as:
+V2 + 5V - V3 = 0 (V3 sees potential drop).

This implies: V2 = V3 - 5V.

I think you made a typo above. Just wanted to confirm if I have it right. Please let me know. Thank you.

2: Please have a look on this link: https://img52.imageshack.us/img52/4862/img0006rw.jpg

The 10A current is distributed along two paths when it reached the point 'V1'. So, aren't V1 and Vo related by this equation:

10 - (V1/6) = Vo/2
20 - V1/3 = Vo
60 - V1 = 3Vo
Vo = (60 - V1)/3

Best wishes
PG
 

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jony130

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1: The quoted KVL equation above refers to the attached circuit diagram.

We start traversing the loop clockwise starting at V2. The V2 sees potential rise. It could be written as:
+V2 + 5V - V3 = 0 (V3 sees potential drop).

This implies: V2 = V3 - 5V.

I think you made a typo above. Just wanted to confirm if I have it right. Please let me know. Thank you.
This KVL equations
V2 = V3 + 5V or V2 - V3 = 5V or V3 = V2 - 5V
represent this circuit
25_1309700292.jpg

And I don't see any error.


2: Please have a look on this link: https://img52.imageshack.us/img52/4862/img0006rw.jpg

The 10A current is distributed along two paths when it reached the point 'V1'. So, aren't V1 and Vo related by this equation:

10 - (V1/6) = Vo/2
20 - V1/3 = Vo
60 - V1 = 3Vo
Vo = (60 - V1)/3

Best wishes
PG

Normally when we deal with supernode and start writing KCL but first we virtually short V1 and Vo and then start to write down KVL equations.

But of course your equation are good.
 

PG1995

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This KVL equations
V2 = V3 + 5V or V2 - V3 = 5V or V3 = V2 - 5V
represent this circuit
25_1309700292.jpg

And I don't see any error.

Hi Jony

Thanks a lot for all the help and sorry for the confusion. I was confusing two different circuits.

Regards
PG
 

PG1995

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Hi again, Jony,

I was wondering one thing. In this post of yours you used the attached clipped diagram. Your attached diagram in your mentioned post has the polarities of voltage supply reversed. My linked scanned diagram has the +ve polarity on the left side. This point really confused me in some of the post. Please clarify this if possible. Thanks a lot.
 

jony130

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But can you tell me withe part of my post confuses you?
As I said for figure 1
4_1309894119.jpg

We don't need to know the polarities of V3 and V2 to be able to write KVL equations.
V2 = V3 + 5V or V2 - V3 = 5V or V3 = V2 - 5V

And if we change the polarity of a 5V voltage source figure2
69_1309894532.jpg

We can write KVL without knowing V2 and V3 polarity.
V3 = V2 + 5V ; V3 - V2 = 5V; V2 = V3 - 5V

And of course if we assume the current directions for I2 and I3 we already (immediately) know the polarity of V2 and V3.
 
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PG1995

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But can you tell me withe part of my post confuses you?

Thanks for the clarification. Very nice of you. There is confusion now. In 5th post I said this:
I think you made a typo above. Just wanted to confirm if I have it right. Please let me know.

Actually what had confused me was that you had used different polarities for the battery. By the way, did you swapped the polarities of the battery because my scan has +ve polarity on the left? Just curious.

Regards
PG
 

jony130

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By the way, did you swapped the polarities of the battery because my scan has +ve polarity on the left? Just curious.

Regards
PG
Yes I swapped the polarities of the battery to show you KVL for a different case then your original example.
 

PG1995

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Thanks a lot for the reply.

Best wishes
PG
 

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