electric-circuitselectric-currentelectrical-resistancehomework-and-exercisessymmetry
I was able to solve the above question by assuming potentials( Nodal Analysis), but the solution was very lengthy as i had three variables and had to solve three equations.
Is it possible to solve this question using symmetry and get a quicker and more elegant solution.
Any help is appreciated.
This is your circuit:
The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, $I_1=I_2$ , So the potential difference across yellow resistors is the same. It means that the potential of point $\mathbf{A}$ is equal to potential of point $\mathbf{B}$ :
$$I_1=I_2\to V_A=4.4-I_1R \text{ , }V_B=4.4-I_2R\to V_A=V_B$$
So there isn't any potential difference across the blue resistor, and the current through it is 0, and it can be omitted from the circuit without any change in the behavior of the circuit.
The basic principle of symmetry being used here is that if the problem has a certain symmetry, then the solution must also have that symmetry.
In this case, if you rotate the charges by 120 degrees about the center point of the triangle, then you end up with the same configuration of charges. Therefore, the problem has a symmetry under rotations by 120 degrees about the center point of the triangle.
Now, suppose the solution involved the electric field pointing in, say, the "up" direction (toward the charge at the top of the triangle). Now again imagine rotating the configuration of charges by 120 degrees. The electric field vector will now be pointing diagonally downward. But this is a contradiction -- for a given configuration of charges, there is only one possible electric field. Yet, we have shown that there are two allowed electric fields for the same problem. This is a contradiction. The resolution is that the electric field must have been zero.
In other words, the original problem does not change if we rotate the charges by 120 degrees. Therefore, the solution (ie, the electric field) must also not change if we rotate the charges by 120 degrees. In this context, the only possible electric field is zero, since any non-zero vector will change under a 120 degree rotation.
Best Answer
The fact the battery has no resistance does not stop you from using superposition.
Let's say you start by shorting the right battery and finding the currents due to the left battery.
You get 2/12 A through the resistors from A to C and C to E. And 2/8 A along the path from A to B, through the shorted battery from B to F, and then through the 4-ohm resistor from F to E. There is no current through the resistors BD or DF in this partial solution because they are shorted by the right-hand battery. And there is no current in resistor CD because of symmetry.
Now short the left battery and find the currents due to the right battery.
You get 2/6 A through resistor BD and DF, and 2/8 A through resistors AB and EF (but in the opposite direction from the current through these resistors in the first part of the solution). No current through resistors AC or CE because they're shorted by the left-hand battery. And no current through resistor CD because of symmetry.
Add these up and you get 1/6 A through resistors AC and CE, and 1/3 A through resistors BD and DF.
You get no current through resistor CD because it had no current in either of the partial solutions.
You get no current through resistors AB and EF because the currents from the two partial solutions for these resistors were equal but opposite, so they sum to 0.