Problem 43 - 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. $$ V=50+98 i \quad I=8+5 i $$

Step-by-Step Solution

Answer

The impedance \( Z \) is approximately \( -1.011 + 6i \).

Step 1: Understand the Given Values

We have the voltage, \( V = 50 + 98i \), and the current, \( I = 8 + 5i \). We need to find the impedance \( Z \) using the formula \( Z = \frac{V}{I} \).

Step 2: Set Up the Division

To find \( Z \), we need to perform the division of two complex numbers. We express \( Z \) as: \[ Z = \frac{50 + 98i}{8 + 5i} \].

Step 3: Multiply by the Conjugate

To simplify the division, multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of \( 8 + 5i \) is \( 8 - 5i \). Therefore, the equation becomes: \[ Z = \frac{(50 + 98i)(8 - 5i)}{(8 + 5i)(8 - 5i)} \].

Step 4: Calculate Denominator

Calculate the denominator: \( (8 + 5i)(8 - 5i) = 8^2 - (5i)^2 = 64 - (-25) = 64 + 25 = 89 \).

Step 5: Calculate Numerator

Compute the numerator by expanding \( (50 + 98i)(8 - 5i) \):- For the real part: \( 50 \times 8 - 98 \times 5 = 400 - 490 = -90 \).- For the imaginary part: \( 50 \times (-5) + 98 \times 8 = -250 + 784 = 534 \).The resulting numerator is \( -90 + 534i \).

Step 6: Divide to Find Impedance

Now divide the resulting real and imaginary parts by the denominator to find \( Z \):\[ Z = \frac{-90 + 534i}{89} = -1.011 + 6i \].

Key Concepts

Impedance Calculation

In electrical circuits, impedance is a vital concept that indicates the opposition a circuit presents to the flow of alternating current (AC). It's similar to resistance in direct current (DC) circuits but extends this idea to include reactance, thereby making it a complex number. Impedance, denoted by \(Z\), is expressed in ohms and combines both resistance (real part) and reactance (imaginary part).

The formula to find impedance in a circuit is \(Z = \frac{V}{I}\), where \(V\) is the voltage, and \(I\) is the current. Both voltage and current are also represented as complex numbers. By knowing these values, you can calculate impedance, which helps in understanding how the circuit will behave under various conditions, especially involving AC.

In our problem, we calculated the impedance \(Z\) by dividing the complex voltage \((50 + 98i)\) by the complex current \((8 + 5i)\). This involves using principles from complex number algebra, as we'll see in the next sections.

Voltage and Current Representation

In electrical engineering, representing voltage and current as complex numbers allows us to efficiently handle phase shifts and magnitude changes. Each complex number depicts a magnitude and a phase angle, which are crucial in AC circuit analysis.

Voltage, denoted as \(V = 50 + 98i\) in our problem, represents both the real component (50) and the imaginary component (98i) that affects how the voltage vector rotates in the complex plane. Similarly, current represented by \(I = 8 + 5i\) has its real (8) and imaginary parts (5i), symbolizing its own magnitude and phase characteristics.

These representations allow engineers to model how AC voltage and currents vary sinusoidally. Instead of dealing with oscillating sinusoidal functions, we use complex numbers to simplify calculations of power, impedance, and other circuit behaviors.

Complex Number Division

One of the key operations in this exercise is dividing two complex numbers to find the impedance \(Z\). This involves a few steps that ensure the resulting number is in a standard form that combines a real part and an imaginary part.

To divide complex numbers, such as \(Z = \frac{50 + 98i}{8 + 5i}\), we use the concept of multiplying by the conjugate. The conjugate of \(8 + 5i\) is \(8 - 5i\). By multiplying both the numerator and denominator by this conjugate, we convert the denominator into a real number. This step helps in preventing division by a complex number directly.

After executing the multiplication and simplifying both the numerator and the denominator, you end up with a simple division of a complex number by a real number, turning it into a more manageable form. This division results in \(-1.011 + 6i\), which makes the behavior of the impedance clear by separating the real and imaginary effects impacting the circuit.