Problem 43
Question
Solve the physics formula \(\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{R}\) for \(R_{2}\)
Step-by-Step Solution
Verified Answer
The physics formula solved for \(R_{2}\) is \(R_{2} = \frac{1}{R - R_{1}}\)
1Step 1: Rewrite the Equation
First we start by rewriting the equation to remove the fraction: \(R = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}\)
2Step 2: Separate the Denominator
Next, separate the denominator so that \(R_2\) is isolated on one side: \(R = \frac{1}{\frac{1}{R_{1}}} + \frac{1}{R_{2}} = R_{1} + \frac{1}{R_{2}}\)
3Step 3: Isolate \(R_{2}\)
Move \(R_{1}\) to the other side of the equation: \(\frac{1}{R_{2}} = R - R_{1}\)
4Step 4: Solve for \(R_{2}\)
Finally, isolate \(R_{2}\) by taking the reciprocal of both sides: \(R_{2} = \frac{1}{R - R_{1}}\)
Key Concepts
Understanding ReciprocalsSolving for a VariableFraction Manipulation Made Easy
Understanding Reciprocals
A reciprocal is essentially the inverse of a number or a fraction. When you take the reciprocal of a number, you are flipping its numerator and denominator. For example, the reciprocal of \( \frac{3}{4} \) is \( \frac{4}{3} \). If you have a whole number, like 5, its reciprocal is \( \frac{1}{5} \).
Reciprocals are useful in mathematics because multiplying a number by its reciprocal always results in 1. That is, \( a \times \frac{1}{a} = 1 \). In the context of our formula, taking the reciprocal is used to rearrange terms and solve equations more easily.
In the physics formula \( \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{R} \), each term on the left-hand side is already in reciprocal form. To solve for \( R_{2} \), taking the reciprocal of both sides of an equation helps in isolating \( R_{2} \), making it easier to find its value.
Reciprocals are useful in mathematics because multiplying a number by its reciprocal always results in 1. That is, \( a \times \frac{1}{a} = 1 \). In the context of our formula, taking the reciprocal is used to rearrange terms and solve equations more easily.
In the physics formula \( \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{R} \), each term on the left-hand side is already in reciprocal form. To solve for \( R_{2} \), taking the reciprocal of both sides of an equation helps in isolating \( R_{2} \), making it easier to find its value.
Solving for a Variable
Solving for a variable means rearranging an equation to make the variable the subject or main focus. For example, if you need to find \( R_{2} \) from the equation \( \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{R} \), our goal is to rearrange this equation so \( R_{2} \) is isolated on one side.
Steps to solve for \( R_{2} \):
Steps to solve for \( R_{2} \):
- First, rewrite the original equation in a more manageable form. In this case, rewrite it considering the reciprocal operation.
- Isolate \( R_{2} \) by moving all other terms to the opposite side of the equation.
- As a result, you derive \( \frac{1}{R_{2}} \) as a function of \( R \) and \( R_{1} \).
- Finally, take the reciprocal of the isolated term to find \( R_{2} \).
Fraction Manipulation Made Easy
Fraction manipulation involves changing the form of a fraction without altering its value. This includes operations like addition, subtraction, multiplication, division, and simplification of fractions.
In the given physics problem, you have fractions like \( \frac{1}{R_{1}} \) and \( \frac{1}{R_{2}} \). Understanding how to manipulate these fractions is crucial for solving the equation.
Useful tips for fraction manipulation:
In the given physics problem, you have fractions like \( \frac{1}{R_{1}} \) and \( \frac{1}{R_{2}} \). Understanding how to manipulate these fractions is crucial for solving the equation.
Useful tips for fraction manipulation:
- To add fractions, they must have a common denominator. If they don't, find the least common multiple and convert each to equivalent fractions with the same denominator.
- When simplifying fractions, like moving \( R_{1} \) to the other side, subtraction is key. Subtract \( \frac{1}{R_{1}} \) from both sides to begin simplifying the equation.
- Always remember that variables in the denominator can be isolated by adopting reciprocal operations.
Other exercises in this chapter
Problem 43
Solve. $$\frac{x+1}{x^{2}+x-2}=\frac{x+2}{x^{2}-1}+\frac{3}{x+2}$$
View solution Problem 43
Multiply. $$\frac{x^{4} y^{2}}{x^{2}+3 x-28} \cdot \frac{x^{2}-49}{x y^{4}}$$
View solution Problem 43
Write the fractions in terms of the LCM of the denominators. $$\frac{x}{x^{2}+x-6}, \frac{2 x}{x^{2}-9}$$
View solution Problem 43
Simplify. $$\frac{x^{-2}-y^{-2}}{x^{-2} y^{-2}}$$
View solution