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I am having trouble reconciling two seemingly contradictory results. Consider two conductors, X and Y, made from the same material with the same length and carrying the same current. If X has twice the cross-sectional area of Y, then from the formula $I=nevA$, we deduce that the drift velocities for X and Y have the ratio $$v_X=\frac{v_Y}{2}.$$ Now, consider the resistance of X and Y. Since resistance is inversely proportional to current, and current is proportional to drift velocity, resistance must be inversely proportional to drift velocity. This gives the result: $$R_X=2R_Y.$$ However, this would seem to contradict the formula for the resistivity of a wire, $$R=\frac{\rho l}{A}$$ which states that resistance is inversely proportional to area. Since X has twice the area of Y (and being made from the same material with equal length, $\rho$ and $l$ are constant) we obtain the contradictory result $$R_X=\frac{R_Y}{2}.$$

A Possible Reconciliation

I have tried to reconcile this by substituting one formula into the other. Rearranging the resistivity formula to make $A$ the subject $$A=\frac{\rho l}{R}$$ and substituting this into $I=nevA$ we get $$R=\frac{n\rho l e}{I} v.$$ Since $n, \rho, l, e$ and $I$ are all constant, we now see that $R\propto v$. This resolves the contradiction, but I am left confused.

The Intuition

My intuition tells me that $R\propto 1/v$ seems more reasonable. If the resistance increases, then it should decrease the average velocity of the electrons along the wire. So, my question is, why does drift velocity increase with resistance in this case?

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1 Answer 1

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Drift velocity depends on the applied voltage bias, whereas resistance does not. The first ration between the drift velocities is obtained that the two conductors carry the same current, which means that the potential differences applied to the two conductors are different.

You could get a deeper understanding of this by reading about the Drude model.

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  • $\begingroup$ Thank you for the link, but I still don't understand why a higher resistance would lead to higher drift velocity. The main point of the model seems to be that the current density is proportional to the electric field. How does this relate to the resistance? Also, I'm not quite sure which ratio you are saying is correct. Are you saying $R_X = 2R_Y$ after all? $\endgroup$
    – DJTS
    Dec 3, 2020 at 13:58
  • $\begingroup$ @DJTS That current density is proportional to the electric field is known as Ohm's law: $j=\sigma E$, can be converted to the current and voltage bias as: $I=jA=\sigma A/l V=V/R$. Perhaps this resolves the confusion? $\endgroup$
    – Roger V.
    Dec 3, 2020 at 14:04
  • $\begingroup$ Yes, I can now see the relationship between Drude's model and resistance, thanks. However, what I am left confused by is the intuition of the relationship $R\propto v$. How do these formulae explain the rise in drift velocity with an increase in resistance? Do they imply a larger electric field acting on the electrons? $\endgroup$
    – DJTS
    Dec 3, 2020 at 14:09
  • $\begingroup$ Resistance is independent on the drift velocity - it is a characteristic of a conductor. If you have two conductors with different resistance and you want to the same current flowing in both, you need to apply higher potential difference to the one with higher resistance. $\endgroup$
    – Roger V.
    Dec 3, 2020 at 14:12
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    $\begingroup$ Thank you. So I am assuming that this relationship $R\propto v$ is just a manifestation of the present example and not a fundamental relationship between resistance and drift velocity. This makes sense. So what is happening is a larger potential difference is being applied across the larger resistance, in order to keep the current constant. This, in turn, increases the drift velocity, which lead me to the fallacious conclusion that resistance is dependent on drift velocity. In that case, the correct answer should be $R_X = R_Y / 2$, from the resistivity formula. Is that correct? $\endgroup$
    – DJTS
    Dec 3, 2020 at 14:20

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