Algebra in Physics

Question

The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. A wire with a length of 200 inches and a diameter of one-quarter of an inch has a resistance of 20 ohms. Find the electrical resistance in a 500 inch wire with the same diameter.

Answer 1

This is an example of a physics concept used in an algebra problem. Even though the units are not in scientific measurements, for problems such as this, do not worry about converting unless otherwise stated.

This problem makes use of the Physic concept of resistiviy expressed as the equation:

R = ρL/A

where R is the resistance in ohms, ρ is the resistivity, L is the length of a wire in meters, and A is the cross-sectional area of the wire in m².

Given the problem above, it is essentially asking to solve for the electrical resistance. As the resistivity is needed, the problem provides a means to solve this. In situations with variation problems, you are most likely be dealing with a science concept expressed in variables in relation to value. A constant will be used and the problem will supply a way to solve the constant that will be used to solve the problem. For these problems, don't think about the units. Simply go with what is provided in the problem.

First, we need to come up the equation that relates the electrical resistance, r, to the length, l, and diameter, d,and a contant, k:

r = kl/d²

Then, we need to find the constant, k. This can be accomplished by using the values from the second sentence:

20 = k200/(0.25)²

k = (20 * (0.25)²⁾/ 200 = 0.00625

Then we find the electrical resistance, r:

r = (0.0625(500))/(0.25)^{2 =} 50 ohm