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.

Answer 1

\[\text { Drift velocity of free electrons is given by }: \]

\[ v_d = \frac{I}{Ane}\]

\[or, v_d \propto \frac{I}{A}\]

\[\text { where A is the area of the conductor and I is the current flowing through the conductor } . \]

\[(i) \text { When the two conductors are connected in series, the current is same through the two wires }\]

\[\text { Therefore, required ratio of drift velocity of free electrons in wire } P_1 to P_2 is\]

\[\frac{v_{d_{P_1}}}{v_{d_{P_2}}} = \frac{I \times A_2}{I \times A_1} = A_2 : A_1 \]

\[\text { where, v_{d_{P_1}} is the drift velocity of electrons in wire} P_1 \text { and } v_{d_{P_2}} \text { is the drift velocity of electrons in wire } P_2 \]

\[(ii) \text { When the two conductors are connected in parallel, the potential drop aross both the conductor is same } . \]

\[ \Rightarrow v_{d_{P_1}} = \frac{V}{R_1 A_1} = \frac{V}{\frac{\rho l}{A_1} A_1} = \frac{V}{\rho l} \text { and } v_{d_{P_2}} = \frac{V}{R_2 A_2} = \frac{V}{\frac{\rho l}{A_2} A_2} = \frac{V}{\rho l}\]

\[\text { Therefore, required ratio of drift velocity of free electrons in wire } P_1 to P_2 is\]

\[\frac{v_{d_{P_1}}}{v_{d_{P_2}}} = \frac{\frac{V}{\rho l}}{\frac{V}{\rho l}} = 1: 1\]