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Chapter 18: Problem 46

Answer the given questions by setting up and solving the appropriate proportions. If \(c / d\) is in inverse ratio to \(a / b,\) then \(a / b=d / c\) (see Exercises 29 and 30). The current \(i\) (in A) in an electric circuit is in inverse ratio to the resistance \(R\) (in \(\Omega\) ). If \(i=0.25\) mA when \(R=2.8 \Omega\), what is \(i\) when \(R=7.2 \Omega ?\)

Short Answer

Expert verified
The current \( i \) is approximately 0.0972 mA when \( R = 7.2 \Omega \).

Step by step solution

01

Understand the Problem Statement

We are given that the current \( i \) is in inverse ratio to the resistance \( R \). This means that as \( R \) increases, \( i \) decreases, and vice versa. We need to find the new current when the resistance changes from \( 2.8 \Omega \) to \( 7.2 \Omega \).
02

Set Up the Inverse Proportion

Since \( i \) and \( R \) are in inverse proportion, we can set up the equation: \( i_1 \times R_1 = i_2 \times R_2 \), where \( i_1 = 0.25 \) mA, \( R_1 = 2.8 \Omega \), and \( R_2 = 7.2 \Omega \). We need to find \( i_2 \).
03

Solve for the Unknown Current

Substitute the known values into the equation: \( 0.25 \times 2.8 = i_2 \times 7.2 \). Calculate the left-hand side: \( 0.25 \times 2.8 = 0.7 \). So, the equation becomes \( 0.7 = i_2 \times 7.2 \).
04

Calculate the New Current

Rearrange the equation to solve for \( i_2 \): \( i_2 = \frac{0.7}{7.2} \). Calculate \( i_2 \): \( i_2 \approx 0.0972 \) mA.

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Key Concepts

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

Proportions
Proportions are mathematical expressions that assert two ratios are equivalent. They are significant in solving various real-world problems, like determining relationships between quantities.
A simple example of a proportion is when you say that the ratio of apples to oranges is the same in two different groups of fruits. It looks like this: \( \frac{a}{b} = \frac{c}{d} \). Here, \( a \), \( b \), \( c \), and \( d \) are real numbers, and none of the denominators, \( b \) or \( d \), should be zero.
Proportions help in predicting or understanding relationships.
  • They allow you to find missing values in problems when three out of four values are known.
  • The equality statement can be used to form equations which then can be solved for unknown variables.
You can frequently find proportions used in scale drawings, maps, cooking recipes, and many other areas.
Current and Resistance
The flow of electricity through a circuit is defined by two primary concepts: current and resistance. Current, symbolized as \( i \), is the flow of electric charge and is measured in amperes (A), whereas resistance, symbolized as \( R \), is a measure of the opposition to the flow, measured in ohms (\( \Omega \)).
To understand how current and resistance interact, consider a simple electric circuit:
  • Current is akin to the flow of water through a pipe.
  • Resistance is like the narrowing of the pipe, hindering the flow.
The higher the resistance, the lower the current, provided the voltage remains the same. This relationship is a key component of Ohm's Law, which states \( V = i \times R \), where \( V \) is voltage.
Inverse Proportion
An inverse proportion, or inverse relationship, occurs when one value increases while the other decreases.
Mathematically, two quantities \( x \) and \( y \) are in inverse proportion if \( x \times y = k \), where \( k \) is a constant.
This implies that as one value doubles, the other is halved, maintaining a consistent product.
In the context of electric circuits, the current \( i \) and resistance \( R \) are said to be in inverse proportion. This is represented by the equation: \( i \times R = k \). As the resistance \( R \) increases, the current \( i \) decreases, and vice versa, keeping their product the same, i.e., constant.
  • Inverse proportions are crucial in understanding real-world phenomena.
  • They are applicable in areas such as speed and travel time, supply and demand, and physics, especially when one property opposes the other.
Solving Equations
Solving equations is a fundamental aspect of mathematics. It involves finding the unknown value that satisfies the equality of the equation. Different equations require different techniques.
In problems involving proportions or inverse relations, equations are set up and then manipulated. Consider the electric circuit problem where you need to find a new current. The steps are as follows:
  • First, formulate the equation from the inverse relation: \( i_1 \times R_1 = i_2 \times R_2 \).
  • Next, substitute the known values into the equation.
  • Then, rearrange to isolate the unknown variable; this may involve dividing or multiplying both sides of the equation.
  • Finally, solve for the unknown, which in this case, would be the new current \( i_2 \).
With practice, solving such equations becomes intuitive and straightforward, allowing for efficient problem-solving in both academic and real-world situations.

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Answer the given questions by setting up and solving the appropriate proportions. A motorist traveled \(5.0 \mathrm{km}\) (accurately measured) in \(4 \mathrm{min} 54 \mathrm{s}\) and the speedometer showed \(58 \mathrm{km} / \mathrm{h}\) for the same interval. What is the percent error in the speedometer reading?
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