Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 425: 39

a) The graph is shown below b) $I\propto \frac{1}{R}$ c) $I=\frac{6}{R}$

(a) $I\propto \frac{1}{R}$ for the given data. (b) We see that the value of the resistance decreases with the increase in the value of the current; they are inversely proportional to each other. Thus, $I\propto \frac{1}{R}$   (c) Since,  $I\propto \frac{1}{R}$ We have, $\ \ I=\frac{V}{R}$ Here, V is the parameter constant. For, $\begin{align} & I=0.5 \\ & R=12 \\ \end{align}$ $\begin{align} & \ \ I=\frac{V}{R} \\ & 0.5=\frac{V}{12} \\ & \ \ V=6 \\ \end{align}$ Now $I=\frac{6}{R}$ $\begin{align} & I=1 \\ & R=6 \\ \end{align}$ $\begin{align} & I=\frac{6}{R} \\ & 1=\frac{6}{6} \\ & 1=1 \\ \end{align}$ Hence, verified. For, $\begin{align} & I=1\cdot 5 \\ & R=4 \\ \end{align}$ $\begin{align} & I=\frac{6}{R} \\ & 1\cdot 5=\frac{6}{4} \\ & 1\cdot 5=1\cdot 5 \\ \end{align}$ Hence, verified. For, $\begin{align} & I=2 \\ & R=3 \\ \end{align}$ $\begin{align} & I=\frac{6}{R} \\ & 2=\frac{6}{3} \\ & 2=2 \\ \end{align}$ Hence, verified. For $\begin{align} & I=2\cdot 5 \\ & R=2\cdot 4 \\ \end{align}$ $\begin{align} & I=\frac{6}{R} \\ & 2\cdot 5=\frac{6}{2\cdot 4} \\ & 2\cdot 5=2\cdot 5 \end{align}$ Hence, verified. For, $\begin{align} & I=3 \\ & R=2 \end{align}$ $\begin{align} & I=\frac{6}{R} \\ & 3=\frac{6}{2} \\ & 3=3 \end{align}$ Hence, verified. For, $\begin{align} & I=4 \\ & R=1\cdot 5 \end{align}$ $\begin{align} & I=\frac{6}{R} \\ & 4=\frac{6}{1\cdot 5} \\ & 4=4 \\ \end{align}$ Hence, verified. For, $\begin{align} & I=5 \\ & R=1\cdot 2 \end{align}$ $\begin{align} & I=\frac{6}{R} \\ & 5=\frac{6}{1\cdot 2} \\ & 5=5 \end{align}$ Hence, verified.