The resistance of a cylindrical conductor is given by the formula:
R=ρ*L/A
where:
( R ) is the resistance,
( \rho ) is the resistivity of the material,
( L ) is the length of the conductor, and
( A ) is the cross-sectional area of the conductor.
Given that the two conductors are made of the same material and are connected in parallel, the equivalent resistance (( R_{eq} )) can be calculated using the formula for parallel resistors:
1���=1�1+1�2Req1=R11+R21
We know that ( R_2 = 469 , \Omega ), and that the first conductor is 6 times the length of the second, and has half the radius. Therefore, the resistance of the first conductor (( R_1 )) can be calculated as:
�1=�6�14��2=24�2R1=ρ41πd26L=24R2
Substituting ( R_1 = 24 R_2 ) into the formula for ( R_{eq} ), we get:
1���=124�2+1�2=2524�2Req1=24R21+R21=24R225
Solving for ( R_{eq} ), we find:
���=24�225Req=2524R2
Substituting ( R_2 = 469 , \Omega ) into the equation, we can calculate the equivalent resistance.