Question #51a07

1 Answer
Dec 6, 2017

Increased resistance decreases the drift velocity which in turn decreases the current.


The relevant equation to consider here is:
#I = nAvq#
The terms are: I = current, n = number of charge carriers per unit volume, A = cross-sectional area, v = drift velocity (of charge carriers) and q = charge on carriers. These terms are the factors that affect the magnitude of current.

Resistance doesn’t feature in the equation. But we know that an increased resistance does decrease current for a constant potential difference (#I=V/R#). So, according to our equation above, increasing the resistance must be decreasing one of the factors of current – n, A, v or q.

If we consider a metal wire we can eliminate some of those factors. Resistance in metals is affected by impurities and temperature (I’m not here discussing changes to the physical dimensions of the conductor, just changes to the actual medium of the current). Changes to neither of those will change A or q – the wire stays the same size (for moderate changes to temperature) and the charge on electrons is unchanged.

Arguably n is not affected by much. Adding more impurities probably does decrease the number of charge carriers but I don’t think it would decrease it significantly. If we were comparing a conductor with a semi-conductor then this is a very significant factor.

However, changes to both the number of impurities and temperature do affect v, the drift velocity. Both factors cause more obstructions to the flow of electrons through the conductor. So an increase in resistance causes v to decrease which in turn decreases the current.