Electronic – Where did I go wrong in formulating the mesh equations

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:

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.