I am new to theory of circuits and I just learned about mesh analysis. When I was doing some reviewing, I stumbled across this question and wonder if you can use it here:
I plugged the equation into my calculator and it just returned i1=i2=i3=0. Shouldn’t I be able to solve for i3 which is Io?
Your first equation (\$-13I_1 +7I_2=0\$) is incorrect because it doesn't account for the voltage across the current source. (But because there's no way to determine the voltage across the current source from the voltage across it, you can't use a mesh equation for a mesh that contains a current source. That's okay because you know right away that \$I_1 = 13\ {\rm A}\$ so you don't need an equation for mesh 1 anyway)
I unintentionally solved your problem in a comment for The Photon. At that time, I was trying to make and clarify my point to him. And to do that concretely and firmly I had to solve it so there was no remaining question about where I was coming from. Since I already messed that up (but now deleted, since he can just look here) I decided I may as well codify it in an answer. So here it is.
To solve the mesh equations all you need to do is to assign an unknown voltage across the \$I_S\$ current source and call it \$V_S\$. Then quite simply proceed to write out your standard mesh equations as you would in any other case.
I may as well just lay out what I provided in LTspice to start, because I've labeled everything there:
The mesh equations are then:
$$\begin{align*} 0\:\text{V}+V_S-R_1\cdot I_A-R_2\cdot\left(I_A-I_B\right)&=0\:\text{V} \\\\ 0\:\text{V}-R_2\cdot\left(I_B-I_A\right)-R_3\cdot I_B-R_4\cdot\left(I_B-I_C\right)&=0\:\text{V} \\\\ 0\:\text{V}-R_4\cdot\left(I_C-I_B\right)-R_5\cdot I_C-R_6\cdot I_C&=0\:\text{V} \end{align*}$$
Nothing fancy there. You also happen to know that \$I_A=15\:\text{A}\$. So you can either just plug that into the above equations and solve for \$V_S\$, \$I_B\$, and \$I_C\$ (3 equations and 3 unknowns) or else you can just add one more equation, \$I_A=15\:\text{A}\$ and solve for \$V_S\$, \$I_A\$ (which will be obvious), \$I_B\$, and \$I_C\$ (4 equations and 4 unknowns, with one of them given, so to speak.) It's all the same thing, really.
And yes, you do need to recognize the fact that there is some voltage across the current source. It might be anything. So it's an unknown to solve for. But there's no reason why you cannot just recognize that fact and include it in your mesh loop equation set.
Mesh works just fine.
Let's solve the above using SymPy and take all four equations:
eq1 = Eq( 0 + vs - 6*ia - 7*(ia-ib), 0 )
eq2 = Eq( 0 - 7*(ib-ia) - 2*ib - 4*(ib-ic), 0 )
eq3 = Eq( 0 - 4*(ic-ib) - 3*ic - 5*ic, 0 )
eq4 = Eq( ia, 15 )
solve( [ eq1, eq2, eq3, eq4 ], [ ia, ib, ic, vs ] )
{ia: 15, ib: 9, ic: 3, vs: 132}
That's it! 4 equations and 4 unknowns.
Now, we could have substituted in for \$I_A\$ and solved the following:
eq1 = Eq( 0 + vs - 6*15 - 7*(15-ib), 0 )
eq2 = Eq( 0 - 7*(ib-15) - 2*ib - 4*(ib-ic), 0 )
eq3 = Eq( 0 - 4*(ic-ib) - 3*ic - 5*ic, 0 )
solve( [ eq1, eq2, eq3 ], [ ib, ic, vs ] )
{ib: 9, ic: 3, vs: 132}
That's the 3 equations and 3 unknowns variation. Either way, you get the same results.
Take your pick.
