A wire of initial length Lo and radius rO has a measured resistance of 1.0 ohm. the wire is drawn under tensil stress and to a new uniform radius of 0.25rO. what is the new resistance of the wire? i know the answer is 256 ohm but i have no clue how to get the answer can anyone help me?
The cross sectional ara is reduced by a factor (1/4)^2 = 1/16. The length must increase by a factor of 16 to keep the total volume constant.
(since Volume = Area x Length)
The resistance is proportional to Length/Area^2, so that increases by a factor of 16/(1/16) = 16 * 16 = ?
To find the new resistance of the wire, we can use the formula for resistance in a wire:
R = ρ * L / A
where:
- R is the resistance
- ρ (rho) is the resistivity of the material the wire is made of
- L is the length of the wire
- A is the cross-sectional area of the wire
In this case, we know the initial resistance, R₀, so we can set up the equation:
R₀ = ρ * L₀ / A₀
where:
- R₀ = 1.0 ohm (given)
- L₀ = initial length of the wire (Lo)
- A₀ = initial cross-sectional area of the wire (π * r₀²)
We are given that the wire is drawn to a new uniform radius of 0.25r₀. So, the new length, L, will be:
L = Lo (because the length remains the same)
The new cross-sectional area, A, can be calculated using the new radius, r:
A = π * r²
To calculate the new resistance, R, we need to find the new values of L and A. Let's substitute the values into the equations:
L = Lo = initial length of the wire
A = π * (0.25r₀)² = π * (0.25)² * r₀² = π * 0.0625 * r₀² = 0.19635 * r₀²
We can now substitute the new values of L and A into the resistance equation:
R = ρ * L / A = ρ * Lo / (0.19635 * r₀²)
Since R₀ = 1.0 ohm, ρ = R₀ * (0.19635 * r₀²) / Lo. Plugging this value into the equation:
R = (R₀ * (0.19635 * r₀²) / Lo) * Lo / (0.19635 * r₀²)
R = R₀
So, the new resistance, R, is equal to the initial resistance, R₀, which is 1.0 ohm. Therefore, the answer of 256 ohms seems to be incorrect. The correct new resistance should be the same as the initial resistance, which is 1.0 ohm.