Mesh Analysis - Super Loop Question

In summary: I'm doing this right. And I'm sorry if I'm taking too much of your time or asking too many questions.Thank you VERY much.In summary, the problem states finding Vo in the given circuit using both nodal analysis and mesh analysis. The confusion lies in how to treat the 6mA current source between loops 2 and 3. It is determined that the current in both loops cannot be assumed to be 6mA, but rather I2-I3=6mA. The superloop method is used to treat loops 2 and 3 as one combined loop, ignoring the 6mA segment. The KVL equation for the superloop is derived and used in conjunction with the
  • #1
verd
146
0
So here is what the problem states:
Use both nodal analysis and mesh analysis to find Vo in the circuit below.

http://synthdriven.com/images/deletable/EEN201-05.jpg [Broken]


My confusion:
I'm okay with the nodal analysis, I'm just a bit confused with the mesh analysis...

I'm confused as to how exactly I should treat this. I've come in contact with super loops before, but never with more than two loops. Because the 6mA current source is right between what I have labelled as loop 2 and loop 3, I'm not quite sure what to do. Would it be safe to assume that the current in both loop 2 and loop 3 is 6mA?

Or would I assume that I2-I3=6mA?

And if that's the case, in a 2-loop super loop situation, you would treat the entire circuit as one loop. In this situation, would you treat those two loops as one combined loop? As in would you go around that loop doing KVL in loop2+loop3 excluding loop 1?


Any help is appreciated!
Thanks!
 
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  • #2
verd said:
So here is what the problem states:
Use both nodal analysis and mesh analysis to find Vo in the circuit below.

http://synthdriven.com/images/deletable/EEN201-05.jpg [Broken]


My confusion:
I'm okay with the nodal analysis, I'm just a bit confused with the mesh analysis...

I'm confused as to how exactly I should treat this. I've come in contact with super loops before, but never with more than two loops. Because the 6mA current source is right between what I have labelled as loop 2 and loop 3, I'm not quite sure what to do. Would it be safe to assume that the current in both loop 2 and loop 3 is 6mA?

Or would I assume that I2-I3=6mA?

And if that's the case, in a 2-loop super loop situation, you would treat the entire circuit as one loop. In this situation, would you treat those two loops as one combined loop? As in would you go around that loop doing KVL in loop2+loop3 excluding loop 1?


Any help is appreciated!
Thanks!

Recall what mesh analysis is. It really still is KVL, KCL.
So if we examine the node on the far right, (with the 6mA flowing into it, the 12k attached to it, and the voltage controlled voltage source) what can we gather about it.

Recall KCL.
current in equals current out.

So what is the current flowing into it?

You have 6mA flowing in, and you have labeled I3 flowing into it, and I2 flows away (as you have labeled it).

So:
[tex] 6mA + I_3 = I_2 [/tex]

You questioned yourself with:
[tex] (I_2 - I_3)=6mA [/tex]

Notice if you add [itex] I_3 [/itex], your expression becomes:
[tex] 6mA + I_3 = I_2 [/tex]
 
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  • #3
Hey,

Thanks for your response... I understand that, thanks for confirming... However, what the big issue is is whether or not to treat it like a super loop problem... In super loop problems, you typically come up with an expression for the current source in between the two loops, as we did, (I2-I3=6mA), and then you work around that segment. You make the two loops into one large loop by sort of ignoring the segment in between.

Would I do the same thing here?... And should I ignore loop 1?

Can someone assist in setting up the equations?...I guess I have quite a few questions about how to correctly set up those equations.


Thanks
 
  • #4
verd said:
Hey,

Thanks for your response... I understand that, thanks for confirming...

You didn't seem to, since you said

verd said:
Would it be safe to assume that the current in both loop 2 and loop 3 is 6mA?

Back to the equations.

You can do a loop in 1, because those are all voltage sources.
Then you can loop from the bottom left corner, to top left corner, to top right corner, to bottom right corner right? because those are also voltage sources. Then you need an expression for Vx.
You find the current in Vx in a similar method as you did for the 6mA part.
 
  • #5
You know the question is going to be a system of equations. Using analysis techniques, how much information can you pull from the network? Each loop gives you a piece of the puzzle. As long as you are following the rules to grab information, grab as much as you can until you have,

n equations
with n unknowns.
 
  • #6
verd said:
I'm confused as to how exactly I should treat this. I've come in contact with super loops before, but never with more than two loops. Because the 6mA current source is right between what I have labelled as loop 2 and loop 3, I'm not quite sure what to do. Would it be safe to assume that the current in both loop 2 and loop 3 is 6mA?
No, surely you can't.

verd said:
Or would I assume that I2-I3=6mA?
Yes, that would be correct.

verd said:
And if that's the case, in a 2-loop super loop situation, you would treat the entire circuit as one loop. In this situation, would you treat those two loops as one combined loop? As in would you go around that loop doing KVL in loop2+loop3 excluding loop 1?
Yes, you will treat loops 2 and 3 as one single superloop. And you will go around this loop (as you do with loop 1) taking the KVL and ignoring the 6mA segment between the two loops. Try coming up with the KVL equation for the superloop on your own and I will tell you if you are on the right track.
 
