How are we able to flip the triangular resistor to the opposite breaking its connection with the lower 2r resistor?
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$\begingroup$ There is no real question, what do you mean? Does the current remains constant? Does it open a door to a world in a different dimension, can you make your question clear? $\endgroup$– iharobJul 4, 2015 at 18:27
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$\begingroup$ @iharob click on the question to view the image. $\endgroup$– satyatechJul 4, 2015 at 18:31
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2$\begingroup$ I saw the image, it's of no help at all. $\endgroup$– iharobJul 4, 2015 at 18:32
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2$\begingroup$ Perhaps you could add some more context to the question. $\endgroup$– ragnarJul 4, 2015 at 18:43
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$\begingroup$ Questions here get closed if the entire question is not included. Instead of posting just a link, please put all information relevant to the question here so we can read it. $\endgroup$– DanielSankJul 4, 2015 at 20:31
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2 Answers
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Since the circuit is fully symmetrical (left/right symmetry) the potential at C is exactly half the potential between A and B. This means that there is no current flowing across the point (from the "tip of the V" to the middle of the two resistors at point C), and you can break it without changing the underlying equations describing the current flow.
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I assume you mean, how do we know we can transform the circuit this way without changing any of the node voltages or branch currents?
Let's use your node "B" as the reference node, and assign it a potential of 0 V. Then we can see that $V_A$ is just the battery voltage.
From symmetry, we can tell that the voltage at "C" must be $\frac{V_A}{2}$.
Similarly, in the transformed circuit, symmetry tells us that the voltage at "C" will still be $\frac{V_A}{2}$, as will the voltage at the junction between the two "r"-valued resistors at the bottom of the schematic. Since this voltage doesn't change, we know that none of the currents through the resistors changed either.
answered Jul 4, 2015 at 21:13
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$\begingroup$ exactly what i wanted thanks to your brain!! haha!! $\endgroup$ Jul 5, 2015 at 1:07