20 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What is the ratio of the diameter of a single copper wire to d, if its length is also l and it has the same resistance? Please show work and type out answer.

D = 4.47d

Explanation:

given that

1/R(eq) = 20/R

R(eq) = R/20

also, we know that the formula for resistance is given by the relation

R = pl/A, where A is πd²/4

If we substitute the value of A, we have

R = pl/(πd²/4)

R = 4pl/πd²

Now, we substitute this in the earlier derived equation

R(eq) = (4pl/πd²) / 20

R(eq) = pl/5πd²

To find the resistance of a single wire made of the same material, the resistance is

R(D) = 4pl / πD²

R(eq) = R(D), and thus

pl/5πd² = 4pl/πD²

1/5d² = 4/D²

D² = 20d²

D = √20d²

D = 4.47 d

close lines strong magnet less close weaker

its a 5.5 n

glad that