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

• Two conductors A and B
• Length in ratio (l) = 4:3
• Area of cross section in ratio(A) = 1:2
• Resistivity in ratio (ρ) = 2:3

Find out

• Ratio of resistance(R)

Solution

As we know that

➞ ρ = RA/l

➞ R = ρ×l/A

Now, let

• ρ₁ = 2
• ρ₂ = 3
• A₁ = 1
• A₂ = 2
• l₁ = 4
• l₂ = 2

*According to the given condition*

➞ R₁/R₂ = ρ₁ × l₁/A₁/ρ₂ × l₂/A₂

➞ R₁/R₂ = 2 × 4/1/3 × 3/2

➞ R₁/R₂ = 8/1/9/2

➞ R₁/R₂ = 8 × 2/9

➞ R₁/R₂ = 16/9

➞ R₁:R₂ = 16:9

Hence,

• Ratio of resistance
• R₁:R₂ = 16:9

$\rule{200}3$

Answer 2

Answer:

Explanation:Given

Two conductors A and B

Length in ratio (l) = 4:3

Area of cross section in ratio(A) = 1:2

Resistivity in ratio (ρ) = 2:3

Find out

Ratio of resistance(R)

Solution

As we know that

➞ ρ = RA/l

➞ R = ρ×l/A

Now, let

ρ₁ = 2

ρ₂ = 3

A₁ = 1

A₂ = 2

l₁ = 4

l₂ = 2

*According to the given condition*

➞ R₁/R₂ = ρ₁ × l₁/A₁/ρ₂ × l₂/A₂

➞ R₁/R₂ = 2 × 4/1/3 × 3/2

➞ R₁/R₂ = 8/1/9/2

➞ R₁/R₂ = 8 × 2/9

➞ R₁/R₂ = 16/9

➞ R₁:R₂ = 16:9

Hence,

Ratio of resistance

R₁:R₂ = 16:9

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