In this circuit, I tried to count the 8 & 12 Voltage sources as one Supernode, but the answers were different from a simulation I ran on LTSpice, The only way to get the same numbers as the simulation is by applying KCL at the Ground node, though am not sure I should hence the book specifically said not to apply KCL at the Ground Node, what should I do ?
Thanks

 

There's nothing inherently wrong with applying KCL at the ground node, it just usually accomplishes nothing. You already know the voltage of that node relative to ground.

As for what you should do, the best advice would be to show your work. We can't possibly tell you what you did wrong until you show us what you did.

Remember also that the choice of the common reference (i.e., "ground") node is completely arbitrary. You can pick any node you want as your reference, do the analysis, and then simply shift the results by the appropriate amount to move the reference to a different node.

If that diagram didn't have the center node marked as the reference, which node just screams out demanding to be the reference node?

 

At First, I drew a Supernode out of the 2 sources

(V1)/4+(V2)/10+(V3)/5=0.25, V2-V1=8, V3-V1=12
The second Solution was by applying KCL at the reference node:
(V1)/4+(V2)/10+(V3)/5=0, V2-V1=8, V3-V1=12

Now I made the V1 node the reference if I understood correctly, and the previous ref is now v1
V2=8V, V3= 12V
12/5 + 8/10 + (V1)/6 = 0
V1 = 19.2V
And that's it not sure if this is right or how to continue

Sorry for the delay & Thank You for your help.

 

At First, I drew a Supernode out of the 2 sources
View attachment 141711
(V1)/4+(V2)/10+(V3)/5=0.25, V2-V1=8, V3-V1=12
The second Solution was by applying KCL at the reference node:
(V1)/4+(V2)/10+(V3)/5=0, V2-V1=8, V3-V1=12

Now I made the V1 node the reference if I understood correctly, and the previous ref is now v1
V2=8V, V3= 12V
12/5 + 8/10 + (V1)/6 = 0
V1 = 19.2V
And that's it not sure if this is right or how to continue

Sorry for the delay & Thank You for your help.

Look at the right hand side of your very first equation. Why do you have 0.25 A there?

 

Look at the right hand side of your very first equation. Why do you have 0.25 A there?

From The 0.25A Current Source as it is accessing the Supernode through V2

Edit: And exiting from V3??

 

Now I made the V1 node the reference if I understood correctly, and the previous ref is now v1
V2=8V, V3= 12V
12/5 + 8/10 + (V1)/6 = 0
V1 = 19.2V
And that's it not sure if this is right or how to continue

Sorry for the delay & Thank You for your help.

Calling a different node v1 but then using the same diagram with the original v1 is a bad idea -- and I think it caused you to mess up.

Your KCL equation appears to be the current flowing INTO the center node. But (V1)/6 Ω would appear to be the current flowing in from the original V1 node, not the new V1 node. Also, where did the 6 Ω come from?

Also, 12 V / 5 Ω is NOT the current flowing from V3 to the center node. Because you didn't draw a new diagram that matched your new set up, you still used the center node as 0 V when you did KCL. Always draw a diagram that is going to match your work.

You also want to ask if your answer makes sense. You have nodes V2 and V3 at 8 V and 12 V respectively. You then have the center node (let's call it V4) that is connected between these two nodes and 0 V. Does it make sense that V4 could be greater than 12 V?

Another thing you want to do is get in the practice of ALWAYS tracking your units properly. Most mistakes (not all) that you will make will mess up the units IF the units are there to get messed up. If they aren't, then you will sail right on past the error and spend lots of time and effort pursuing an answer that is guaranteed to be wrong.

Finally, get in the habit of setting up your equations, double checking them to make sure they are correct, and then set about solving them. The set up is where most, if not all, of the electrical engineering takes place. If you make a mistake here, the rest of it will have a hard time catching it because everything else is just math and math doesn't care about physics.

