Question

Two wires A and B of different metals have their lengths in the ratio 1:2 and their radii in the ratio 2:1 respectively. The I-V graphs for them are shown in the figure. Find the ratio of their:
(i) Resistances \( \frac{R_A}{R_B} \)
(ii) Resistivities \( \frac{\rho_A}{\rho_B} \)

Hint:

Resistance depends on length, area, and resistivity. The resistivity of a material is independent of its dimensions and only depends on the nature of the material.

Answer 1

Calculating Resistance and Resistivity Ratios

1: Resistance Formula
The resistance of a wire is given by: \[ R = \frac{\rho L}{A} \] where:
- \( \rho \) = resistivity,
- \( L \) = length of the wire,
- \( A = \pi r^2 \) = cross-sectional area. Given:
- Length ratio: \( L_A : L_B = 1:2 \),
- Radius ratio: \( r_A : r_B = 2:1 \),
- Cross-sectional area ratio: \[ A_A : A_B = \pi (2r)^2 : \pi (r)^2 = 4:1 \]
2: Ratio of Resistances \[ \frac{R_A}{R_B} = \frac{\rho_A L_A / A_A}{\rho_B L_B / A_B} \] \[ = \frac{\rho_A (1) / 4}{\rho_B (2) / 1} \] \[ = \frac{\rho_A}{\rho_B} \times \frac{1}{4} \times \frac{1}{2} \] \[ = \frac{\rho_A}{\rho_B} \times \frac{1}{8} \] Thus, \[ \frac{R_A}{R_B} = \frac{\rho_A}{8 \rho_B} \] From the I-V graph, the slope represents resistance \( R \). Using the given graph values, \[ \frac{R_A}{R_B} = 2 \] Thus, \[ 2 = \frac{\rho_A}{8 \rho_B} \]
3: Ratio of Resistivities Rearranging: \[ \rho_A = 16 \rho_B \] \[ \frac{\rho_A}{\rho_B} = 16 \]
4: Conclusion
- Ratio of resistances: \( \frac{R_A}{R_B} = 2 \)
- Ratio of resistivities: \( \frac{\rho_A}{\rho_B} = 16 \)