# Another (easy) Mesh analysis question

Discussion in 'General Electronics Chat' started by inkyvoyd, Nov 26, 2012.

1. ### inkyvoyd Thread Starter New Member

Dec 6, 2011
25
0

I'm pretty sure this one is a conceptual problem though. In the same pdf (http://ocw.mit.edu/courses/electric...pring-2006/lecture-notes/nodal_mesh_methd.pdf), on page 17, I'm having trouble understanding how to calculate node voltages (v1 and v2) - as well as a few mesh currents, particularly the one flowing through the 4 ohm resistor.

I have drawn three mesh currents - and the equations are as follow
6(I_a-I_b)+4(I_a-I_c)=12
2I_b+2(I_b-I_c)+6(I_b-I_a)=0
3I_c+4(I_c-I_a)+2(I_c-I_b)=0
I have defined I_a to be the left loop current, I_b to be the top right loop current, and I_c to be the bottom right loop current.

Solving for this on my calculator yields
I_a=129/35 I_b=93/35 I_c=78/35
It's evident to me that i_1=129/35, and i_2=i_c-i_b=-3/7
However, I'm not particularly sure how to calculate the voltage at nodes, - perhaps I just don't understand this subject well enough.

Thanks in advance for any help!

Last edited: Nov 26, 2012
2. ### inkyvoyd Thread Starter New Member

Dec 6, 2011
25
0
This is by no means urgent - the Science Olympiad competition I am prepping for is a few months away.

I would greatly appreciate it, however, if anyone explained to me what I don't seem to be grasping. Should I do node analysis instead for this problem?

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
326
There are 3 networks on page 17 of the referenced pdf file. Which one are you trying to analyze?

You should capture an image of the one you're working with and annotate it to show the nodes and loop currents you are using in your equations.

4. ### inkyvoyd Thread Starter New Member

Dec 6, 2011
25
0
Sorry, I'm having trouble with the second problem on that page - I guess this is sorta stupid, but honestly I'm not very good with forums and I'm not very sure how to do what you described.

5. ### electron_prince Member

Sep 19, 2012
93
3
Okay, check the attachments. I've solved it for you. I am quite bored right now and i made several mistakes while solving it. sorry for it.

Note: in attachment 3

V1 = 12 - (i_1' - i_2')6 instead of V1 = 12 - (i_1' - i_2')/6

because of the fact that V = IR

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• ###### 27112012157.jpg
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Last edited: Nov 26, 2012
6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
326
The voltage at V1 is given by 4(I_a-I_c). You have that expression in your first equation. The mesh method works by adding the voltage drops around a mesh and the voltage across the 4Ω resistor is one of those voltages; it's also the voltage V1. Do you see the expression for V2 (the drop across the 3Ω resistor) in your third equation?

7. ### inkyvoyd Thread Starter New Member

Dec 6, 2011
25
0
I am reading this right now - thank you for the information!

It's 3I_c isn't it?

8. ### electron_prince Member

Sep 19, 2012
93
3
I just want to point out two laws

1. Ohms Law: V = IR

where V = potential difference across the resistor
I = current through the resistor
R = Resistance of the resistor.

2. Current flows from higher potential to lower potential.

If you know the current through the 6 ohm resistor then you can calculate the potential difference across that 6 ohm resistor by applying ohms law.

One end of the resistor is at +12V Potential. And as you know the potential difference you can get the Potential for resistor's other end.