Hi,I guess the nodal analysis is not really that hard either, it gives the same solution:

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Yeah if you know a simpler way you could always use that, but if the circuit is more complicated then you need some form of analysis that in itself is organized so it is easier to apply.

One of the things i always like about it is that we are always forming simple parts like:

(v2-v1)/R1

and summing terms like that. So we dont have to concentrate on anything but one circuit element at a time, usually.

Now here is a point about Nodal for that circuit. It's a minor point for this circuit, but for a more complicated circuit Nodal may be the best way to go about solving it.

The solution i got with my program was:

[v1=(E3+4*E2+4*E1)/11,v2=(3*E3+E2+E1)/11]

where E1,E2,E3 are Vin1, Vin2, Vin3, and v2 is the first node on the left and v1 is the output node.

What is interesting is that we got both node voltages at the same time. IF these were both outputs, we would need that information. What else is interesting is we find that there are no resistors required in the solution and that becomes apparent from the solution, when all resistors are the same value. Did you happen to notice that with the other solution? Maybe you did, but again with a more complicated solution it may be harder to figure out.

BTW we set up 'little parts' like (v2-v1)/R1 one at a time but here are even simpler ways to write out the equations where we only look at a single current contribution at a time, so instead of writing v2 and v1 in the same term, we can break them down further in to one term +v2/R1 and the other term -v1/R1. Seems like a moot point but when we go to write the equations we only have to look at ONE end of the resistor at a time rather than both at once.

Also a little interesting is with E1=11 and E2=22 and E3=33, we get outputs of 12v and 15v, both integers. There would of course be other integer solutions.

Your circuit was interesting because it was a little simple but not too simple.

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