A steel wire of length LL and radius r1r1 has a resistance RR. A second steel wire has the same length but a radius r2r2 and a resistance of 2RR. Find the ratio r1/r2r1/r2. Express your answer to three significant figures.

r₁ / r₂ = 1.41

The resistance, R, of a wire is related to its resistivity, ρ, its crossectional area, A, and its length, L, as follows;

R = ρL/A         (i)

Where;

A = π x r²      [r = radius of the wire]

Therefore, equation (i) can be re-written as;

R =  ρL / (π x r²)         (ii)

(i) Now, For the first wire;

R = R

r = r₁

L = L

Substitute these values into equation (ii) as follows;

R =  ρL / (π x r₁²)

Make r₁ subject of the formula;

r₁² =  ρL / (π x R)

r₁ = √{ρL / (πR)]

(ii) Also, for the second wire;

R = 2R

r = r₂

L = L

Substitute these values into equation (ii) as follows;

2R =  ρL / (π x r₂²)

Make r₂ subject of the formula;

r₂² =  ρL / (π x 2R)

r₂ = √{ρL / (2πR)]

(iii) Now let's find the ratio of r₁ / r₂

r₁ = √{ρL / (πR)]

r₂ = √{ρL / (2πR)]

=> r₁ / r₂ = √ [  {ρL / (πR)]   / {ρL / (2πR)]   ]

=> r₁ / r₂ = √ [  {ρL / (πR)]   ÷  {ρL / (2πR)]   ]

=> r₁ / r₂ = √ [  {ρL / (πR)]   x  {2πR / (ρL)]   ]

=> r₁ / r₂ = √ [  {2πR / (πR)]  ]

=> r₁ / r₂ = √ [2]

=> r₁ / r₂ = √2

=> r₁ / r₂ = 1.41421

=> r₁ / r₂ = 1.41

Therefore, the ratio r₁ / r₂ = 1.41

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

im guessing the west of station b