Shaskolskaya & Eltsin Problem 260

Physics

Problem

Since wires MACN and MBDN are connected in parallel between points M and N, the fall of potential is one and the same for the whole length of these wires. The fall of potential is distributed uniformly along each of these wires. Therefore if length MA equals length MB, the potentials at A and B will be the same and current will not pass along wire AB. The same thing can be said of sections MC and MD. If they are of the same length as one another, current will not pass along CD. But wires AB and 224 SOLUTIONS CD will be at different potentials and if points E and F upon these wires are connected, current will pass along EF and consequently also along AE, BE, FC and FD. This will happen wherever points E and F are chosen along wires AB and CD, since the potential at E will always be higher than that at F.

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