Two spherical conducting wires A and B are connected to the same potential difference. Wire A is three times as long as wire B, with a radius double that of wire B and resistivity doubles that of wire B. What is the power delivered to wire A to the power delivered to wire B?
To find the power delivered to wire A and wire B, we need to compare their power using the given information.
The power in a conductor can be calculated using the formula:
P = (V^2) / R,
where P is the power, V is the potential difference, and R is the resistance of the conductor.
Let's start by comparing the resistances of wire A and wire B. The resistance of a conductor can be calculated using the formula:
R = (ρ * L) / A,
where R is the resistance, ρ (rho) is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire.
Given that wire A is three times longer than wire B, with a radius double that of wire B, and a resistivity double that of wire B, we can write:
L(A) = 3 * L(B), (1)
ρ(A) = 2 * ρ(B), (2)
R(A) = (ρ(A) * L(A)) / A(A), (3)
R(B) = (ρ(B) * L(B)) / A(B). (4)
Now, we can compare the cross-sectional areas of wire A and wire B. The cross-sectional area of a wire is given by:
A = π * r^2,
where A is the cross-sectional area, and r is the radius of the wire.
Given that the radius of wire A is double that of wire B, we can write:
r(A) = 2 * r(B). (5)
To find the power delivered to wire A and wire B, we need to calculate their resistance first. Using equations (2), (3), (4), and (5), we can substitute the values and simplify the equations.
For wire A:
L(A) = 3 * L(B),
ρ(A) = 2 * ρ(B),
r(A) = 2 * r(B).
Substituting these values into equation (3):
R(A) = (2 * ρ(B) * 3 * L(B)) / (π * (2 * r(B))^2,
R(A) = (6 * ρ(B) * L(B)) / (4 * π * r(B)^2),
R(A) = (3/2) * (ρ(B) * L(B)) / (π * r(B)^2).
Similarly, for wire B:
R(B) = ρ(B) * L(B) / (π * r(B)^2).
Now, we can find the ratio of the power delivered to wire A and wire B:
P(A) / P(B) = (V^2 / R(A)) / (V^2 / R(B)),
P(A) / P(B) = R(B) / R(A),
P(A) / P(B) = [(π * r(B)^2) * (ρ(B) * L(B))] / [(3/2) * (ρ(B) * L(B))],
P(A) / P(B) = 2/3.
Therefore, the power delivered to wire A is two-thirds (2/3) of the power delivered to wire B.