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Drift Velocity in a Resistive Network

In the circuit shown, the battery is connected across AB. Will the drift velocity of electrons be maximum in the smallest resistance $R$ as there would be least resistance for the electrons to move in $R$? (all resistors are made of wires with identical cross sectional area and identical material)


Answer

The given circuit is actually balanced Wheatstone bridge as $\frac {4R}{2R} = \frac {8R}{4R}$ which means no current flows through the resistance $R$. So, even though the resistance $R$ is smallest, the drift velocity is not going to be maximum in $R$. In fact the drift velocity would be zero in $R$. 

In the top we would have $2R+4R=6R$ and in the bottom we would have $4R+8R=12R$. $6R$ and $12R$ would be in parallel, so the higher current would flow through $6R$. So, the current would be maximum in $2R$ and $4R$ both. 

Now, $I=nAev_d$

So, drift velocity is same for resistive wires having identical current, cross-sectional area and identical material. Hence, the drift velocity would be maximum in $2R$ and $4R$ both.