Help solving circuit with Mesh Analysis and Current Source

Question

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:

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.

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\$.)

Therefore:

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

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.)