Problem 55 - Technical Mathematics with Calculus

Question

To find two resistors that will give an equivalent resistance of \(400 \Omega\) when wired in series and \(75 \Omega\) when wired in parallel, we must solve the quadratic equation $$R^{2}-400 R+30,000=0$$ Factor the left side of this equation. Factor the left side of this equation.

Step-by-Step Solution

Answer

The factored form of the quadratic equation is \((R-200)(R-200)=0\), which has a repeated root, so the two resistors are both \(200\Omega\).

Step 1: Apply the Quadratic Formula

To factor the quadratic equation given by \(R^{2}-400R+30,000=0\), we can apply the quadratic formula \(R = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a = 1\), \(b = -400\), and \(c = 30,000\). However, in this case, factoring can be done directly without using the quadratic formula if the factors are readily identifiable.

Step 2: Identify Factors

The quadratic equation can also be factored directly. We look for two numbers that multiply to give \(ac = 30,000\) and add to give \(b = -400\). The numbers that satisfy these conditions are \(-200\) and \(-200\), since \(-200 \times -200 = 30,000\) and \(-200 - 200 = -400\).

Step 3: Write the Factored Form

Using the factors identified in Step 2, we can write the quadratic as \((R-200)(R-200)=0\).

Step 4: Set Each Factor Equal to Zero

To find the roots of the quadratic equation, set each factor equal to zero: \(R-200=0\).Thus, the solution for R is \(R=200\) Ohms.

Key Concepts

Factoring Quadratic Equations

Factoring quadratic equations is a fundamental skill in algebra. It involves finding two binomial expressions that, when multiplied together, produce the original quadratic equation.

Quadratic equations take the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients and \(x\) is the variable we aim to solve for. The exercise we're reviewing involves factoring the equation \(R^2-400R+30,000=0\) to find the values of electrical resistance in series and parallel circuits.

The first step in factoring is to find two numbers that multiply to \(ac\) and add to \(b\). In our example, we identified \( -200\) and \( -200\) because \( -200 \times -200 = 30,000\) (the product of \(a\) and \(c\)) and \( -200 - 200 = -400\) (the value of \(b\)). The quadratic equation thus factors to \( (R-200)(R-200)=0\), leading to the solution \( R=200 \) Ohms.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations where factoring is not straightforward. The formula is \( x = \frac{-b \(pm\) \sqrt{b^2-4ac}}{2a} \) and provides a solution for any quadratic in the standard form \( ax^2 + bx + c = 0\).

In the context of our problem, the quadratic formula would be used as a fallback if the factors of the equation were not easily recognizable. With coefficients \(a=1\), \(b=-400\), and \(c=30,000\), the formula could be applied to find the resistance values. However, because we could directly factor the equation, the use of the quadratic formula was unnecessary in this case.

Electrical Resistance

Electrical resistance, measured in ohms (\( \Omega \)), is a measure of the opposition to the flow of current in an electrical circuit. It's a crucial concept in both theoretical and practical electronics. Knowing how to solve for resistance is essential when designing circuits for desired performance.

In our exercise, we calculated resistance values for two resistors that, in series, have a combined resistance of \(400 \Omega\) and, in parallel, \(75 \Omega\). Understanding how resistors combine in different configurations is essential for circuit design and analysis. The quadratic equation arose from the rules governing series and parallel circuits, demonstrating the overlap between algebra and electrical engineering.

Series and Parallel Circuits

There are two fundamental ways to connect resistors in a circuit: series and parallel. In a series circuit, resistors are connected end-to-end, and the total resistance is the sum of the individual resistances. Conversely, in a parallel circuit, resistors are connected across the same two potential points, and the total resistance is found using the formula \( \frac{1}{R_{total}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ... + \frac{1}{R_{n}} \).

The problem we're exploring involves finding resistor values for both configurations to achieve specific resistance measurements. Solving quadratic equations like the one presented is a practical application of algebra in electrical engineering, where understanding the behavior of resistors in series and parallel is vital.