Respuesta :
Answer:
3/4
Explanation:
Here is the complete question
Consider two cylindrical conductors made of the same ohmic material. If ρ2 = ρ1 , r2 = 2 r1 , ℓ2 = 3 ℓ1 , and V2 = V1 , find the ratio R2 R1 of the resistances
Resistance, R = ρl/A where ρ = resistivity, l = length and A = area
For cylinder 1, R₁ = ρ₁l₁/A₁
For cylinder 2, R₂ = ρ₂l₂/A₂
ρ₂ = ρ₁, l₂ = 3l₁, r₂ = 2r₁ where r₁ and r₂ are the radii of the cylinders.
R₂/R₁ = ρ₂l₂/A₂ ÷ ρ₁l₁/A₁
= ρ₂l₂A₁/ρ₁l₁A₂
= ρ₂l₂πr₁²/ρ₁l₁πr₂²
= ρ₂l₂r₁²/ρ₁l₁r₂²
substituting the required values
= ρ₁ × 3l₁ × r₁²/ρ₁ × l₁ × (2r₁)²
R₂/R₁ = 3/4
The ratio of r2 to r1 of the two cylindrical conductors made of the same ohmic material is 1/4.
What is the resistance?
Resistance is obstacle for the current flow in the circuit. It is the measure of reverse act to current flow in through a material.
It can be given as,
[tex]R=\dfrac{\rho L}{A}[/tex]
Here, ([tex]\rho[/tex]) is the specific resistance (l) is the length of the wire and (A) is the cross sectional area of the wire.
Here, two cylindrical conductors are given which are made of the same ohmic material.
Lets denote the cylinder first with subscript 1. Then resistance by te cylindrical conductor 1 is,
[tex]R_1=\dfrac{\rho_1 L_1}{A_1}\\R_1=\dfrac{\rho_1 L_1}{\pi r^2_1}[/tex]
Lets denote the second cylinder with subscript 2. Then resistance of the cylindrical conductor 2 is,
[tex]R_2=\dfrac{\rho_2 L_2}{A_2}\\R_2=\dfrac{\rho_2 L_2}{\pi r^2_2}[/tex]
Here, given in the problem that,
[tex]r_2=2r_1\\\rho_2=\rho_1\\v_2=v_1\\L_2=2L_1[/tex]
Thus, put these values in the above expression as,
[tex]R_2=\dfrac{(\rho_1) (2L_1)}{\pi 2r_1^2}[/tex]
Thus the ratio of second resistance to first is,
[tex]\dfrac{R_2}{R_1}=\dfrac{(\rho_1) (2L_1)}{\pi (2r_1)^2}\div \dfrac{(\rho_1) (L_1)}{\pi r_1^2}\\\dfrac{R_2}{R_1}=\dfrac{1}{4}[/tex]
Hence, the ratio of r2 to r1 of the two cylindrical conductors made of the same ohmic material is 1/4.
Learn more about the resistance here;
https://brainly.com/question/17563681
