Why does current flow in a short circuited wire? I understand that it offers negligible resistance to the flow of charges, but two points on the short circuited wire will have the same potential, so shouldn't current through it be zero? Since ohm's law states that I=(delta V)/R
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Two points on a short-circuited wire are not at the same potential���an ideal battery has a voltage difference between its terminals by definition, and that voltage difference has to be dropped somewhere. When there are no proper resistors in the circuit between the battery terminals, the wire itself is now the "resistor," just with a voltage drop equal to that of the battery and thus very high current. This approximation isn't good for long, because either something (either the wire, or a contact, or internal resistance of the battery) will get hot enough to fail or the battery won't be able to provide the whole current.
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$\begingroup$ Okay. Is that why in a balanced Wheatstone setup, the "middle wire" is considered as an open circuit and not a short circuit? $\endgroup$ Commented Apr 12 at 5:12
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2$\begingroup$ @ThrisshaArcot, in a balanced Wheatstone bridge none of the currents and potentials change when the "middle wire" is connected or disconnected. You can treat it as an open circuit with no potential difference or a short circuit with no current. $\endgroup$– PeterCommented Apr 12 at 5:42