Question:
A uniform resistive wire of length L and resistance R is connected to a constant voltage source of V volts. The wire is made of a material with resistivity ρ. The wire is then bent into a circular loop with radius r = L/(2π). Calculate the resistance R' of the new loop in terms of R and r.
Explanation:
To calculate the new resistance R' of the circular loop, we consider the fact that the resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. We can use the equation R = (ρ * L) / A, where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area.
The original wire has length L and resistance R, so its cross-sectional area is given by A = (ρ * L) / R.
When the wire is bent into a circular loop, the new length of the loop is equal to the circumference of the circle, which is 2πr. Therefore, the length of the wire in the loop is L' = 2πr.
The new loop has radius r = L / (2π), so its circumference is 2πr = L and its new length is L' = L.
Given that voltage source is constant V volts, we use Ohm's Law V = I * R, where V is the voltage, I is the current, and R is the resistance.
Since the voltage is constant, the current flowing through the wire in the loop is the same as the current in the straight wire. Therefore, V = I * R'.
Combining steps 1, 2, 3, 4, and 5, we can calculate the new resistance R' as follows:
R = (ρ * L) / A (equation 1)
A = (ρ * L) / R (equation 2)
L' = 2πr (equation 3)
V = I * R (equation 4)
V = I * R' (equation 5)
Substituting equation 2 into equation 1, we have:
R = (ρ * L) / ((ρ * L) / R)
R = R (canceling ρ and L terms)
Substituting equation 3 into equation 1, we have:
R' = (ρ * L') / ((ρ * L) / R)
R' = (ρ * 2πr) / ((ρ * L) / R)
R' = (2πrR) / L
Substituting equation 4 into equation 5, we have:
V = I * R
V = I * R'
I * R = I * (2πrR) / L
R = (2πrR) / L
Therefore, the resistance R' of the new loop can be expressed as (2πrR)/L.