So it should look something like this:

V1 = 0 V
V2 = 8 V
V3 = 12 V

KCL @ V4: (V4 - V1)/(4 Ω) + (V4 - V2)/(10 Ω) + (V4 - V3)/(5 Ω) = 0

That's all the EE there is. The equation is set up and written so that it is easy to check against the problem and verify that it is correct. Each term is the current flowing away from V4 to one of the other connected nodes and all three of the other nodes are accounted for.

Now you can move on to solving the problem.

 

From The 0.25A Current Source as it is accessing the Supernode through V2

Edit: And exiting from V3??

YES! It is entering and exiting the supernode, so you need TWO terms in your equation, +0.25 A - 0.25 A, which cancel out.

One way to help avoid missing this is count how branches are connected to your supernode (i.e., cross the boundary)? There are five. But your equation only has four. So one is missing.

Get in the habit of doing every little sanity check you can think of as you do the work. You WILL make silly, stupid mistakes on a regular basis -- we ALL do. But a good engineer diligently checks their work to catch and correct those mistakes.

 

From The 0.25A Current Source as it is accessing the Supernode through V2

Edit: And exiting from V3??

If the current is both entering and exiting then perhaps your equation should be:

(V1)/4+(V2)/10+(V3)/5=0.25-0.25, V2-V1=8, V3-V1=12

Or, in other words: (V1)/4+(V2)/10+(V3)/5=0, V2-V1=8, V3-V1=12

If I solve the equation as you had it, I get:



But with the suggested change:



You don't need to change your reference node; you just need to notice that the .25 amp both enters and leaves the supernode. In other words, it does nothing.

 

You don't need to change your reference node; you just need to notice that the .25 amp both enters and leaves the supernode. In other words, it does nothing.

Just to make the context clear, he was doing two separate things at my suggestion. The first was to show his supernode equation. The second was to approach the problem without a supernode at all by choosing a different reference.

 

Just to make the context clear, he was doing two separate things at my suggestion. The first was to show his supernode equation. The second was to approach the problem without a supernode at all by choosing a different reference.

I know that. I was simply pointing out how close his supernode attempt was to being correct, which you also pointed out. I was still working on my post when you had already posted about the his failure to account properly for the current source so I didn't see yours. There was nothing wrong with your suggestion to choose another reference node; just that his first approach was A-ok with a small mistake.

 

Okay -- I had thought that you had thought the two were part of the same effort, because taken on its own it does tend to read that way.

 

So it should look something like this:

V1 = 0 V
V2 = 8 V
V3 = 12 V

KCL @ V4: (V4 - V1)/(4 Ω) + (V4 - V2)/(10 Ω) + (V4 - V3)/(5 Ω) = 0

Wow my equations were totally wrong, I resolved it with this method & it checked out.

YES! It is entering and exiting the supernode, so you need TWO terms in your equation, +0.25 A - 0.25 A, which cancel out.

One way to help avoid missing this is count how branches are connected to your supernode (i.e., cross the boundary)? There are five. But your equation only has four. So one is missing.

Get in the habit of doing every little sanity check you can think of as you do the work. You WILL make silly, stupid mistakes on a regular basis -- we ALL do. But a good engineer diligently checks their work to catch and correct those mistakes.

Thank you so very much for your help, you really opened my mind to new ways of thinking.

If the current is both entering and exiting then perhaps your equation should be:

(V1)/4+(V2)/10+(V3)/5=0.25-0.25, V2-V1=8, V3-V1=12

Or, in other words: (V1)/4+(V2)/10+(V3)/5=0, V2-V1=8, V3-V1=12

If I solve the equation as you had it, I get:

View attachment 141737

But with the suggested change:

View attachment 141739

You don't need to change your reference node; you just need to notice that the .25 amp both enters and leaves the supernode. In other words, it does nothing.

Resolved it and my answers checked out with yours, don't know how I didn't see the current flowing out, Thank you so much for your help.