Problem 122 - Elementary and Intermediate Algebra

Question

Electronics. The impedance \(Z\) in an AC (alternating current) circuit is a measure of how much the circuit impedes (hinders) the flow of current through it. The impedance is related to the voltage \(V\) and the current \(I\) by the formula \(V=I Z .\) If a circuit has a current of \((0.5+2.0 i)\) amps and an impedance of \((0.4-3.0 i)\) ohms, find the voltage.

Step-by-Step Solution

Answer

The voltage is \( 6.2 - 0.7i \) volts.

Step 1: Understand the Formula

The relationship between voltage \( V \), current \( I \), and impedance \( Z \) is given by the formula \( V = I \cdot Z \). This means you need to multiply the current by the impedance to find the voltage.

Step 2: Identify the Components

Substitute the known values into the formula: the current \( I = 0.5 + 2.0i \) and the impedance \( Z = 0.4 - 3.0i \). We now need to find \( V = (0.5 + 2.0i) \cdot (0.4 - 3.0i) \).

Step 3: Perform Complex Multiplication

Use the distributive property to multiply the complex numbers: \((0.5 + 2.0i)(0.4 - 3.0i) = 0.5 \cdot 0.4 + 0.5 \cdot (-3.0i) + 2.0i \cdot 0.4 + 2.0i \cdot (-3.0i) \).Calculate each term: - \( 0.5 \cdot 0.4 = 0.2 \)- \( 0.5 \cdot (-3.0i) = -1.5i \)- \( 2.0i \cdot 0.4 = 0.8i \)- \( 2.0i \cdot (-3.0i) = -6i^2 = 6 \), knowing that \( i^2 = -1 \).

Step 4: Simplify the Expression

Combine the real and imaginary parts of the expression:\[ 0.2 + 6 + (-1.5i + 0.8i) \]- Real part: \( 0.2 + 6 = 6.2 \)- Imaginary part: \( -1.5i + 0.8i = -0.7i \)Thus, the voltage \( V = 6.2 - 0.7i \).

Key Concepts

Complex Numbers Multiplication

To solve problems involving complex numbers multiplication, we need to understand how to work with these numbers. Complex numbers have a real part and an imaginary part. When multiplying complex numbers, each part of one number must be multiplied by each part of the other number.
For example, consider multiplying two complex numbers, such as \[ (a + bi) \] and \[ (c + di) \]. The product is determined by applying the distributive property: \[ (a + bi)(c + di) = ac + adi + bci + bdi^2. \]

Remember, the imaginary unit is denoted by \( i \), where \( i^2 = -1 \). Calculating further, \[ ac \] is the product of the real parts, \( adi \) and \( bci \) are the products of the real part with the imaginary part, and \( bdi^2 \) contributes to the real part because \( i^2 = -1 \), making it \(-bd\).
Thus, in expanding the terms, regrouping the real and imaginary parts provides the result.

• The real part becomes \[ (ac - bd). \]
• The imaginary part becomes \[ (ad + bc)i. \]

Use this method, and you'll find that multiplying complex numbers becomes intuitive with practice.

Ohm's Law for AC Circuits

Ohm's Law is a foundational principle in electrical circuits, describing the relationship among voltage, current, and impedance. Specifically for AC circuits, it is expressed as
\( V = I \cdot Z \)
where:

• \( V \) is the voltage across the circuit.
• \( I \) is the current flowing through the circuit.
• \( Z \) is the impedance, which can include resistance and reactance, expressed as a complex number.

In AC circuits, impedance acts somewhat like resistance but also accounts for phase differences between the current and voltage. This is vital as alternating currents oscillate, and understanding the phase shift helps correctly design circuits that can efficiently transfer power.
For the given problem, we see that by multiplying the current value, a complex number, by the impedance, another complex number, we calculate the voltage. This highlights how Ohm's Law for AC circuits not only involves scalar values but extends to complex numbers, which requires a grasp of complex arithmetic.

Complex Impedance Calculation

Impedance in AC circuits is a complex concept (pun intended!). It accounts for both the resistive and reactive elements in a circuit. When calculating complex impedance, we usually express it as
\( Z = R + jX \)
where:

• \( R \) is the resistance, measured in ohms.
• \( X \) is the reactance, also measured in ohms, representing the effects of capacitance and inductance.
• \( j \) is the imaginary unit, used to signify that reactance and resistance are not entirely in phase.

Reactance introduces a phase shift between voltage and current, crucial for systems operating with alternating current. This can affect how the circuit responds to signals at different frequencies. Adjusting the resistive and reactive components accordingly allows engineers to tune circuits for particular applications.
By analyzing the given problem with \( I = 0.5 + 2.0i \) and \( Z = 0.4 - 3.0i \),we see how these concepts translate to finding the voltage as a complex number. The resulting voltage \( V = 6.2 - 0.7i \) shows both magnitude and phase shift, integral for interpreting AC behavior in real-world circuit analysis.