A wire has a resistance of 27.2 ohms. It is melted down, and from the same volume of metal a new wire is made that is 3 times longer than the original wire. What is the resistance of the new wire?
Resistance and length of wire are proportional, but what I am unsure of is if the width has to be factored in, if so, I have no idea how to do so. As you increase the length from the same volume, the width would decrease. If you don't factor this in, it's just (27.2)(3)= 81.6 ohms. Please help!
If it is three times longer, it must have 1/3 the cross sectional area of the original wire. The new resistance will be nine times higher.
To determine the resistance of the new wire, you need to consider its length as well as its cross-sectional area. Resistance is directly proportional to length but inversely proportional to cross-sectional area.
Given that the new wire is three times longer than the original wire, we can call the original length "L" and the new length "3L".
Now, let's assume that the width (or the cross-sectional area) of the original wire is "A", and the width of the new wire is "xA". Since the volume of the metal used remains the same, we can say:
(L * A) = (3L * xA)
From this equation, we can determine that x = 1/3. This means that the width of the new wire is one-third the width of the original wire.
Using the formula for resistance (R), which is R = (ρ * L) / A, where ρ is the resistivity of the material, we can substitute the given values:
For the original wire: R1 = (ρ * L) / A
For the new wire: R2 = (ρ * 3L) / (1/3A)
To simplify further, we can cancel out the common factors:
R2 = 3 * (ρ * L) / (1/3A)
R2 = 9 * (ρ * L) / A
Therefore, the resistance of the new wire (R2) is nine times the resistance of the original wire (R1). In this case, R2 = 9 * R1.
Given that the resistance of the original wire is 27.2 ohms, the resistance of the new wire is:
R2 = 9 * R1
R2 = 9 * 27.2
R2 = 244.8 ohms
So, the resistance of the new wire is 244.8 ohms.