Question

Two wires made of the same material have the same length \( l \) but different cross-sectional areas \( A_1 \) and \( A_2 \). They are connected together with a cell of voltage \( V \). Find the ratio of the drift velocities of free electrons in the two wires when they are joined in: (i) series, and (ii) parallel.

Hint:

In series circuits, current is constant, so drift velocity varies inversely with cross-sectional area. In parallel circuits, voltage is constant, and current divides in proportion to area, keeping drift velocities equal.

Answer 1

The drift velocity of electrons is given by: \[ v_d = \frac{I}{nAe} \] where \( v_d \) = drift velocity, \( I \) = current, \( n \) = number density of electrons, \( A \) = cross-sectional area, \( e \) = elementary charge. (i) Series connection: In a series circuit: - The current \( I \) is the same through both wires. - So, the drift velocities are: \[ v_{d1} = \frac{I}{nA_1e}, \quad v_{d2} = \frac{I}{nA_2e} \Rightarrow \frac{v_{d1}}{v_{d2}} = \frac{A_2}{A_1} \] (ii) Parallel connection: In a parallel circuit: - The voltage \( V \) across each wire is the same. - Resistance \( R = \frac{\rho l}{A} \Rightarrow I \propto A \) - Therefore, \[ I_1 \propto A_1, \quad I_2 \propto A_2 \Rightarrow v_{d1} = \frac{I_1}{nA_1e} = \frac{A_1}{nA_1e} = \frac{1}{ne} \] \[ v_{d2} = \frac{I_2}{nA_2e} = \frac{A_2}{nA_2e} = \frac{1}{ne} \Rightarrow \frac{v_{d1}}{v_{d2}} = 1 \]