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Mesh Current Analysis Problem

I think I've got my mesh equation right for mesh 1,3,4, and 5. But I'm dumbfounded at mesh 2 since the 20 mA current source is included in the mesh, when I write it all and place all the unknown currents in the left side, and the voltage at right side, there exist a 20 mA and 6 V. Hence, I cannot readily apply system of equation to solve for the unknown currents.

Edit: Here are my mesh equations:

Mesh (1) $$300(i1)-100(i3) = 12V$$

Mesh (2) $$5000(i2)-2000(i3)-1000(i5) = 0.02 A - 6 V$$

Mesh (3) $$-100(i1)-2000(i2)+2100(i3)+1x10^5i(5)=0$$

Mesh (4) $$-1x10^5(i5)-4(i2) = -0.02A$$

Mesh (5) $$-996(i2)+(11000)i5 = 0$$

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  • \$\begingroup\$ I'm in the process of answering this but I'd like to save time and not redo all of your work. because you can't get the answer without doing all the other meshes first. Can you provide the work you did on the other branches first and I will show you how to finish it off? \$\endgroup\$ – Jeffrey Phillips Freeman Sep 8 '20 at 6:10
  • \$\begingroup\$ Are you asking us to provide a solution to your homework problem? You must show the work you’ve done, and you may then get some guidance. \$\endgroup\$ – Chu Sep 8 '20 at 6:48
  • \$\begingroup\$ Ok my bad, I'll send it. \$\endgroup\$ – Abdul Rahman Hadji Ibrahim Sep 8 '20 at 7:17
  • \$\begingroup\$ Is it necessary to use mesh analysis? Nodal analysis, for example, is better for this particular circuit. You must use the voltages across the current sources for mesh analysis, you can't add voltages and currents! This is why mesh analysis isn't the best choice for this circuit. Also, a minor point, you haven't shown the polarity of Vx across R6. \$\endgroup\$ – Chu Sep 8 '20 at 7:49
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    \$\begingroup\$ For nodal analysis there are just two equations in two unknowns. \$\endgroup\$ – Chu Sep 8 '20 at 11:22
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Any time you have a current source that exists as part of more than one mesh you must do supermesh analysis in order to get an answer. To do that you first remove the branch with the current source and work out the current, then you put the branch back in and then apply nodal analysis to finish it off.

I am going to work through the whole problem so it can be easier to follow all the steps. One important step when doing Mesh and Nodal Analysis is to define an arbitrary current flow reference direction for each component, which also implies the reference direction for the voltage drop, as well as a reference point as ground. In this case ground will be the node at the bottom of the schematic, our reference current will flow downward on any vertical component and to the right for any horizontal component. This would imply that for any component the side where the current flows in is the positive side and where it flows out is the negative side.

First let us do the equation for Mesh 1.

\$0 = V_2 + V_{R1} + V_{R2}\$

\$0 = 12V + V_{R1} + V_{R2}\$

\$-12V = V_{R1} + V_{R2}\$

\$-12V = I_{R1} \cdot R_1 + I_{R2} \cdot R_2\$

\$-12V = 200\Omega \cdot I_{R1} + 100\Omega \cdot I_{R2}\$

Now Mesh 3, since its not labeled the voltage source in this mesh I will call \$V_1\$:

\$0 = V_1 + V_{R4} + V_{R2}\$

\$0 = 10 \cdot V_x + V_{R4} + V_{R2}\$

\$0 = 10 \cdot V_{R6} + V_{R4} + V_{R2}\$

\$0 = 10 \cdot I_{R6} \cdot R_6 + I_{R4} \cdot R_4 + I_{R2} \cdot R_2\$

\$0 = 10 \cdot I_{R6} \cdot 10000\Omega + I_{R4} \cdot 2000\Omega + I_{R2} \cdot 100\Omega\$

\$0 = 100000\Omega \cdot I_{R6} + 2000\Omega \cdot I_{R4} + 100\Omega\ \cdot I_{R2}\$

Now here is where things get tricky, we have two current sources shared among the remaining three meshes. Therefore we need to use supermesh analysis and remove these. That gives us one big supermesh between meshes 2, 3, and 5. Here is what the supermesh looks like:

enter image description here

For that super mesh we have the following KVL:

\$0 = V_1 + V_{R4} + V_3 + V_{R3} + V_{R6}\$

\$0 = 10 \cdot V_x + V_{R4} + 6V + V_{R3} + V_{R6}\$

\$-6V = 10 \cdot V_x + V_{R4} + V_{R3} + V_{R6}\$

\$-6V = 100000\Omega \cdot I_{R6} + I_{R4} \cdot R_4 + I_{R3} \cdot R_3 + I_{R6} \cdot R_6\$

\$-6V = 100000\Omega \cdot I_{R6} + 2000\Omega \cdot I_{R4} + 2000\Omega \cdot I_{R3} + 10000\Omega \cdot I_{R6}\$

Next we do nodal analysis on all the nodes which were modified or removed when we did the supermesh. In nodal analysis remember all current flowing into a node should equal 0. Let's start with the node that is at the junction of meshes 4, 3, and 2. Remember that because we want current flowing into the node to equal zero any time our current convention is the opposite of that we need to make it negative.

\$0 = I_{R4} + -I_{V1} + -I_1\$

\$0 = I_{R4} + -I_{V1} + -0.2A\$

\$0.2A = I_{R4} + -I_{V1}\$

Now lets do our ground node at the bottom of the circuit. We will call the new current source here that isn't labeled \$I_2\$. Note that \$I_2\$ is particularly confusing here, our current convention points out of the node, which would normally make \$I_2\$ negative, however the direction that the current source is facing is also opposite of our convention, which would make it negative a second time. As such this cancels out and we want \$I_2\$ to be positive.

