AC Circuits and Phasors Explained

The electricity in every Pakistani home is alternating current at 230 V and 50 Hz, constantly swinging back and forth. Analysing circuits with sine waves directly means wrestling with trigonometric identities and differential equations, but a brilliant shortcut called the phasor turns the whole job into ordinary algebra with complex numbers. Once you understand phasors, AC circuits become almost as easy as DC ones.

Describing a Sinusoid

An AC voltage can be written as

$$ v(t) = V_m \cos(\omega t + \phi) $$

where \( V_m \) is the peak (amplitude), \( \omega = 2\pi f \) is the angular frequency in rad/s and \( \phi \) is the phase angle. Meters and nameplates usually quote the RMS value, which for a sinusoid is

$$ V_{\text{rms}} = \frac{V_m}{\sqrt{2}} $$

So the 230 V mains has a peak of about \( 230\sqrt{2} \approx 325 \) V.

What Is a Phasor?

Euler's formula, \( e^{j\theta} = \cos\theta + j\sin\theta \), lets us write the sinusoid as the real part of a rotating complex number:

$$ v(t) = \operatorname{Re}\left\{ V_m e^{j\phi} e^{j\omega t} \right\} $$

In a linear circuit driven at one frequency, every voltage and current rotates at the same \( \omega \). So the factor \( e^{j\omega t} \) carries no new information, and we keep only the complex number

$$ \mathbf{V} = V_m e^{j\phi} = V_m \angle \phi $$

This is the phasor: a fixed arrow whose length is the amplitude (or RMS value) and whose angle is the phase. Engineers use \( j \) for \( \sqrt{-1} \) because \( i \) is reserved for current.

The magic is that differentiation becomes multiplication: \( \dfrac{d}{dt} \to j\omega \). Differential equations become algebraic equations.

Impedance: Ohm's Law for AC

With phasors, each component obeys \( \mathbf{V} = \mathbf{I}\mathbf{Z} \), where the impedance \( \mathbf{Z} \) is:

• Resistor: \( Z_R = R \) (voltage and current in phase)
• Inductor: \( Z_L = j\omega L = jX_L \) (current lags voltage by 90°)
• Capacitor: \( Z_C = \dfrac{1}{j\omega C} = -jX_C \) (current leads voltage by 90°)

with reactances \( X_L = \omega L \) and \( X_C = 1/(\omega C) \), both in ohms. A handy memory aid is "ELI the ICE man": in an inductor (L), E (voltage) leads I; in a capacitor (C), I leads E.

Impedances combine exactly like resistors, in series and parallel, and Kirchhoff's laws still apply, but with complex numbers.

Power in AC Circuits

Only resistance consumes real power. If the current lags or leads the voltage by angle \( \theta \), the average (real) power is

$$ P = V_{\text{rms}} I_{\text{rms}} \cos\theta $$

The factor \( \cos\theta \) is the power factor. Utilities penalise large industrial consumers with low power factors, because they draw extra current that heats cables without doing useful work.

Worked Example 1: An Inductive Load on the 50 Hz Mains

A motor winding is modelled as \( R = 30 \ \Omega \) in series with \( L = 0.1 \ \text{H} \), connected to 230 V RMS, 50 Hz. Find the current, its phase, the power factor and the real power.

Step 1: Angular frequency and reactance.

$$ \omega = 2\pi(50) = 314.16 \ \text{rad/s}, \qquad X_L = \omega L = 314.16 \times 0.1 = 31.42 \ \Omega $$

Step 2: Impedance.

$$ \mathbf{Z} = 30 + j31.42 \ \Omega $$

$$ |\mathbf{Z}| = \sqrt{30^2 + 31.42^2} = \sqrt{1886.96} = 43.44 \ \Omega, \qquad \theta = \tan^{-1}\left(\frac{31.42}{30}\right) = 46.3^\circ $$

Step 3: Current. Taking the voltage as reference, \( \mathbf{V} = 230\angle 0^\circ \) V:

$$ \mathbf{I} = \frac{230\angle 0^\circ}{43.44\angle 46.3^\circ} = 5.29\angle -46.3^\circ \ \text{A} $$

The current lags the voltage by 46.3°, as expected for an inductive load.

Step 4: Power factor and power.

$$ \text{pf} = \cos 46.3^\circ = 0.691 \ \text{(lagging)} $$

$$ P = 230 \times 5.29 \times 0.691 = 841 \ \text{W} $$

Sanity check. Real power is only consumed by the resistor: \( P = I^2R = (5.29)^2 \times 30 = 841 \) W. The two methods agree. Note also that the resistor voltage (\( 5.29 \times 30 = 159 \) V) and inductor voltage (\( 5.29 \times 31.42 = 166 \) V) add to more than 230 V arithmetically, but as phasors at right angles they combine to \( \sqrt{159^2 + 166^2} \approx 230 \) V.

Worked Example 2: A Series RLC Circuit

A series circuit has \( R = 10 \ \Omega \), \( L = 50 \ \text{mH} \) and \( C = 100 \ \mu\text{F} \), supplied with 100 V RMS at 50 Hz. Find the current and its phase, and the resonant frequency.

Step 1: Reactances.

$$ X_L = 314.16 \times 0.05 = 15.71 \ \Omega $$

$$ X_C = \frac{1}{314.16 \times 100 \times 10^{-6}} = 31.83 \ \Omega $$

Step 2: Total impedance.

$$ \mathbf{Z} = 10 + j(15.71 - 31.83) = 10 - j16.12 \ \Omega $$

$$ |\mathbf{Z}| = \sqrt{10^2 + 16.12^2} = 18.97 \ \Omega, \qquad \theta = \tan^{-1}\left(\frac{-16.12}{10}\right) = -58.2^\circ $$

Step 3: Current.

$$ \mathbf{I} = \frac{100\angle 0^\circ}{18.97\angle -58.2^\circ} = 5.27\angle 58.2^\circ \ \text{A} $$

The current leads the voltage, because the capacitive reactance dominates at 50 Hz.

Step 4: Resonant frequency, where \( X_L = X_C \):

$$ f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.05 \times 100 \times 10^{-6}}} = \frac{1}{2\pi(2.236 \times 10^{-3})} = 71.2 \ \text{Hz} $$

Sanity check. Since 50 Hz is below \( f_0 \), we expect \( X_C > X_L \) and a leading current, which is exactly what we found. At resonance the impedance would drop to just \( R = 10 \ \Omega \) and the current would rise to 10 A.

Did You Know?

• Charles Proteus Steinmetz popularised complex-number (phasor) methods for AC circuits in a paper presented in 1893, transforming electrical engineering practice.
• In a series RLC circuit near resonance, the voltage across the capacitor can be many times larger than the supply voltage.
• Pakistan, the UK and most of Europe use 50 Hz, while the USA uses 60 Hz, a legacy of early competing power companies.

Key Takeaways

• A phasor represents a sinusoid by its amplitude and phase as a complex number.
• Impedances \( R \), \( j\omega L \) and \( 1/(j\omega C) \) let you use Ohm's and Kirchhoff's laws in AC circuits.
• Inductive loads make current lag; capacitive loads make it lead.
• Real power is \( VI\cos\theta \); the power factor \( \cos\theta \) measures how usefully current is drawn.
• Resonance occurs at \( f_0 = 1/(2\pi\sqrt{LC}) \), where impedance is purely resistive.

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