Phasors and AC Impedance

Analyzing Alternating Current (AC) circuits using time-domain sinusoids requires solving difficult differential equations. To simplify this, electrical engineers transform time-domain sinusoids into the phasor domain, where variables are treated as complex numbers. This allows AC circuits to be solved using simple algebra, exactly like DC circuits.

Minimal Complex Number Notation

To understand phasors, we use a basic complex number notation. In electrical engineering, the imaginary unit is denoted by $j$ (where $j^2 = -1$) instead of $i$, to avoid confusion with electrical current. A phasor $\mathbf{V}$ can be written in polar form as a magnitude and an angle: $\mathbf{V} = V_m \angle \phi$

The Phasor Concept

A sinusoid in the time domain, $v(t) = V_m \cos(\omega t + \phi)$, is transformed into a phasor $\mathbf{V} = V_m \angle \phi$. (Note: By convention, we reference phasors to the cosine function. If a sine function is given, it is converted to cosine by subtracting $90^\circ$.)

Impedance ($Z$)

Impedance ($\mathbf{Z}$) is the AC equivalent of resistance. It represents the total opposition a component offers to alternating current. Like resistance, it is measured in Ohms ($\Omega$). Because AC circuits involve phase shifts, impedance is a complex number.

The fundamental relationship in the phasor domain is Ohm's Law for AC: $\mathbf{V} = \mathbf{I} \mathbf{Z}$

Impedance of Basic Components

For the three ideal circuit elements, the impedances are defined as follows:

1. Resistor (R): $\mathbf{Z}_R = R$ (The voltage and current are exactly in phase).

2. Inductor (L): $\mathbf{Z}_L = j \omega L$ (The current lags the voltage by $90^\circ$).

3. Capacitor (C): $\mathbf{Z}_C = \frac{1}{j \omega C} = -j \frac{1}{\omega C}$ (The current leads the voltage by $90^\circ$).

The inverse of impedance is Admittance ($\mathbf{Y} = 1/\mathbf{Z}$), measured in Siemens (S).

Worked Example

Problem: A $10 \text{ mH}$ inductor is connected to an AC voltage source. The frequency of the source is $50 \text{ Hz}$. Calculate the impedance of the inductor.

Solution:

1. Identify the given values: $L = 10 \times 10^{-3} \text{ H}$, $f = 50 \text{ Hz}$.
2. Calculate the angular frequency $\omega$: $\omega = 2 \pi f = 2 \pi (50) \approx 314.16 \text{ rad/s}$
3. Calculate the impedance $\mathbf{Z}_L$: $\mathbf{Z}_L = j \omega L = j (314.16) (10 \times 10^{-3})$ $\mathbf{Z}_L = j 3.1416 \, \Omega$

The magnitude of the impedance is $3.1416 \, \Omega$, and the phase angle is exactly $+90^\circ$.

Engineering Check

Notice that the impedance of an inductor ($j \omega L$) increases with frequency, acting like an open circuit at very high frequencies. Conversely, the magnitude of a capacitor's impedance ($1/\omega C$) decreases with frequency, acting like a short circuit at high frequencies. This fundamental behavior is the basis for all electronic filtering (e.g., low-pass and high-pass filters).\n