how to find power equation on the 2Ω resistor as a function of R1, R2, R3, V1 and V2?
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No need for another loop. You need one independent equation per mesh. The beauty of mesh current analysis is that by writing one loop equation for each mesh, in terms of the mesh currents, you are applying KVL around just the correct number of independent loops that are needed, and doing it in such a way that KCL is automatically satisfied.Is the way I solved this correct? Should I add another loop, or is it not needed?
I think you are going down a rabbit hole. Whatever you pick for those component values, I can almost guarantee that I can pick different values that will increase the power in that resistor. Most likely, all I would need to do is just take whatever value you have for V1 and double it.I need to determine the values of resistors (R1, R2, R3) and voltages (V1, V2) that result in the maximum power (Pmax) across the 2Ω resistor using a singleobjective optimization algorithm in MATLAB. Therefore, I believe I should derive the power equation on the 2Ω resistor as a function of R1, R2, R3, V1, and V2.
You know that the power in a resistor is in terms of the current flowing through it.It is an optimization algorithm problem. However, I need the power equation on the 2Ω resistor so that I can apply the algorithm in MATLAB.
Thank youYou know that the power in a resistor is in terms of the current flowing through it.
You know what the current flowing through it is in terms of your mesh currents.
Solve for those currents and plug them into the equation for the power in that equation, any you have your power equation in terms of those five unknowns.
And then you will almost certainly find that there is no single combination of them that results in maximum power in that resistor because the power will increase without bound as you change one or more of those unknown values.
The direction that you go for each loop is completely arbitrary  you can flip a coin for each one.Can you confirm whether taking the direction of voltages for these four loops is correct or not?
the equations you wrote arent correct, you might wanna orient the polarity of each component then apply KVL correctly, also for maximum im not quite sure what you are looking forI followed your suggestion by eliminating everything except the upperleft mesh and replacing the 2 Ω resistor and the 4 V source with short circuits. Additionally, I set V1 equal to 14 V. Through inspection, I found that I1 becomes +2. However, when I removed the 2 Ω resistor and the 4 V source in the mesh equation, I1 became 2. Therefore, I believe I should multiply voltage sources in the equations by a minus sign. Can it be inferred that in mesh current analysis, if we enter the minus part of a voltage source and exit from the positive part, we should use V; otherwise, we should use +V?
You just need to consider whether the voltage increases or decreases as you go through the component in the direction of your mesh current.I followed your suggestion by eliminating everything except the upperleft mesh and replacing the 2 Ω resistor and the 4 V source with short circuits. Additionally, I set V1 equal to 14 V. Through inspection, I found that I1 becomes +2. However, when I removed the 2 Ω resistor and the 4 V source in the mesh equation, I1 became 2. Therefore, I believe I should multiply voltage sources in the equations by a minus sign. Can it be inferred that in mesh current analysis, if we enter the minus part of a voltage source and exit from the positive part, we should use V; otherwise, we should use +V?
You just need to consider whether the voltage increases or decreases as you go through the component in the direction of your mesh current.
Remember, your mesh equation is nothing more than KVL applied around the mesh. You can decide whether you want to sum up the voltage gains, or sum up the voltage drops. Totally arbitrary. But once you decide, you need to be consistent for each mesh.
So, looking at the original circuit and that upperleft mesh, let's sum up the voltage drops (which is typically what people do).
As you go from the negative terminal of a voltage source to the positive terminal, you GAIN whatever the output voltage is, which means that the voltage DROP is the negative of that. So the voltage drop across the source is V1.
As you go across a resistor in the direction of the current, you DROP a voltage that is equal to the current times the resistance. So the voltage drop across the 7 Ω resistor is (I1)(7 Ω).
Similarly, the voltage drop across the 4 V source is 4 V, because you are actually gaining voltage.
The 2 Ω resistor has two currents superimposed on it. I1 drops a voltage while I2 gains voltage, so that drop is (I1I3)(2 Ω).
The resulting equation is therefore:
V1 + I1(7 Ω)  4 V + (I1I3)(2 Ω) = 0
If you had chosen to sum up the voltage gains, the only difference is that each term in this equation would be multiplied by 1.
In post #8 and #9 it was suggested that there is no maximum power solution other than power increasing without bound. Did you find anything else?Thank you,
Your responses greatly assisted me in comprehending mesh analysis and solving the problem accurately.
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