Suppose in a circuit, a silver wire of length L and cross—sectional area A is replaced by an aluminiumwire of length 5L and cross—sectional area 9A. The resistance of the circuit will ________.

(Given thatρ_silver = 1.6 × 10^-8 Ωm and ρ_aluminium =2.6 × 10^-8Ωm)

Correct option is C

To solve this problem, we will calculate the resistance of the silver wire and the aluminum wire separately using the formula for resistance:

​$$R = \rho \frac{L}{A} ​$$​​

Where:

• Ris resistance
• ρ is resistivity
• L is length

For the silver wire:

​$$R_{\text{silver}} = \rho_{\text{silver}} \frac{L}{A}$$​​

Given:

​$$\rho_{\text{silver}} = 1.6 \times 10^{-8} \, \Omega\text{m}, \, L = L, \,$$A=A​

​$$R_{\text{silver}} = \frac{1.6 \times 10^{-8} \times L}{A}$$​​

For the aluminum wire:

​$$R_{\text{aluminium}} = \rho_{\text{aluminium}} \frac{L_{\text{aluminium}}}{A_{\text{aluminium}}}$$​​

Given:

​$$ρ_{aluminium}=2.6×10^−8Ωm,L_{aluminium}=5L,A_{aluminium}=9A\, \\ \ \\R_{\text{aluminium}} = \frac{2.6 \times 10^{-8} \times 5L}{9A}$$​​

The resistance of the circuit will decrease to 0.9 times of itself.

The resistance of the circuit decreases because aluminum has a higher resistivity than silver but the effect of increasing the length and cross-sectional area results in an overall decrease in resistance by 10%.

● Resistance depends on material properties (resistivity), length, and cross-sectional area.
● Increasing the length increases resistance linearly.
● Increasing the cross-sectional area reduces resistance inversely.
● Aluminum is less conductive than silver but is commonly used due to cost and availability.
● The ratio of resistivities and geometric changes determines the new resistance.