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Question
Two metallic wires, P1 and P2 of the same material and same length but different cross-sectional areas, A1 and A2 are joined together and connected to a source of emf. Find the ratio of the drift velocities of free electrons in the two wires when they are connected (i) in series, and (ii) in parallel.
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Solution
\[\text { Drift velocity of free electrons is given by }: \]
\[ v_d = \frac{I}{Ane}\]
\[or, v_d \propto \frac{I}{A}\]
\[\text { where A is the area of the conductor and I is the current flowing through the conductor } . \]
\[(i) \text { When the two conductors are connected in series, the current is same through the two wires }\]
\[\text { Therefore, required ratio of drift velocity of free electrons in wire } P_1 to P_2 is\]
\[\frac{v_{d_{P_1}}}{v_{d_{P_2}}} = \frac{I \times A_2}{I \times A_1} = A_2 : A_1 \]
\[\text { where, v_{d_{P_1}} is the drift velocity of electrons in wire} P_1 \text { and } v_{d_{P_2}} \text { is the drift velocity of electrons in wire } P_2 \]
\[(ii) \text { When the two conductors are connected in parallel, the potential drop aross both the conductor is same } . \]
\[ \Rightarrow v_{d_{P_1}} = \frac{V}{R_1 A_1} = \frac{V}{\frac{\rho l}{A_1} A_1} = \frac{V}{\rho l} \text { and } v_{d_{P_2}} = \frac{V}{R_2 A_2} = \frac{V}{\frac{\rho l}{A_2} A_2} = \frac{V}{\rho l}\]
\[\text { Therefore, required ratio of drift velocity of free electrons in wire } P_1 to P_2 is\]
\[\frac{v_{d_{P_1}}}{v_{d_{P_2}}} = \frac{\frac{V}{\rho l}}{\frac{V}{\rho l}} = 1: 1\]
