I've been for some hours trying to solve this thing with Mesh Analysis but I don't seem to be able to understand how to do it! Could anyone please explain how I can do this?

Disclaimer: I've never been told about Supermeshes or to "invent" a voltage V across the source - I found out about them a hour ago or so (still didn't help, as I remain confused...). What I was told was that the number of mesh equations I needed were Branches-Current_Sources-(Nodes-1) = B-C-(N-1)=5-1-(3-1)=2 mesh equations. The one with the source on it would have 10 A flowing.

I have a resolution of this, but the mesh has the opposite direction to the current source. I'd say that wouldn't matter. I didn't look at the resolution until I finished my attempt - in which I put the direction with the current source direction. That gives me about 6 A on both mesh currents, and that's wrong, according with the resolution and with LTspice, which are both giving the same values (1st mesh, +-1.25 A; 2nd mesh, +-4.4 A; +- depending on your mesh direction). But I don't get why I must put the direction opposite to the current source!

This is really confusing, I've no idea what I'm doing wrong on this. This is probably really easy, but I don't get there xD. Any help would be appreciated!

Circuit: enter image description here

My resolution (meshes clockwise, except the 3rd one, which goes with the current source):

$$\left[ {\begin{array}{*{20}{c}}6&{ - 4}\\{ - 4}&8\end{array}} \right]\left[ \begin{array}{l}{I_{11}}\\{I_{22}}\end{array} \right] = \left[ \begin{array}{l}10\\10 \times 3 = 30\end{array} \right]$$

That gives: $${I_{11}} = 6.25\;A$$ $${I_{22}} = 6.88\;A$$

My teacher has -30 V in the last matrix (3rd mesh opposite to the current source direction), which results in -1.25 A and -4.4 A, which is correct... Not sure what I did wrong. Already looked at this various times and I don't find the mistake.

Note: the resolution starts with the matrix. We don't write equations, since we were told to just go for the matrices which seems to be faster than being writing the equations and get to the matrices anyways. If needed, I can write the equations and get to the matrices.

  • 1
    \$\begingroup\$ Just pick a loop direction. Lots of people choose clockwise. Lots choose counter-clockwise. And still others just make random choices or whatever feels good at the time. It does not matter. The analysis always produces the same, correct result in the end. See this example I did a while back. And yes, you can choose to go with the current source or against it. Doesn't matter. Likely it is just your use of signs. That's what gets everyone, in the end. \$\endgroup\$
    – jonk
    Commented Jul 19, 2021 at 7:01
  • 2
    \$\begingroup\$ I am not sure what's the big deal here. There are three meshes and all look basic. One doesnt even need to know what a "supermesh" is to solve this one. Perhaps you should write down each equation and show us where are you stuck. \$\endgroup\$
    – Mitu Raj
    Commented Jul 19, 2021 at 7:02
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    \$\begingroup\$ @DADi590 If you show your efforts (just one of them, not all), I would imagine we could quickly point out where you made your mistake. (Pick one that doesn't get the right answer.) \$\endgroup\$
    – jonk
    Commented Jul 19, 2021 at 7:05
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    \$\begingroup\$ Not asked, but the "write matrices"-principle is useful for the future. One day you will meet something which cannot be kept together without matrices. People who already can read, write and speak matrices do not see that day as anything special. Those who have avoided matrices can see that day traumatic. \$\endgroup\$
    – user136077
    Commented Jul 19, 2021 at 8:07
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    \$\begingroup\$ That's right. But that's just the implication/derivation from the basic concept of mesh analysis. \$\endgroup\$
    – Mitu Raj
    Commented Jul 19, 2021 at 10:54

1 Answer 1


There's a schematic editor available to you, here. You should probably use it. It numbers the parts.


simulate this circuit – Schematic created using CircuitLab

$$\begin{align*} 0\:\text{V} - R_1\cdot I_A + 10\:\text{V}-R_3\cdot\left(I_A-I_B\right) &= 0\:\text{V}\\\\ 0\:\text{V} -R_3\cdot\left(I_B-I_A\right) - R_2\cdot I_B - R_4\cdot\left(I_B-I_C\right) &= 0\:\text{V}\\\\ I_C&=-10\:\text{A} \end{align*}$$

(Note that \$I_C\$ is negative because the current source is opposite to my choice of direction for \$I_C\$.)


$$\begin{align*} - \left(R_1+R_3\right)\cdot I_A +R_3\cdot I_B &= -10\:\text{V}\\\\ R_3\cdot I_A - \left(R_2+R_3+R_4\right)\cdot I_B &= R_4\cdot 10\:\text{A} \end{align*}$$

So the matrix is easy:

$$\left[\begin{smallmatrix} -6 & 4\\4 & -8 \end{smallmatrix}\right] \left[\begin{smallmatrix}I_A\\I_B\end{smallmatrix}\right]=\left[\begin{smallmatrix}-10\\30\end{smallmatrix}\right]$$

From which I do get \$I_A=-\frac54\:\text{A}\$ and \$I_B=-\frac{35}8\:\text{A}\$. I don't know if that's just a coincidence in my choice of clockwise currents for \$I_A\$ and \$I_B\$ with respect to your reference, though. Since those currents are both negative, it just means that my choice of direction is opposed to the circulation direction of conventional current.

Here's the final result:


simulate this circuit

(Which I hope matches up with your LTspice results.)

  • \$\begingroup\$ It does match..... Since you wrote it I'm looking at it trying to understand what's wrong with my thought haha. I'll be back when I figure it out... My sign convention is the opposite of yours and my last mesh is opposite (which as far as I know, doesn't matter at all), so I'm taking a bit to process that and see also why writing the equations myself it seems to work fine, but going directly to the matrices it doesn't. "Processing data... Please stand by." xD. \$\endgroup\$
    – Edw590
    Commented Jul 19, 2021 at 8:02
  • \$\begingroup\$ I'm seeing "answered 40 minutes ago" - that's the time I've been trying to understand yours and comparing to mine and the same time hahaha. Finally got it!! As a start there's a mistake in the resolution (must have copied wrong from the board), which you address there: the current with the opposite mesh direction is 10 A... Shouldn't be, it seems (it's the only solved example I have of this type of exercises). So that was problem 1. Problem 2 was again my confusion with signs when going in a loop summing voltages... Can't let this go xD. 3 years and I still confuse the signs. Wow. Thank you!!! \$\endgroup\$
    – Edw590
    Commented Jul 19, 2021 at 8:19
  • \$\begingroup\$ Yep, I was able of making a mess out of such a simple thing. I'm proud of myself 😂. Finally got this perfectly. Took 3 years haha - we never again got to touch on these kinds of circuits with current sources and being asked specifically for a mesh analysis. But we're allowed to improve our past grades sometimes, so here I am trying to understand it again, and finally being able to with some help haha. Thanks again!!! \$\endgroup\$
    – Edw590
    Commented Jul 19, 2021 at 8:37
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    \$\begingroup\$ @DADi590 I'm so glad it helped. And thanks so much for letting me know, too. It makes all this worth every moment. :) \$\endgroup\$
    – jonk
    Commented Jul 19, 2021 at 9:48

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