Problem 56 - Elementary Algebra

Question

. In an experiment, an electronics student finds that the total resistance of a series circuit with two resistors is 110 ohms. The total resistance of a parallel circuit with the same resistors is \(9.09\) ohms. The resistance of each resistor in ohms is a solution of the equation \(x^{2}-110 x+1000=0\). Find the resistance of each resistor.

Step-by-Step Solution

Answer

The resistances of the two resistors are 100 ohms and 10 ohms.

Step 1: Understand the given equation

The problem provides the quadratic equation to find the resistance of two resistors: Equation: \[ x^2 - 110x + 1000 = 0 \] Here, we need to solve for the values of x that satisfy this equation.

Step 2: Use the quadratic formula

The quadratic formula to find the roots of the equation \(ax^2 + bx + c = 0\) is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the given equation, \( a = 1 \), \( b = -110 \), and \( c = 1000 \).

Step 3: Substitute the values into the quadratic formula

Substituting the given values into the quadratic formula: \[ x = \frac{-(-110) \pm \sqrt{(-110)^2 - 4 \cdot 1 \cdot 1000}}{2 \cdot 1} \] Simplifying step-by-step: \[ x = \frac{110 \pm \sqrt{12100 - 4000}}{2} \] \[ x = \frac{110 \pm \sqrt{8100}}{2} \] \[ x = \frac{110 \pm 90}{2} \]

Step 4: Solve for both possible values

Now calculate the two potential values for x: 1. \[ x = \frac{110 + 90}{2} = \frac{200}{2} = 100 \] 2. \[ x = \frac{110 - 90}{2} = \frac{20}{2} = 10 \]

Step 5: Verify the results

The roots of the equation are 100 and 10. Therefore, the resistance values for the resistors are 100 ohms and 10 ohms, respectively. These values satisfy the conditions for both series and parallel circuits as described in the problem.

Key Concepts

Quadratic Formula

When solving quadratic equations like the one given in the problem (\[ x^2 - 110x + 1000 = 0 \]), we often use the quadratic formula. This formula is a method used to find the solutions, or roots, of quadratic equations, which are in the form \( ax^2 + bx + c = 0 \).
The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation.

• For our exercise, \(a = 1\), \(b = -110\), and \(c = 1000\).
• We substitute these values into the formula to find the resistance values of the resistors in the circuits.

By simplifying step-by-step, we calculate the roots:
\[ x = \frac{110 \pm 90}{2} \]
This results in two values: 100 and 10, which are the resistance values of the resistors.

Resistance in Series Circuits

In a series circuit, resistors are connected end-to-end so that the current flows through each resistor one after the other. The total resistance in a series circuit is the sum of the individual resistances.
For example, if you have two resistors,
\( R_1 \) and \( R_2 \), the total resistance \( R_{total} \) is:
\[ R_{total} = R_1 + R_2 \]
In our problem, the total resistance of the series circuit with two resistors is given as 110 ohms.

• This means that if we add the resistances of the two resistors (100 ohms and 10 ohms), we get 110 ohms, which confirms the setup of the series circuit in the problem.
• This demonstrates how the individual resistances sum up in a series connection.

Resistance in Parallel Circuits

In a parallel circuit, resistors are connected so that the current can flow through multiple paths. The total resistance of parallel resistors is found using the reciprocal formula:
\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \]
This means each resistor's value inversely affects the total resistance.
For our problem, the total resistance of the parallel circuit is given as 9.09 ohms.

• By solving for individual resistors of 100 ohms and 10 ohms, we can verify using the formula:
• \( \frac{1}{R_{total}} = \frac{1}{100} + \frac{1}{10} \)
• \( \frac{1}{R_{total}} = 0.01 + 0.1 = 0.11 \)

Thus, \( R_{total} = \frac{1}{0.11} \approx 9.09 \) ohms, matching the given value and illustrating how parallel circuit resistance works.