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For any given resistor following Ohm's law say a carbon resistor, it's resistance is independent of the value of current and voltage.

However, say we had a 5 ohm resistor in a circuit with a 10V battery. Ideally, the 10V potential would be dropped entirely across the resistor. This 10V potential is the work that would be done to move a 1C charge around the circuit by the definition of potential.

But, if say I connected another 5 ohm resistor in series with this, the drop across each resistor would reduce to 5V. How is it possible that for a given resistance whose value doesn't change (at relatively low current values) the work done to move a unit positive charge changes? After all resistance is the obstruction offered to the flow of current and hence the energy required to overcome this opposition for a given amount of charge should always be fixed?

What aspect of this reasoning is incorrect? The only thing I could thing of would be that we are dealing with potential energy instead of potential since that would make since due to varying current leading to a reduced amount of charge and hence reduced potential energy. But I'm pretty sure that goes against the definition.

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If you put two 5ohm resistors in series, then the current flowing will be half of what it was with a single resistor.

With half the current, it takes twice as long to move 1C through the circuit, that is, to have the battery expend 10J of energy through the resistors. The battery is working at half the power, power = rate of doing work.

Each resistor dissipated 5J, that is 5v was required to push 1C through them at the lower current.

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  • \$\begingroup\$ What factor is responsible for the lower voltage required? Higher relaxation time? Or just lesser number of effective collisions? \$\endgroup\$ – LeroyJD Mar 11 '17 at 6:53
  • \$\begingroup\$ It's rather difficult to assign 'cause', try not to get hung up on it. Does the current cause the voltage, or the other way around? The point is they are both parameters of the same interaction, the dissipation of energy that was in the battery, into the resistor. The definition of voltage you quote is just that, a definition, it's not a model. The main factor responsible for the lower voltage is that you chose to put a second resistor in series with the first, and connect both to the original battery, so each ended up with half the voltage on it. Set experimental conditions, see what occurs. \$\endgroup\$ – Neil_UK Mar 11 '17 at 7:06

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