Chapter 8.2, Problem 2P

To determine

To calculate: Resistance when current is $\text{8 A}$ .

Answer to Problem 2P

The current will be $\text{8 A}$ for a resistance of $\text{15 }\Omega$ .

Explanation of Solution

Given information:

The following variation:

The current flowing in a wire is inversely proportional to the resistance of the wire.

When $\text{current}=5\text{ A}$ , $\text{resistance}=24\Omega$

Calculation:

Let I be current flowing in the wire and R be the resistance. Then, I varies inversely as R , that is,

  $I=k\frac{1}{R}$

where k is a constant.

Substituting $I=5\text{ A}$ and $R=24\Omega$ ,

  $\begin{array}{l}I=k\frac{1}{R}[\text{Equation]}\\ 5=\frac{k}{24}[\text{Substitute]}\\ k=\left(5\right)\left(24\right)[\text{Multiply both sides by 24]}\\ k=120[\text{Simplify]}\end{array}$

Therefore, an equation of the inverse variation is $I=120\frac{1}{R}$

To find R when $I=8\text{ A}$ , substitute in $I=120\frac{1}{R}$ .

  $\begin{array}{l}I=120\frac{1}{R}[\text{Equation]}\\ 8=\frac{120}{R}[\text{Substitute]}\\ 8R=120[\text{Multiply both sides by }R\text{]}\\ R=\frac{120}{8}[\text{Divide both sides by 8]}\\ R=15[\text{Simplify]}\end{array}$