Chapter 18, Problem 8Q

To determine

To explain:

The side lengths of a rectangular solid made of carbon, are given to be a, 2a and 3a. Determine the direction that will provide the least resistance and the one that will provide the greatest resistance.

Answer to Problem 8Q

Solution:

Correct answer is (a)along side of length a, (b)along side of length 3a

Explanation of Solution

Given:

The side lengths of a rectangular solid made of carbon, are given to be a, 2a and 3a. Let us assume that the three sides of lengths are along x, y and z axes, respectively.

Formula used:

    • The resistance (R)of a resistor is related with the resistivity of it’s material (ρ), it’s length (l)and the cross-sectional area (A)as,

           $R=\frac{\rho \times l}{A}=\frac{\rho \times l}{a\times b}$ where a and b are sides of cross-sectional face of the resistor.

Calculation:

     a)For the resistance through which the current that passes in the solid along x-direction, the cross-sectional face will have the dimensions 2a and 3a and the length would be a. Therefore it will offer a resistance:

             $R_x=\frac{\rho \times l}{A}=\frac{\rho \times l}{a\times b}=\frac{\rho \times a}{2a\times 3a}=\frac{\rho }{6a}$

     b)For the resistance through which the current that passes in the solid along y-direction, the cross-sectional face will have the dimensions a and 3a and the length would be 2a. Therefore it will offer a resistance:

             $R_y=\frac{\rho \times l}{A}=\frac{\rho \times l}{a\times b}=\frac{\rho \times 2a}{a\times 3a}=\frac{2\rho }{3a}$

     c)For the resistance through which the current that passes in the solid along z-direction, the cross-sectional face will have the dimensions a and 2a and the length would be 3a. Therefore it will offer a resistance:

             $R_z=\frac{\rho \times l}{A}=\frac{\rho \times l}{a\times b}=\frac{\rho \times 3a}{a\times 2a}=\frac{3\rho }{2a}$