Problem 58 - Precalculus: Functions and Graphs

Question

The formula occurs in the indicated application. Solve for the specified variable. \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\) for \(R_{2}\) (three resistors connected in parallel)

Step-by-Step Solution

Answer

\( R_2 = \frac{1}{{\frac{1}{R} - \frac{1}{R_1} - \frac{1}{R_3}}} \)

Step 1: Identify the Goal

Our goal is to solve the equation \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \) for the variable \( R_2 \). We need to express \( \frac{1}{R_2} \) separately in the equation.

Step 2: Isolate the Term with \( R_2 \)

Subtract \( \frac{1}{R_1} + \frac{1}{R_3} \) from both sides of the equation to isolate \( \frac{1}{R_2} \):\[\frac{1}{R} - \left(\frac{1}{R_1} + \frac{1}{R_3}\right) = \frac{1}{R_2}\]Simplifying the right-hand side gives:\[\frac{1}{R} - \frac{1}{R_1} - \frac{1}{R_3} = \frac{1}{R_2}\]

Step 3: Solve for \( R_2 \)

Take the reciprocal of both sides to solve for \( R_2 \):\[ R_2 = \frac{1}{{\frac{1}{R} - \frac{1}{R_1} - \frac{1}{R_3}}} \]This expression gives the value of \( R_2 \) in terms of \( R, R_1, \) and \( R_3 \).

Key Concepts

Resistors in Parallel

When dealing with electrical circuits, resistors can be connected in different configurations. One common configuration is resistors in parallel. Imagine resistors being paths for electricity, similar to how roads are paths for cars. In a parallel configuration, multiple pathways are created for the current to flow. This means electricity can travel through different resistors at the same time.
This setup is quite different from resistors in series, where electricity is forced to follow one single path. In parallel circuits, the total resistance decreases as more resistors are added. This happens because electricity has more routes to follow, reducing the overall congestion. For beginners, think of it like opening more lanes on a highway, allowing more cars to pass through without causing a traffic jam.
The formula used to find the total resistance in a parallel circuit involves adding the reciprocals of each individual resistor's resistance. If you have resistors represented as

• \(R_1\)
• \(R_2\)
• \(R_3\)

the equation for total resistance \(R\) looks like this:\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\] This fundamental understanding of how resistors work in parallel is crucial for designing effective electrical circuits.

Algebraic Manipulation

Solving equations often requires the skill of algebraic manipulation, which is simply the fancy term for moving things around in an equation to isolate a specific variable. The key to mastering algebraic manipulation is understanding that you can perform the same operation on both sides of an equation without changing its balance. In the context of solving for the resistor \(R_2\), we begin by wanting to isolate \(\frac{1}{R_2}\) on one side of the equation. We can do this by subtracting \(\frac{1}{R_1}\) and \(\frac{1}{R_3}\) from both sides. Think of these subtractions like dissolving unnecessary clutter around \(\frac{1}{R_2}\) to see it more clearly:\[\frac{1}{R} - \left(\frac{1}{R_1} + \frac{1}{R_3}\right) = \frac{1}{R_2}\]Breaking down any equation into manageable pieces makes the abstraction of higher-level math more understandable. Once you get used to shifting elements—whether it be adding, subtracting, multiplying, or dividing—you will have a much easier time tackling complex algebraic problems.

Reciprocals

The concept of reciprocals is a core part of the puzzle in dealing with resistors in parallel and similar problems. A reciprocal is simply the inverse of a number. If you have a number \(x\), its reciprocal is \(\frac{1}{x}\). This technique is particularly useful in equations involving fractions.In our resistor equation, each term is already expressed in terms of reciprocals of the resistances. By the time we’ve isolated \(\frac{1}{R_2}\), the final algebraic step requires taking the reciprocal of our entire expression to solve for \(R_2\) itself. This looks like:\[R_2 = \frac{1}{{\frac{1}{R} - \frac{1}{R_1} - \frac{1}{R_3}}}\]Think of taking a reciprocal akin to flipping a fraction upside down. It gives us what we need by shifting the mathematical perspective of the problem. In practical terms, understanding reciprocals helps in reverse-engineering the product of complex fractions to its original value—an invaluable skill for electrical engineering and algebra alike.