Question Resistance and Resistivity: The length of a certain wire is doubled while its radius is kept constant. What is the new resistance of this wire?

Answer: Explanation: The formula for calculating the resistance of a material in terms of its resistivity is expressed as [tex]R = \rho L/A[/tex] where; R is the resistance of the material [tex]\rho[/tex] is the resistivity of the material L is the length of the wire A is the area = πr² with r being the radius [tex]R = \rho L/\pi r^{2}[/tex] If the length of a certain wire is doubled while its radius is kept constant, then the new length of the wire L₁ = 2L The new resistance of the wire R₁ will be expressed as [tex]R_1 = \frac{\rho L_1}{A_1}[/tex] since the radius is constant, the area will also be the same i.e A = A₁ and the resistivity also will be constant. The new resistance will become [tex]R_1 = \frac{\rho(2L)}{A}[/tex] [tex]R_1 = \frac{2\rho L}{\pi r^2}[/tex] Taking the ratio of both resistances, we will have; [tex]\frac{R_1}{R} = \frac{2\rho L/\pi r^2}{\rho L/ \pi r^2} \\\\\frac{R_1}{R} = \frac{2\rho L}{\pi r^2} * \frac{\pi r^2}{ \rho L} \\\\\frac{R_1}{R} = \frac{2}{1}\\\\R_1 = 2R[/tex] This shoes that the new resistance of the wire will be twice that of the original wire Log in to Reply

Answer:Explanation:The formula for calculating the resistance of a material in terms of its resistivity is expressed as [tex]R = \rho L/A[/tex] where;

R is the resistance of the material

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the area = πr² with r being the radius

[tex]R = \rho L/\pi r^{2}[/tex]

If the length of a certain wire is doubled while its radius is kept constant, then the new length of the wire L₁ = 2L

The new resistance of the wire R₁ will be expressed as [tex]R_1 = \frac{\rho L_1}{A_1}[/tex]

since the radius is constant, the area will also be the same i.e A = A₁ and the resistivity also will be constant. The new resistance will become[tex]R_1 = \frac{\rho(2L)}{A}[/tex]

[tex]R_1 = \frac{2\rho L}{\pi r^2}[/tex]

Taking the ratio of both resistances, we will have;

[tex]\frac{R_1}{R} = \frac{2\rho L/\pi r^2}{\rho L/ \pi r^2} \\\\\frac{R_1}{R} = \frac{2\rho L}{\pi r^2} * \frac{\pi r^2}{ \rho L} \\\\\frac{R_1}{R} = \frac{2}{1}\\\\R_1 = 2R[/tex]

This shoes that the new resistance of the wire will be twice that of the original wire