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Why doesn't the electric potential constantly decrease in a circuit as the current moves from the positive to the negative terminal?

Details and Assumptions: 1. The wires show no resistance.

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    $\begingroup$ That totally depends on the circuit. In a circuit with a capacitor the potential will decrease, in a circuit with a power supply that has a virtually infinite capacity to supply energy to the circuit, it does not have to. $\endgroup$ – CuriousOne Apr 29 '15 at 0:30
  • $\begingroup$ Current only moves from positive to negative in the power source, not in an external circuit. Maybe the battery is going dead. $\endgroup$ – Optionparty Apr 29 '15 at 1:57
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    $\begingroup$ It does........ $\endgroup$ – user253751 Feb 11 '19 at 4:27
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In a normal electric circuit, with copper wires at ambient temperature, electric potential does decrease around the circuit, as there will be some resistance and so there has to be a voltage/potential gradient for the current to flow. Copper wires DO have some resistance, so your assumption is invalid in that case.

However, your assumption would be valid for superconducting wires below their critical temperature. Here, the resistance is zero, which means that you can have a current flowing without needing to have any potential gradient or EMF to push it along.

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Leaving aside superconductors, wires will always have some resistance, and the potential always does reduce continuously as the current moves around the circuit. However, the assumption reflects the fact that the wires connecting resistors have low resistance so you can ignore them for the purpose of your calculations. In other words, is a practical approximation to reality that makes the sums easier without an appreciable lack of precision in the answer.

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It does as others mentioned. For example, suppose you have a source with two poles, one pole being at potential 0 (zero) and the other one at a value V. if you insert two resistors in series between the two poles, with resistance R1 and R2, the current in the circuit is I = V/(R1+R2). There is a potential drop of V1 = R1 I across the first resistor and a drop of V2 = R2 I across the second resistor. Note that the total potential drop is V = V1 + V2. If the resistor with resistance R1 is the one that is connected to the 0 potential pole, then the potential at the junction between the two resistors is V1 = R1 V/(R1+R2), which is of course smaller than V, but larger than 0. I ignored wire resistance and the source internal resistance, as per the assumption of an ideal situation. Of course, the total potential between the two poles does not change. However, if you measure the voltage between the 0 pole and the junction of the two resistors, you will see that the value is smaller than V.

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