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23. July 2023 12:40
by Admin
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Complex Number Examples

23. July 2023 12:40 by Admin | 0 Comments

Let's explore a real-life example of using complex numbers to solve a physical problem: Electrical AC circuits and impedance.

In electrical engineering, alternating current (AC) circuits are prevalent, and they involve oscillating voltages and currents. When dealing with AC circuits, complex numbers are used to represent voltages, currents, and impedance, making calculations more straightforward.

Let's consider a simple AC circuit with a resistor and a capacitor in series. We want to find the total impedance of the circuit, which is the effective resistance to the flow of AC current.

The impedance, Z, of a resistor (R) and a capacitor (C) in series can be calculated using the formula:

Z = √(R^2 + (1 / ωC)^2),

where ω is the angular frequency of the AC signal, and C is the capacitance of the capacitor.

Now, let's use complex numbers to solve this problem step by step:

Step 1: Convert the elements to their complex representations.

Let's assume the voltage across the capacitor is represented by Vc, and the current through the capacitor is represented by Ic. We can express these quantities using complex numbers:

Vc = Vc0 * e^(jωt), where Vc0 is the peak voltage of the capacitor and ω is the angular frequency. Ic = Ic0 * e^(j(ωt + φ)), where Ic0 is the peak current of the capacitor, and φ is the phase angle between the voltage and current.

Step 2: Write down the relationship between voltage and current.

For a capacitor, the current is related to the voltage by the following equation:

Ic = jωC * Vc,

where j is the imaginary unit, and ω is the angular frequency, and C is the capacitance.

Step 3: Replace Ic and Vc in the impedance formula.

Substitute the complex representation of current (Ic) into the impedance formula:

Z = √(R^2 + (1 / (jωC))^2).

[ j here is an engineering term for the complex number i so  j^2+1=0 as it can get confused with i's used in engineering...]

Step 4: Simplify the expression.

To simplify, multiply both the numerator and denominator by -j:

Z = √(R^2 + (-j / (ωC))^2).

Now, -j / (ωC) is the complex representation of the impedance of the capacitor.

Step 5: Evaluate the impedance.

Now, we can calculate the total impedance by taking the square root:

Z = √(R^2 - 1 / (ω^2C^2)).

The above expression represents the impedance of the series combination of the resistor and capacitor in the AC circuit.

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