categories: physics, automation & robotics, electrical engineering
Hello everyone, this is the first post in a short series where I want to summarize the physics concepts I keep reaching for the most as an automation/robotics/CS engineering student — not full theory dumps, just the practical core you actually use when you're sizing a motor, reading a datasheet, or debugging a circuit. Today: why AC circuits are analyzed with complex numbers, and how you actually push through a real calculation.
Why complex numbers at all? In DC, a resistor is the only thing you deal with, and Ohm's law is just $$U = I \cdot R$$. In AC, coils and capacitors introduce a phase shift between voltage and current — the current doesn't peak at the same instant as the voltage anymore. A single real number can't hold both "how big" and "how shifted in time" at once, but a complex number can: the magnitude is the size, the angle is the phase shift. That's the whole reason impedance $$\underline{Z} = R + jX$$ exists.
The three impedances you need to know cold:
Resistor: $$\underline{Z}_R = R$$ (no phase shift, purely real)
Inductor (coil): $$\underline{Z}_L = j \omega L = jX_L$$ (current lags voltage by 90°)
Capacitor: $$\underline{Z}_C = -j\frac{1}{\omega C} = -jX_C$$ (current leads voltage by 90°)
where $$\omega = 2\pi f$$ is the angular frequency. Notice the coil and the capacitor have opposite signs — that's why they can cancel each other out in a circuit (more on that below, it's the whole trick behind power-factor correction).
Worked example: motor + compensation capacitor
Take a real setup: a 240 V, 50 Hz supply feeds a line with resistance $$R_l = 6.67\ \Omega$$. At the end of the line, two branches sit in parallel: a motor (resistance $$R_m = 12\ \Omega$$, inductance $$L_m \approx 50.93\ \text{mH}$$) and a compensation capacitor ($$C \approx 127.32\ \mu F$$) wired in to fix the motor's poor power factor.
Step 1 — branch impedances. With $$\omega = 2\pi \cdot 50 = 100\pi\ \text{rad/s}$$: $$\underline{Z}_m = 12 + j(100\pi \cdot 0.16/\pi) = 12 + j16\ \Omega$$ $$\underline{Z}_C = -j\frac{1}{100\pi \cdot 400\cdot10^{-6}/\pi} = -j25\ \Omega$$
Step 2 — combine the parallel branches, the same way you'd combine parallel resistors, just with complex algebra: $$\underline{Z}_p = \frac{\underline{Z}_m \cdot \underline{Z}_C}{\underline{Z}_m + \underline{Z}_C} = \frac{(12+j16)(-j25)}{12-j9} = \frac{100}{3}\ \Omega$$ Then add the line in series: $$\underline{Z}_{tot} = \underline{Z}_l + \underline{Z}_p = \frac{20}{3} + \frac{100}{3} = 40\ \Omega$$. Notice this came out purely real — the capacitor's negative reactance exactly cancelled the motor's positive reactance. That's not a coincidence, it's the capacitor doing its job.
Step 3 — main current via complex Ohm's law: $$\underline{I} = \frac{\underline{U}}{\underline{Z}_{tot}} = \frac{240e^{j0^\circ}}{40} = 6\ \text{A}$$, in phase with the source voltage.
Step 4 — power. Complex power is $$\underline{S} = \underline{U} \cdot \underline{I}^*= 240 \cdot 6 = 1440\ \text{VA}$$, giving active power $$P = 1440\ \text{W}$$ and reactive power $$Q = 0\ \text{var}$$. Zero reactive power at the source is exactly what you want — it means the utility line only has to carry current that does actual work, not current that just sloshes back and forth between the coil and capacitor.
Exponential ↔ rectangular form, the part everyone forgets
Euler's formula is the bridge: $$re^{j\varphi} = r\cos\varphi + jr\sin\varphi = a + jb$$. Think of it as GPS directions vs. a grid: "walk 10 m at a bearing of -53.13°" (exponential form) is the same instruction as "6 steps east, 8 steps south" (rectangular form) — you're just choosing a different, equally valid way to describe the same vector. Concretely, $$10e^{-j53.13^\circ} = 10\cos(-53.13^\circ) + j \cdot 10\sin(-53.13^\circ) = 6 - j8$$.
Three-phase in one line
If a load is symmetric (identical on all three phases — the normal case for a three-phase motor), you never draw all three phases. You solve one phase as a single-phase equivalent circuit, then multiply the resulting power by 3. The two ways to wire a three-phase load:
Star (Y): each element sees the phase voltage $$U_f = U_L/\sqrt{3} \approx 230\ \text{V}$$ (for a 400 V line), line current equals phase current, draws 1/3 the power of the same load in delta.
Delta (Δ): each element sees the full line voltage $$U_L = 400\ \text{V}$$, line current is $$\sqrt{3}$$ times the phase current, draws 3× the power of the same load in star.
Star is used for motor start-up (lower inrush current) and for feeding single-phase household loads off the neutral; delta is used for continuous full-load running and for long-distance transmission (no neutral wire needed).
That's the practical core of AC complex-number analysis: turn every component into a complex impedance, combine them exactly like resistors (series adds, parallel uses the product-over-sum rule), then read off current and power at the end. Everything else is bookkeeping. Thank you for reading! Next post: electromagnetic induction and where the Lorentz force actually comes from.
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