A three quarter inch wire has 12 ohms resistance. How much resistance has the same length of half-inch wire, if resistance varies inversely as the square of the diameter?
1/2 is 2/3 as big as 3/4
So, R will be (3/2)^2 = 9/4 * 12 = 27 ohms
To find the resistance of a half-inch wire with the given information, we need to use the concept of inverse variation and the formula for resistance of a wire.
Let's define some variables:
- Let R1 be the resistance of the three quarter inch wire.
- Let R2 be the resistance of the half-inch wire.
- Let d1 be the diameter of the three quarter inch wire.
- Let d2 be the diameter of the half-inch wire.
According to the problem, the resistance varies inversely as the square of the diameter. This means that if we double the diameter, the resistance will be reduced to one-fourth (since the square of 2 is 4).
So, we can express the relationship between the resistance and the diameter as follows:
R1 ∝ 1/d1^2
R2 ∝ 1/d2^2
Now, let's use the given information: a three quarter inch wire with a resistance of 12 ohms. We can write it as:
R1 = k/(d1^2)
where k is a constant of variation.
To find k, we can substitute the given values:
12 = k/(3/4)^2
12 = k/(9/16)
12 = k * (16/9)
k = (12 * 9) / 16
k = 6.75
Now we can find the resistance, R2, of the half-inch wire:
R2 = k/(d2^2)
R2 = 6.75/(1/2)^2
R2 = 6.75/(1/4)
R2 = 6.75 * 4
R2 = 27 ohms
Therefore, the resistance of the same length of half-inch wire is 27 ohms.