Problem 49 - Intermediate Algebra

Question

Solve each formula for the specified variable. \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\) for \(R\) (from electronics)

Step-by-Step Solution

Answer

\( R = \frac{R_1R_2R_3}{R_2R_3 + R_1R_3 + R_1R_2} \) is the solved formula for \( R \).

Step 1: Recognize the Formula

Understand that the formula given is for calculating the equivalent resistance, \( R \), of three resistors \( R_1, R_2, \) and \( R_3 \) connected in parallel. We need to solve this for \( R \).

Step 2: Combine the Fractions

Start by combining the fractions on the right-hand side:\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]To add the fractions, find a common denominator, which is \( R_1R_2R_3 \). This gives:\[\frac{1}{R} = \frac{R_2R_3 + R_1R_3 + R_1R_2}{R_1R_2R_3}\]

Step 3: Solve for R by Flip and Rearrange

Take the reciprocal of both sides to solve for \( R \):\[R = \frac{R_1R_2R_3}{R_2R_3 + R_1R_3 + R_1R_2}\]This gives the formula for \( R \) in terms of \( R_1, R_2, \) and \( R_3 \).

Key Concepts

Parallel Resistors

When we talk about resistors in parallel, we're discussing a setup where resistors are arranged along multiple paths for the electric current to travel. This is different from resistors in series, where resistors follow one another on a single path. In a parallel resistor network:

• Each resistor is connected to both the same source and returns points.
• The total resistance is lower than the smallest resistance in the network.
• Current can choose different paths, so the voltage across each resistor is the same.

When connected in parallel, the total (or equivalent) resistance can be calculated using the formula:\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]This formula allows us to find how the resistors share the load, each offering a different path for the current, thus reducing overall resistance compared to any single resistor alone.

Resistor Network Formulas

The world of resistors involves various network configurations, and each has a specific formula to calculate its equivalent resistance. Understanding these formulas is crucial for analyzing and designing circuits. The formula for resistors in parallel, for instance:\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]indicates how the inverse of the total resistance is the sum of the inverses of each resistor's resistance.

• This formula is derived from Kirchhoff's Current Law, which states that the total current entering a junction equals the total current leaving.
• It emphasizes that the more paths you have, the easier it is for the current to flow, thereby reducing the overall resistance.

This expressiveness illustrates the significance of resistor network formulas in understanding the behavior of electric circuits, especially when simplifying complex arrangements.

Algebraic Manipulation

Algebraic manipulation is a powerful tool used to solve equations such as those found in resistor calculations. It involves rearranging equations and might require steps like combining terms or finding common denominators.
In the problem:\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]we first combined fractions with a common denominator to make the equation more manageable:\[\frac{1}{R} = \frac{R_2R_3 + R_1R_3 + R_1R_2}{R_1R_2R_3}\]Then, taking the reciprocal of both sides allowed us to isolate R:\[R = \frac{R_1R_2R_3}{R_2R_3 + R_1R_3 + R_1R_2}\]

• Combining fractions helps simplify the expression.
• Taking reciprocals is a direct method to solve for the desired variable.

Algebraic manipulation in formulas like these is essential for obtaining practical solutions to complex resistance problems.