Let us be two cylindrical conductors connected in parallel, to which a potential difference of V = 170V is applied. The two conductors are made of the same material, but the first is 6 times the length of the second, and 1/5 the radius of the second. The resistance of the second is R2 = 469Ω. Determine the equivalent resistance.
Question
Let us be two cylindrical conductors connected in parallel, to which a potential difference of V = 170V is applied. The two conductors are made of the same material, but the first is 6 times the length of the second, and 1/5 the radius of the second. The resistance of the second is R2 = 469Ω. Determine the equivalent resistance.
Solution 1
The resistance of a cylindrical conductor can be given by the formula:
R = ρL/A
where: R is the resistance, ρ is the resistivity of the material, L is the length of the conductor, and A is the cross-sectional area of the conductor.
Given that the first conductor is 6 times the length of the second, and 1/5 the radius of the second, we can express the resistance of the first conductor (R1) in terms of the resistance of the second conductor (R2) as follows:
R1 = ρ(6L)/(π(1/5r)^2) = 150ρL/πr^2 = 150R2
Given that R2 = 469Ω, we find that:
R1 = 150 * 469Ω = 70350Ω
The two resistors are connected in parallel, so the equivalent resistance (Re) can be found using the formula for resistors in parallel:
1/Re = 1/R1 + 1/R2
Substituting the values we found for R1 and R2 gives:
1/Re = 1/70350Ω + 1/469Ω
Solving this equation gives:
Re ≈ 466.67Ω
So, the equivalent resistance of the two cylindrical conductors connected in parallel is approximately 466.67Ω.
Solution 2
The resistance of a cylindrical conductor can be calculated using the formula:
R = ρL/A
where: R is the resistance, ρ is the resistivity of the material, L is the length of the conductor, and A is the cross-sectional area of the conductor.
Given that the first conductor is 6 times the length of the second and 1/5 the radius, we can substitute these values into the formula to find the resistance of the first conductor (R1).
R1 = ρ(6L)/(π(1/5r)^2) R1 = 150ρL/πr^2
We know that the resistance of the second conductor (R2) is 469Ω, and we can express this in terms of ρ, L, and r:
R2 = ρL/πr^2
We can equate these two expressions to solve for ρL/πr^2:
150ρL/πr^2 = 469Ω ρL/πr^2 = 469/150 ρL/πr^2 = 3.13Ω
Substituting this back into the equation for R1 gives:
R1 = 150 * 3.13Ω R1 = 469.5Ω
The two resistors are connected in parallel, so the equivalent resistance (Re) can be calculated using the formula:
1/Re = 1/R1 + 1/R2 1/Re = 1/469.5Ω + 1/469Ω 1/Re = 0.00426 Re = 234.75Ω
So, the equivalent resistance of the two cylindrical conductors connected in parallel is approximately 234.75Ω.
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