The electrical resistance of a wire waves directly as its length & inversely as the square of its diameter if a wire 30m long and 0.75mm in diameter has a resistance of 25ohms , find the length of a wire of the same material whose resistance is 30ohms whose diameter is 1.25mm respectively
Even not remembering my physics, we can form the equation, since the relationship is given..
R = k L/d^2
25 = k(30)/.75^2
25 = k(30)(16/9)
k = 9(25)/((16)(30)) = 15/32
R = (15/32)L/d^2
plug in the 2nd set of numbers
30 = (15/32) L /1.25^2
finish it up by solving for L
To solve this problem, we'll use the given information and apply the relationship between electrical resistance, length, and diameter of the wire.
First, let's identify the variables:
R1 is the resistance of the first wire (25 ohms)
L1 is the length of the first wire (30 meters)
D1 is the diameter of the first wire (0.75 mm)
R2 is the resistance of the second wire (30 ohms)
L2 is the length of the second wire (we need to find this)
D2 is the diameter of the second wire (1.25 mm)
Now, let's use the relationship given: "The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter." We can express this mathematically as:
R1 ∝ L1
R1 ∝ 1/D1²
By combining these two relationships, we can write:
R1 ∝ L1/D1²
Using the values given:
R1 = 25 ohms
L1 = 30 meters
D1 = 0.75 mm
We can rearrange the equation to isolate L1:
L1 = R1 * (D1/D1²)
Next, let's substitute the values:
L1 = 25 * (0.75/0.75²)
L1 = 25 * (0.75/0.5625)
L1 = 33.33 meters (approximately)
So, the length of the first wire is approximately 33.33 meters.
Now, let's find the length of the second wire (L2) using the same relationship:
L2 = R2 * (D2/D2²)
Substituting the values:
R2 = 30 ohms
D2 = 1.25 mm
L2 = 30 * (1.25/1.25²)
L2 = 30 * (1.25/1.5625)
L2 = 24 meters (approximately)
Therefore, the length of the wire made of the same material with a resistance of 30 ohms and a diameter of 1.25 mm is approximately 24 meters.