Problem 46 - College Algebra with Modeling and Visualization

Question

Electricity \(\quad\) Complex numbers are used in the study of electrical circuits. Impedance \(Z\) (or the opposition to the flow of electricity. voltage \(V\) and current \(I\) can all be represented by complex numbers. They are related by the equation \(Z=\frac{V}{I} .\) Find the value of the missing variable. $$ I=\frac{1}{2}+\frac{1}{4} i \quad Z=8-9 i $$

Step-by-Step Solution

Answer

The voltage \( V \) is \( \frac{25}{4} - \frac{5}{2}i \).

Step 1: Understand the Relationship

The relationship among impedance \( Z \), voltage \( V \), and current \( I \) is given by the equation \( Z = \frac{V}{I} \). We need to find \( V \) since \( Z \) and \( I \) are given.

Step 2: Rearrange the Formula

To find \( V \), rearrange the equation \( Z = \frac{V}{I} \) to solve for \( V \). By multiplying both sides by \( I \), we get \( V = Z \times I \).

Step 3: Substitute the Known Values

Substitute the given complex numbers into the equation: \( Z = 8 - 9i \) and \( I = \frac{1}{2} + \frac{1}{4}i \). The equation becomes \( V = (8 - 9i) \times \left( \frac{1}{2} + \frac{1}{4}i \right) \).

Step 4: Multiply Complex Numbers

Use the distributive property to multiply the complex numbers: \[ V = (8 - 9i) \times \left( \frac{1}{2} + \frac{1}{4}i \right) \]Calculate \( (8 - 9i) \times \frac{1}{2} \) and \( (8 - 9i) \times \frac{1}{4}i \):- \( (8 - 9i) \times \frac{1}{2} = 4 - \frac{9}{2}i \)- \( (8 - 9i) \times \frac{1}{4}i = 2i - \frac{9}{4}i^2 \)Recall that \( i^2 = -1 \), so - Replace \( \frac{9}{4}i^2 \) with \( -\frac{9}{4} \),- Simplify the expressions: \[ V = (4 - \frac{9}{2}i) + (2i - \left(-\frac{9}{4}\right)) \]\[ V = 4 - \frac{9}{2}i + 2i + \frac{9}{4} \]

Step 5: Simplify the Expression

Combine like terms for the real and imaginary parts: \[ Real ext{ }Part: 4 + \frac{9}{4} = \frac{16}{4} + \frac{9}{4} = \frac{25}{4} \]\[ Imaginary ext{ }Part: -\frac{9}{2}i + 2i = -\frac{9}{2}i + \frac{4}{2}i = -\frac{5}{2}i \]Thus, \( V = \frac{25}{4} - \frac{5}{2}i \).

Key Concepts

Impedance in Electrical Circuits

Electrical circuits often use a concept called impedance to describe how much they resist the flow of electric current. Much like resistance, impedance is crucial but it's more encompassing. This is because it applies to both DC (direct current) and AC (alternating current) circuits.

Impedance is represented by the letter \( Z \) and can be measured in ohms. When we consider AC circuits, impedance not only accounts for resistance (from resistors) but also includes the effects of capacitors and inductors. These elements cause additional opposition - some lag and some lead the circuit.

The beauty of using complex numbers to express impedance lies in its ability to capture both resistive and reactive components simultaneously. The real part of the complex impedance represents pure resistance, while the imaginary part represents the reactance.

• Reactance could be capacitive, which leads the voltage, or inductive, which lags it.
• The overall goal in using these values is to create a unified representation of total opposition in a circuit.

Complex Number Multiplication

Multiplying complex numbers is similar to multiplying numbers with algebraic expressions. When multiplying two complex numbers, you use the distributive property, also known as the FOIL method.

• Consider a complex number in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part.
• When you multiply \((a + bi)(c + di)\), you distribute each part: \(a \times c + a \times di + bi \times c + bi \times di\).

Remember that \( i^2 = -1 \). This is pivotal in simplifying the expressions since any term involving \( i^2 \) turns the imaginary unit's square into a real number.

Let’s go over the multiplication in the exercise:
Given \((8 - 9i)(\frac{1}{2} + \frac{1}{4}i)\), distribute as follows:

• Multiply \(8 \times \frac{1}{2}\) and \(8 \times \frac{1}{4}i\).
• Multiply \(-9i \times \frac{1}{2} \) and \(-9i \times \frac{1}{4}i\).
• Simplify by substituting \(i^2 = -1\) to deal with imaginary squares.

After calculating, combine like terms to get a clean and concise complex number.

Algebraic Expressions with Complex Numbers

Complex numbers make algebra more interesting. They unfold another layer of mathematical elegance. When managing algebraic expressions that include complex numbers, it helps to know some core principles.

Whether adding or subtracting, ensure to group and operate on real parts separately from imaginary parts. This would mean:

• If you have \((a + bi)\) added to \((c + di)\), simply combine \(a+c\) for the real part and \(b+d\) for the imaginary part.

In the example problem, we found the expression for voltage by multiplying impedance and current, yielding another complex number. It's important after distributing and simplifying to segregate real and imaginary components to achieve the final solution. Once simplified, restructure the equation into its canonical form \(V = \text{real part} + \text{imaginary part} i\).

By approaching algebraic manipulation systematically and using complex numbers, problems in electrical engineering and other fields become easier to manage.