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I found in this link something I can't understand: enter image description here

Consider a current flowing out of point a below. The current must split up into three equal portions, since all three branches from point a are connected to the same resistance. Thus, the currents in branches ab, ac, and ad must be equal.

This is the premise needed to remove the resistor bd. But why "same resistance"? Why does it say that the current equally splits between the three branches? I think this is also evidently in contrast with the last simplification step. How could be the current same in the branches if their resistances are 2R,2R,R?

I believe that there's an unneeded assumption. To remove the bd resistor we just can say that the current in ab and in ac (not in ad!) are the same because the shortest journey to d is 2R. Is it wrong?

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  • $\begingroup$ A bit late, but if you’d like to look at a problem which does involve some of the (what I assume to be intended) reasoning of this problem, I suggest looking at the resistance across the diagonal of a cube where all of the edges are resistors. $\endgroup$ – Josh McK Apr 6 '18 at 2:30
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The solution provided contradicts itself. The three branches from point a are not connected to the same resistance - path ad requires passage through only 1 resistor, while a path passing through b or c requires passage through at least 2 resistors. Further, the statement that the voltages at points b, c, and d are the same is impossible. If this were the case, no current would flow along bd or cd. Since all currents were already said to be equal, no current could flow along ad either, and no current would flow anywhere.

However, since paths $a\rightarrow b \rightarrow d$ and $a\rightarrow c \rightarrow d$ are equivalent, we can state that the voltages at points b and c are equal. This means that no current will flow along bc, so we can disregard that resistor. The subsequent work and the final effective circuit drawn are correct.

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    $\begingroup$ And the author repeats the same wrong argument in problem n. 10! Gosh! $\endgroup$ – Massimo Ortolano Jan 8 '17 at 21:06
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You're correct. It's clear that the author of the test problem is wrong about "3 equal portions".

The voltage drop from A to D via path ACD or ABD is identical. Because all resistors are equal, the electrons would have to "choose" the best direction to cross the BC connection. 50% will "choose" the BC direction and 50% will "choose" the CB direction and will completely negate each other. Therefore, the BC link is irrelevant to the circuit.

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The given answer is nonsense because it is self-contradictory. First it says the current is the same in ab, ac, and ad. Then it says the voltage is the same at points b, c, and d.

But if the voltage is the same at b c and d, there is no current in bc and bd. So there is nowhere for the current through ab to go, and that must also be zero. Therefore the voltage at a and b is the same.

In other words, the voltage at all four nodes is the same, even though there is a 9V battery connected between two of them!

Conclusion: don't go to the University of Alabama to learn Physics and Astronomy!

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protected by Qmechanic Jan 8 '17 at 23:18

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