\$0 = I_{V2} + I_{R2} + I_{V1} + I_2 + I_{R6}\$

Before we substitute in for \$I_2\$ I want to further point out that \$I_o\$ is a bit confusing as well. Since it is reverse of our convention in the circuit we will have to take its negative as well.

\$0 = I_{V2} + I_{R2} + I_{V1} + 4 \cdot -I_o + I_{R6}\$

Now for the node that lies at the junction of Mesh 2 and Mesh 5 on the rightmost part of the circuit.

\$0 = I_{R3} + -I_{R6} + I_{R5}\$

Finally, the node that was internal to our supermesh which lies at the junction of mesh 2, 4, and 5. Note that again \$I_2\$ is a bit confusing here, even though the current does point into the node the current source is oriented the reverse of our current convention. As such it must still be made a negative value.

\$0 = I_1 + -I_2 + -I_{R5}\$

\$0 = 0.02A + -(4 \cdot -I_o) + -I_{R5}\$

\$-0.02A = 4 \cdot I_o + -I_{R5}\$

Now we should have enough equations to solve for any variable as a set of simultaneous equations, lets collect our simultaneous equations here for clarity.

  1. \$-12V = 200\Omega \cdot I_{R1} + 100\Omega \cdot I_{R2}\$
  2. \$0 = 100000\Omega \cdot I_{R6} + 2000\Omega \cdot I_{R4} + 100\Omega\ \cdot I_{R2}\$
  3. \$-6V = 100000\Omega \cdot I_{R6} + 2000\Omega \cdot I_{R4} + 2000\Omega \cdot I_{R3} + 10000\Omega \cdot I_{R6}\$
  4. \$0.2A = I_{R4} + -I_{V1}\$
  5. \$0 = I_{V2} + I_{R2} + I_{V1} + -4 \cdot I_o + I_{R6}\$
  6. \$0 = I_{R3} + -I_{R6} + I_{R5}\$
  7. \$-0.02A = 4 \cdot I_o + -I_{R5}\$

Lets simplify these a bit and replace \$I_{R3}\$ and \$I_{V3}\$ with \$-I_o\$ since that is the only variable we care about anyway. Since they all lie in series they are all equivalent anyway.

  1. \$-12V = 200\Omega \cdot I_{R1} + 100\Omega \cdot I_{R2}\$
  2. \$0 = 100000\Omega \cdot I_{R6} + 2000\Omega \cdot I_{R4} + 100\Omega\ \cdot I_{R2}\$
  3. \$-6V = 100000\Omega \cdot I_{R6} + 2000\Omega \cdot I_{R4} + 2000\Omega \cdot -I_o + 10000\Omega \cdot I_{R6}\$
  4. \$0.2A = I_{R4} + -I_{V1}\$
  5. \$0 = I_{V2} + I_{R2} + I_{V1} + -4 \cdot I_o + I_{R6}\$
  6. \$0 = -I_o + -I_{R6} + I_{R5}\$
  7. \$-0.02A = 4 \cdot I_o + -I_{R5}\$

So now we have 7 unknown variables and 7 equations. We know we have a solvable set of simultaneous equations. Now we just need to solve our simultaneous equations. Since that would be rather verbose and outside of the context of this answer I will leave that part to you.

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    \$\begingroup\$ Thank you very much for taking the time to answer my question. I've spent hours trying to solve this. Now all that's left for me to do is to understand and analyze your solution. Don't worry, this isn't even an assignment, I just stumbled upon this problem on the internet because I'm preparing for my final exams. \$\endgroup\$ – Abdul Rahman Hadji Ibrahim Sep 8 '20 at 9:30
  • \$\begingroup\$ @AbdulRahmanHadjiIbrahim You are welcome, dont forget to like and accept the answer if you found it helped. I know a lot of people on SE tend to be rather anal about labeling anything that is theoretical formal problem as "homework", but personally I dont make that assumption unless you say it is. I usually just assume its a theoretical problem you want help on. I have been out of school for over a decade and often study new textbooks in my free time and have accused of asking for homework help many times. It isnt productive to make those sort of assumptions. \$\endgroup\$ – Jeffrey Phillips Freeman Sep 8 '20 at 9:33
  • \$\begingroup\$ @AbdulRahmanHadjiIbrahim Also please come back and let me know if you need any further clarity. If appropriate feel free to start a new question and tag me if you need help beyond the scope of your original question. Always happy to help. \$\endgroup\$ – Jeffrey Phillips Freeman Sep 8 '20 at 9:34
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    \$\begingroup\$ @JeffreyPhillipsFreeman It's late for me tonight. So if I do, it will be later -- perhaps a day later or so. There are some textbook approaches that I've really learned to dislike after years of experience. One of them is to stop trying to teach others about super anything. I find it much easier to allow how to use the basic tool in a consistent fashion that does not require new ideas. Any new idea adds another source for mental mistakes. Too many ideas spoil things. I'm using a cell phone from bed, so it will have to wait. \$\endgroup\$ – jonk Sep 8 '20 at 12:47
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    \$\begingroup\$ @JeffreyPhillipsFreeman But perhaps the worst annoyance I have is with how calculus has been taught since Dedekind and Weierstrass ruined what is very simple with their Rube Goldberg limit contraption to avoid infinitesimal variables, which they never did accept. That crap is still taught. Yet Abraham Robinson, in the early 1960s was able to formulate the intuitive fluxions from Newton into solid math with his Non-Standard Analysis book. It's good stuff and makes everything so much easier to understand in one's soul. \$\endgroup\$ – jonk Sep 8 '20 at 12:51

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