Question

A conductor with rectangular across section has dimension ( a × 2a × 4a) as shown in figure. Resistance across AB is X, across CD is y and across EF is z. then ratio of xyz is​

Answer 1

Answer:

$\frac{32a}{ density \: of \: conductor}$

Explanation:

Resistance(R) = density×(length/area)

Let, density be d(constant)

Now for AB cross section,

X=d×(length of AB/area of cross section AB)

X=d×(4a/a×2a)

X=d×(2/a)

Similarly for CD cross section,

Y=d×(length of CD/area of cross section CD)

Y=d×(a/4a×2a)

Y=d×(1/8a)

And for EF cross section,

Z=d×(length of EF/area of cross section EF)

Z=d×(2a/4a×a)

Z=d×(1/2a)

Lastly, X:Y:Z= 32a/d