Question

this is on my math homework. The electrical resistance of a wire varies directly with the length of the wire and inversely with the square of the diameter of the wire. If a wire 433 ft long and 4 mm in diameter has a resistance of 1.22 ohms, find the length of a wire of the same material whose resistance is 1.43 ohms and whose diameter is 5 mm. Round to the nearest tenth.

Answer 1

Given that the electrical resistance of a wire varies directly with the length of the wire and inversely with the square of the diameter of the wire, you can identify that it is a Combined Variation.

By definition Combined Variation Equations have this form:

`z=k((x)/(y))`

Where "k" is the Constant of Variation.

Let be "r" the electrical resistance (in ohms) of the wire, "l" the length of the wire (in feet), and "d" the diameter of the wire (in millimeters).

Knowing the above, you know that the equation that represents this situation has this form:

`r=k((l)/(d^2))`

You know that the resistance is 1.22 ​ohms when the wire is 433 feet long and 4 millimeters in diameter. Therefore, you can substitute values and solve for "k":

`1.22=k((433)/(4^2))`
`(1.22)((4^2)/(433))=k`
`k=0.0451`

Substitute "k" and the following values into the equation and then solve for "l", in order to solve the exercise:

`\begin{gathered} r=1.43 \\ d=5 \end{gathered}`

You get:

`1.43=(0.0451\cdot l)/(5^2)`
`(1.43)((5^2)/(0.0451))=l`
`l\approx792.68`

Hence, the answer is:

`l\approx792.7\text{ }ft`