# Mesh Analysis - Super Loop Question

1. Sep 18, 2006

### verd

So here is what the problem states:
Use both nodal analysis and mesh analysis to find Vo in the circuit below.

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!

2. Sep 18, 2006

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:
$$6mA + I_3 = I_2$$

You questioned yourself with:
$$(I_2 - I_3)=6mA$$

Notice if you add $I_3$, your expression becomes:
$$6mA + I_3 = I_2$$

3. Sep 18, 2006

### verd

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. Sep 18, 2006

You didn't seem to, since you said

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. Sep 18, 2006

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. Sep 19, 2006

### doodle

No, surely you can't.

Yes, that would be correct.

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. Sep 19, 2006

### verd

Phew... Okay. Thank you VERY much doodle...

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

Super loop btwn loop 2&3:
$$I_{2}-I_{3}=6mA \longrightarrow I_{2}=6mA+I_{3}$$

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

loop 1:
$$I_{1}(12k)+(I_{1}-I_{3})8k+(I_{1}-I_{2})4k=0$$
$$4mA(12k+8k+4k)+I_{2}(-4k)+I_{3}(-8k)=0$$

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:

$$I_{2}(4k)+I_{3}(8k)=4mA(24k)$$
$$I_{2}+I_{3}(-1)=6mA$$

Solving with my simultaneous equation solver, I get:
$$I_{2}=12mA$$
$$I_{3}=6mA$$

And finding Vo:
$$V_{0}=I_{2}(12k)=(12mA)(12k)=144V$$

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....

Last edited: Sep 19, 2006
8. Sep 19, 2006

### doodle

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: $24 I_1 - 4 I_2 - 8 I_3 = 0[/tex] Loop 2-3: [itex]-12 I_1 + 16 I_2 + 8 I_3 + 6 V_x = 0 \Longrightarrow 12 I_1 - 8 I_2 + 8 I_3 = 0$
since $V_x = 4(I_1 - I_2)$

6mA: $I_2 - I_3 = 0$

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
$(I_{2}-I_{1})4k+(I_{3}-I_{1})8k+6V_{x}+I_{2}(12k)=0$
If you are astute, you would observe a certain pattern to which how these equations are formed.

9. Sep 19, 2006