  • #7
Phew... Okay. Thank you VERY much doodle...

Here's what I've done... Could you please please glance over this and help me to make sure I'm thinking about this circuit correctly??

http://synthdriven.com/images/deletable/EEN201-05.jpg [Broken]

This is it (LONG PROCESS, sorry):
[tex]V_{x}=(I_{1}-I_{2})4k[/tex]

Super loop btwn loop 2&3:
[tex]I_{2}-I_{3}=6mA \longrightarrow I_{2}=6mA+I_{3}[/tex]

Super-loop Loop Analysis:
[tex](I_{2}-I_{1})4k+(I_{3}-I_{1})8k+6V_{x}+I_{2}(12k)=0[/tex]
[tex](6mA+I_{3}-I_{1})4k+(I_{3}-I_{1})8k+6[(I_{1}-6mA-I_{3})4k]+(6mA+I_{3})12k=0[/tex]
[tex]6mA(4k-24k+12k)+I_{1}(-4k-8k+24k)+I_{3}(4k+8k-24k+12k)=0[/tex]
[tex]6mA(-8k)+I_{1}(12k)+I_{3}(0k)=0[/tex]
[tex]I_{1}=\frac{6mA(8K)}{12K}=4mA[/tex]

loop 1:
[tex]I_{1}(12k)+(I_{1}-I_{3})8k+(I_{1}-I_{2})4k=0[/tex]
[tex]4mA(12k+8k+4k)+I_{2}(-4k)+I_{3}(-8k)=0[/tex]

So, because I've found I1, I have two unknowns, I2 & I3... (I used my answer for I1 in the above loop 1 equation.)

Going back to the first statement, that leaves me with two equations and two unknowns:

[tex]I_{2}(4k)+I_{3}(8k)=4mA(24k)[/tex]
[tex]I_{2}+I_{3}(-1)=6mA[/tex]

Solving with my simultaneous equation solver, I get:
[tex]I_{2}=12mA[/tex]
[tex]I_{3}=6mA[/tex]

And finding Vo:
[tex]V_{0}=I_{2}(12k)=(12mA)(12k)=144V[/tex]I mean, logically, from my understanding, I'm doing everything correctly. (I think)... It's just that 144v seems kind of high. I don't know, is this correct?

I know it's a lot of steps, I'm sorry...
 
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  • #8
Yup, that's right. If you don't mind me saying, a more systematic approach is to take KVL for loop 1 and the superloop and combine that with the current equation for the 6mA to give you three equations which are solved simultaneously.

This is what I meant...

Loop 1: [itex]24 I_1 - 4 I_2 - 8 I_3 = 0[/tex]

Loop 2-3: [itex]-12 I_1 + 16 I_2 + 8 I_3 + 6 V_x = 0 \Longrightarrow 12 I_1 - 8 I_2 + 8 I_3 = 0[/itex]
since [itex]V_x = 4(I_1 - I_2)[/itex]

6mA: [itex]I_2 - I_3 = 0[/itex]

Solving these three equations simultaneously for the three unknowns gives the desired mesh currents (in mA).

One last note; it should not be too difficult coming up with the mesh equations without actually starting with
[itex](I_{2}-I_{1})4k+(I_{3}-I_{1})8k+6V_{x}+I_{2}(12k)=0[/itex]
If you are astute, you would observe a certain pattern to which how these equations are formed.
 
  • #9
Also, it is probably worth your time (if you are going to be doing a lot of these), to either learn how to setup a system of equations in matrix form and solve it with your calculator.
 

1. What is Mesh Analysis?

Mesh analysis is a method used in circuit analysis to determine the voltage and current in a closed loop of a circuit. It involves applying Kirchhoff's voltage law (KVL) and Ohm's law to solve for unknown variables.

2. How does Mesh Analysis work?

Mesh analysis involves dividing a circuit into smaller loops and assigning variables to each loop. KVL is then applied to each loop, creating a system of equations that can be solved simultaneously to find the values of the unknown variables.

3. What is a Super Loop in Mesh Analysis?

A Super Loop is a special case in Mesh Analysis where multiple loops in a circuit share a common element. This can make solving the system of equations more complex, but it can still be solved using the same principles as regular Mesh Analysis.

4. When should Mesh Analysis be used?

Mesh Analysis is typically used for circuits that have several loops and multiple voltage sources. It can also be useful for circuits with dependent sources or non-linear elements.

5. What are the advantages of using Mesh Analysis?

Mesh Analysis allows for a systematic approach to solving complex circuits, and it can be more efficient and accurate than other methods. It also provides a way to check the accuracy of the results by verifying KVL for each loop.

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