Problem 45 - Algebra for College Students

Question

Solve each problem. If two receivers with resistances \(R_{1}\) and \(R_{2}\) are connected in parallel, then the formula relates the total resistance for the circuit \(R\) with \(R_{1}\) and \(R_{2}\). Given that \(R_{1}\) is 3 ohms and \(R\) is 2 ohms, find \(R_{2}\) $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$$

Step-by-Step Solution

Answer

The value of \( R_{2} \) is 6 ohms.

Step 1: Write down the formula

The formula for the total resistance for resistors in parallel is given by: \[ \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \]

Step 2: Substitute the given values

We know that \( R = 2 \) ohms and \( R_{1} = 3 \) ohms. Substitute these values into the formula: \[ \frac{1}{2} = \frac{1}{3} + \frac{1}{R_{2}} \]

Step 3: Isolate \( \frac{1}{R_{2}} \)

To find \( \frac{1}{R_{2}} \), subtract \( \frac{1}{3} \) from both sides of the equation: \[ \frac{1}{R_{2}} = \frac{1}{2} - \frac{1}{3} \]

Step 4: Find a common denominator and simplify

The common denominator for 2 and 3 is 6. Rewrite the fractions with the common denominator: \[ \frac{1}{2} = \frac{3}{6} \] \[ \frac{1}{3} = \frac{2}{6} \] Now perform the subtraction: \[ \frac{1}{R_{2}} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

Step 5: Solve for \( R_{2} \)

Take the reciprocal of both sides to find \( R_{2} \): \[ R_{2} = \frac{6}{1} = 6 \] ohms

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits. It states that the current (I) passing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) between them. The formula for Ohm's Law is:
\[ V = IR \]
In our exercise, however, we are dealing with a parallel circuit. Ohm's Law helps us understand the relationship between current, voltage, and resistance, which is crucial when we substitute the values into our parallel resistance formula.

Equivalent Resistance

Equivalent resistance in a circuit is the single resistance that could replace multiple resistors and still produce the same effect on the overall circuit. For resistors in parallel, the formula to find the equivalent resistance is:
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]
This is derived from the fact that the voltage across all parallel resistors remains the same, but the total current is the sum of the currents through each resistor. In the given exercise, we use this formula to solve for the unknown resistance, \( R_2 \), by substituting known values and simplifying the equation.

Electrical Circuits

Electrical circuits are paths that allow the flow of electrons or electric current. They consist of various components like resistors, capacitors, and inductors, connected by conductive wires. There are two basic types of circuits: series and parallel.

• Series Circuits: Components connected end to end, so there's only one path for current flow.
• Parallel Circuits: Components connected across the same two points, providing multiple paths for current.

The given exercise focuses on a parallel circuit. In parallel circuits, the total resistance decreases because the current has multiple paths to travel, as demonstrated by our calculation for finding the unknown resistance.

Fractions Subtraction

In mathematical operations involving fractions, finding a common denominator is crucial for correctly adding or subtracting them. Here's a step-by-step approach:

• Identify the denominators of the fractions you want to subtract.
• Find the least common denominator (LCD). This is the smallest number that both denominators can divide into evenly.
• Rewrite each fraction as an equivalent fraction with the common denominator.
• Perform the subtraction.

In our exercise, we used this approach when subtracting \( \frac{1}{3} \) from \( \frac{1}{2} \). The common denominator for 2 and 3 is 6, so we rewrote \( \frac{1}{2} \) as \( \frac{3}{6} \) and \( \frac{1}{3} \) as \( \frac{2}{6} \), then performed the subtraction: \( \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \). This simplification was necessary to solve for \( \frac{1}{R_2} \).