Question

Two conductors A and B of have their lengths in ratio 4 : 3, area of cross-section in ratio 1 : 2 and their resistivities in ratio 2 : 3. What will be the ratio of their resistances?

Answer 1

Given that, two conductors A and B of have their lengths in ratio 4:3, area of cross-section in ratio 1:2 and their resistivities (resistivity) in ratio 2:3.

We have to find the ratio of resistances.

We know that resistance of a wire is directly proportional to it's length and inversely proportional to Area of cross-section. Where rho (p = resistivity) is constant.

R = p l/A

Let's denote the resistance for A by R' and for B conductor by R".

For conductor A:

R' = 2 × 4/1

R' = 8 ohm

For conductor B:

R" = 3 × 3/2

R" = 9/2 ohm

As per given condition we have to find R'/R". So,

R'/R" = 8/(9/2)

R'/R" = 16/9

Therefore, the ratio of resistances is 16:9.

Answer 2

Explanation:

⭐ AnswEr ⭐

Given:

Ratio of length(l)=4:3

Ratio of cross section (A)=1:2

Ratio of resistivity (R)=2:3

Solution:

$R_A$(resistance of A)=ρl/a

$\implies \: R_A = 2x \times \frac{4x}{x}$

$\implies \: R_A = 8x$

$R_B$(resistance of B)=ρl/a

$\implies \:R_B = \frac{9}{2} x$

$\implies \: R_B = 3x \times \frac{3x}{2x}$

Ratio of their resistance= 8x×2/9x =16/9=16:9

♣️answer is 16:9