Two conductors are made of the same material and have the same length. They each have circular cross sectional areas. The ratio of their radii is 2:1. they carry the same current. What is the ratio of the electrical potential across their lengths?
i know that you use R=P(L/A) when your trying to find the cross sectional area. But i don't know what to use when there radii is 2:1
The larger diameter wide has four times the cross sectional area of the thinner wire, and 1/4 of its resistance.
If they carry the same current I, the voltage drop V = IR is four times higher for the thinner wire.
When the radii of two conductors are in a ratio of 2:1, we can represent the radii as 2x and x, where x is a constant. Let's assume that the current passing through both conductors is I.
The electrical potential (V) across a conductor is given by Ohm's law:
V = IR
Since the length (L) of both conductors is the same, we can write their potentials as:
V1 = I * R1
V2 = I * R2
Substituting the values of R1 and R2 using the given ratio of radii:
V1 = I * (2x)
V2 = I * x
Dividing V1 by V2 to find the ratio of potentials:
V1/V2 = (I * (2x)) / (I * x)
V1/V2 = 2x/x
V1/V2 = 2
So, the ratio of the electrical potential across their lengths is 2:1.
To find the ratio of the electrical potential across the lengths of the two conductors, we need to consider their resistances. The resistance of a conductor is proportional to its length (L) and inversely proportional to its cross-sectional area (A).
Given that both conductors have the same length, we can focus on the ratio of their cross-sectional areas.
Let's assume the radius of the first conductor is "r" and the radius of the second conductor is "2r" (since the ratio of their radii is given as 2:1).
The cross-sectional area (A) of a conductor can be calculated using the formula for the area of a circle: A = πr^2.
For the first conductor (with radius "r"), its cross-sectional area (A1) is A1 = πr^2.
For the second conductor (with radius "2r"), its cross-sectional area (A2) is A2 = π(2r)^2 = 4πr^2.
Now, we can find the ratio of their cross-sectional areas (A2/A1):
A2/A1 = (4πr^2) / (πr^2) = 4.
This means that the cross-sectional area of the second conductor is four times larger than that of the first conductor.
Since resistance is inversely proportional to cross-sectional area, the ratio of their resistances would be the reciprocal of the ratio of their cross-sectional areas.
So, the ratio of their resistances (R1/R2) would be 1/4.
Since the conductors carry the same current, the ratio of their electrical potentials (V1/V2) would be the same as the ratio of their resistances.
Therefore, the ratio of the electrical potential across the lengths of the two conductors is 1/4.