91Ó°ÊÓ

Use the four-step procedure for solving variation problems given on page 445 to solve Exercises 21–36. The table shows the values for the current, I, in an electric circuit and the resistance, R, of the circuit. $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text {I(amperes) } & {0.5} & {1.0} & {1.5} & {2.0} & {2.5} & {3.0} & {4.0} & {5.0} \\\\\hline R \text { (ohms) } & {12.0} & {6.0} & {4.0} & {3.0} & {2.4} & {2.0} & {1.5} & {1.2} \\\\\hline\end{array}$$ a. Graph the ordered pairs in the table of values, with values of I along the x@axis and values of R along the y@axis. Connect the eight points with a smooth curve. b. Does current vary directly or inversely as resistance? Use your graph and explain how you arrived at your answer. c. Write an equation of variation for I and R, using one of the ordered pairs in the table to find the constant of variation. Then use your variation equation to verify the other seven ordered pairs in the table.

Step by step solution

01

Graphing Ordered Pairs

Plot the ordered pairs with I-values along the x-axis and R-values along the y-axis. The pairs are (0.5, 12.0), (1.0, 6.0), (1.5, 4.0), (2.0, 3.0), (2.5, 2.4), (3.0, 2.0), (4.0, 1.5), (5.0, 1.2). Connect the eight points with a smooth curve.

02

Confirming Variation Relation

Study the relationship between I and R as depicted in the graph. The graph shows that as the current (I) increases, the resistance (R) decreases. This behavior characterizes an inverse variation.

03

Formulating Equation of Variation

The general formula for inverse variation is \( Y = \frac{k}{X}\) where Y is the dependent variable, X is the independent variable and k is the constant of variation. Using this formula and the first ordered pair (0.5, 12.0), the constant of variation k can be solved: \(12 = \frac{k}{0.5}\) Thus, k = 12 * 0.5 = 6. Hence, the equation of variation is \( R = \frac{6}{I}\)

04

Verification of other ordered pairs

Substitute each of the other seven pairs into the equation of variation \(R = \frac{6}{I}\) to ascertain that they satisfy the equation: \((1.0, 6.0), (1.5, 4.0), (2.0, 3.0), (2.5, 2.4), (3.0, 2.0), (4.0, 1.5), (5.0, 1.2)\) Each of these substitutions should result in an equality, thus verifying that they fit the inverse variation equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current and Resistance Relationship

The relationship between current (I) and resistance (R) in an electrical circuit is a fundamental concept in physics. It helps us understand how electric circuits function. In this context, resistance measures how much a material opposes the flow of current. Knowing this relationship is vital.
When examining this connection, we see that as the current in the circuit increases, the resistance decreases. This inverse relationship can be intuitive if we consider a water hose analogy: the wider the hose (less resistance), the more water (current) can flow through it.

• High resistance means less current can pass through.
• Low resistance allows for more current to pass through.

This phenomenon is an example of inverse variation, where two variables move in opposite directions. When one increases, the other decreases, maintaining a constant product. Understanding this inverse relation helps predict how changes in current affect resistance and vice versa.

Equation of Variation

To describe the inverse relationship between current (I) and resistance (R) mathematically, we use the equation of variation. This equation takes the form of \( Y = \frac{k}{X} \), where \( Y \) is the dependent variable, \( X \) is the independent variable, and \( k \) is the constant of variation.
In our exercise, resistance \( R \) is expressed as inversely proportional to current \( I \), which gives us the equation \( R = \frac{k}{I} \).
To find \( k \), we use one of the ordered pairs from the table. For example, using the pair (0.5, 12.0), we substitute into the equation:\[ 12 = \frac{k}{0.5} \]Solving this gives \( k = 6 \). Therefore, the variation equation becomes:\[ R = \frac{6}{I} \]This equation tells us that for every level of current \( I \), the resistance \( R \) can be calculated. Each ordered pair from the table should satisfy this equation, helping verify our constant of variation.

Graphing Ordered Pairs

Graphing the relationship between current and resistance is a visual way to understand their behavior. By plotting the ordered pairs, you can see how these values interact. In our example, current \( I \) is along the x-axis, and resistance \( R \) along the y-axis.
Each ordered pair such as (0.5, 12.0), (1.0, 6.0), represents a point. When plotted on a graph, these points are connected smoothly to illustrate the inverse relationship. As the current increases, the resistance curves downward, showing decreasing values.

• Ordered pairs provide specific coordinates to plot.
• The curve helps illustrate the inverse nature.

Connecting the points helps in visualizing the entire pattern. Such a graph clearly shows that as one value rises, the other descends. It is a handy tool for quickly understanding the inverse variation concept and verifying the mathematical equation derived.