Mesh Analysis Of Following Circuit

Question

I have a proposed circuit as demonstrated in the photo below. Included are full measurements for resistors, LEDs (symbolised by a diode), a current source, mesh loops and current directions.

I am currently in the process of trying to complete a mesh analysis on this circuit but I'm having trouble finalizing the first two loops of the circuit due to the placement of the current source. I first started by working backwards starting with loop 4 and then 3. However, I am not sure if this is correct working because of how I haven't solved the first two loops yet. Below is a demonstration of my work thus far.

Loop 4: $$0.306i_4 + 17.7i_4 + 27.4 + 0.306i_4 - 27.4 + 17.7(i_4-i_3) = 0$$ $$36.012i_4 - 17.7i_3 = 0$$

Loop 3: $$0.198i_3 + 17.7(i_3-i_4) + 27.4 + 0.198i_3 - 27.4 + 17.7(i_3-i_2) = 0$$ $$35.796i_3 - 17.7i_4 - 17.7i_2 = 0$$

I was first wondering if anyone could help me in solving the first two loops, mainly by explaining how the current source affects the analysis and the working to solve it, and then providing confirmation on whether the working for Loops 3 & 4 will work for solving the rest of the analysis and are correct.

** I have also provided an alternate 'design' of this circuit which swaps the position of the resistor and current source in the first loop. My reason for doing so is I believe this would make the circuit much easier to solve but I am not sure if it is considered as the 'same' as the previous circuit. If someone could also confirm this that would be much appreciated.**

Proposed Circuit

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Redesigned Circuit

Answer 1

This is a physics problem. The source can be thought as a variable voltage source $-V=350 i_1$ (in the first diagram), with $i_1-i_2=0.48A$. The two diagrams are equivalent, just the notation of some currents is different. In the first one, the current through the source is $i_1-i_2$, in the second it's $i_1$.

Now, in order to solve the problem, I'm going to use the first diagram. You have four unknown currents, and one unknown voltage, so you need 5 equations. One equation is $i_1-i_2=0.48A$. In loop 1, $-V=350 i_1$. In loop 2, $V=3.42i_2+17.7(i_2-i_3)+27.4+3.42i_2$. You have the other two equations. You can see from the equations from loops 1 and 2 that you can ignore the voltage $V$ by writing one loop over the outer edges of loops 1 and 2. The equation for this is $$350 i_1+3.42i_2+17.7(i_2-i_3)+27.4+3.42i_2=0$$

Answer 2

If you’re building the equations “by hand,” then a typical way to deal with an ideal current source like this is to pretend that its branch isn’t there and push the current out to the nodes to which it connects, making them current sources and sinks—the sums of the currents at those nodes will be non-zero. For the way you’ve set up the problem, this gives you an equation relating $i_1$ and $i_2$, and so you need one more independent equation for a solution. You can derive one from any mesh that’s independent of the two that you’ve already used, such as the one labeled $I_2$ in the second diagram. This is what Andrei does in his answer.

In a graph-theoretical framework, you would assign these current sources to existing parallel resistive branches. Effectively, you give these branches an excess current. Algebraically, these “excess” currents are collected into a term $\mathbf K$, which is the current counterpart of the $\mathbf W$ term that represents voltage sources. If the branch currents and voltage drops are given by $\mathbf I$ and $\mathbf V$, respectively, then the current through the resistor with resistance $z_\alpha$ in branch $\alpha$ is $$(V^\alpha-W^\alpha)=z_\alpha(I_\alpha-K_\alpha)$$ or in vector form, $\mathbf V-\mathbf W=Z(\mathbf I-\mathbf K)$. In this formalism, Kirchhoff’s laws are expressed as $$\mathbf I=\operatorname\sigma\mathbf J\\ \mathbf V=-\operatorname{[d]}\mathbf\Phi.$$ Here, $\mathbf J$ is the mesh current vector, $\mathbf\Phi$ is the node potentials, $\operatorname{[d]}$ is the coboundary operator, basically the incidence matrix of the network with columns corresponding to nodes chosen to be at ground deleted, and $\sigma$ is the map from the quotient space of cycles to the space of 1-chains. From these equations we can derive $$\mathbf J=(sZ\sigma)^{-1}s(Z\mathbf K-\mathbf W)$$ from which we get the solution $$\mathbf I=\sigma\mathbf J=\sigma(sZ\sigma)^{-1}s(Z\mathbf K-\mathbf W)\tag{*}$$ assuming, of course, that $sZ\sigma$ is nonsingular.

Numbering the nodes in the second diagram clockwise from top left and the branches clockwise from the left and then across the middle, we have for your circuit $$\mathbf K=(0.48,0,0,0,0,0,0,0)^T \\ \mathbf W=(0,0,0,27.4,0,0,27.4,27.4)^T \\ Z=\operatorname{diag}(350,3.42,0.198,0.306+17.7+0.306,0.198,3.42,17.7,17.7).$$ For simplicity, I’ve combined the three resistances along the right into one. The matrix for $\sigma$ is fairly easy to construct. We get a maximal tree by deleting three branches, so the mesh space is three dimensional. If we choose the three bottom branches, we end up with the basis meshes labeled $I_2$, $I_3$ and $I_4$ in the second diagram, so using these meshes and the orientations in that diagram we get $$s = \left[\begin{array}{r}1&1&0&0&0&1&1&0 \\ 0&0&1&0&1&0&-1&1 \\ 0&0&0&1&0&0&0&-1 \end{array}\right]$$ and $\sigma$ is just its transpose. Plugging these values into equation (*) yields $$\mathbf I=(0.387351, 0.387351, 0.253027, 0.124363, 0.253027, 0.387351, 0.134324, 0.128663)^T$$ which jibes with the values that you get by solving the hand-built system of linear equations.

You can see that this approach of assigning the current source to another branch is basically the same as the method described at top by making the following observation: Kirchhoff’s current law can be stated as $\operatorname\partial\mathbf I=0$, where $\operatorname\partial$ is the boundary map from 1-chains to 0-chains for the network. The action of $\operatorname\partial$ on a branch current is to push it out, with opposite signs, to the two end nodes of the branch. So, if there’s an excess current in a branch, from the point of view of the nodes it looks like a source-sink pair.