Phasors — why we “cheat” (and why it’s allowed)

1) The problem we’re pretending not to have

AC systems are time-dependent. Voltage and current are functions of time:

\[ v(t) = V_\text{peak}\sin(\omega t + \theta) \]

That’s fine for a single waveform on a napkin. It becomes a swamp when you have:

• multiple phases (120° separation)
• inductors and capacitors (phase shifts)
• power (averages, products, and “why is there a double-frequency term?”)

2) Why we don’t “just use sine functions directly”

In theory, you can solve everything in the time domain with trig + calculus. In practice, it’s repetitive, fragile, and hard to audit.

Inductors and capacitors drag calculus into everything

For an inductor: \(v(t) = L\frac{di(t)}{dt}\). For a capacitor: \(i(t) = C\frac{dv(t)}{dt}\). So even a “simple” circuit wants differentiation/integration immediately.

The system is linear. Our first approach is… melodramatic.

3) An intentionally ugly example

Take a simple RL load driven by a sine:

\[ v(t) = V_m \sin(\omega t) \]

The steady-state current is:

\[ i(t) = \frac{V_m}{\sqrt{R^2 + (\omega L)^2}} \sin\left(\omega t - \tan^{-1}\left(\frac{\omega L}{R}\right)\right) \]

That’s already “fine but annoying”. Now compute instantaneous power:

\[ p(t)=v(t)i(t) \]

You’ll get a constant term (average power) plus a term oscillating at \(2\omega\) (the “double-frequency wiggle”). And that’s before you do three phases.

// Time-domain reality check (not code you want to maintain):
v(t) = Vm * sin(ωt)
i(t) = (Vm / |Z|) * sin(ωt - φ)
p(t) = v(t) * i(t)  // expands into DC + cos(2ωt - φ) terms

4) The key realization: steady state doesn’t care about time

In sinusoidal steady state, everything oscillates at the same frequency \(\omega\). Only amplitude and phase differ.

So we factor out the shared time dependence and keep the meaningful parts.

5) Phasors: time removed, meaning retained

We represent a sinusoid by its RMS magnitude and phase:

\[ v(t) = V_\text{peak}\sin(\omega t + \theta) \quad\Longrightarrow\quad \underline{V} = V_\text{RMS}\angle\theta \]

Impedance becomes algebra

\[ Z_R = R,\quad Z_L = j\omega L,\quad Z_C = \frac{1}{j\omega C} \]

Now Ohm’s law is still Ohm’s law:

\[ \underline{V} = \underline{I}\,Z \]

6) Why RMS is the amplitude that survives contact with reality

Peak values are visually dramatic. They are also not what power cares about. Power is energy per time, and for periodic signals we care about an average effect.

RMS (root mean square) is defined as:

\[ V_\text{RMS} = \sqrt{\frac{1}{T}\int_0^T v^2(t)\,dt} \]

RMS of a sinusoid (proof sketch)

Let \(v(t)=V_m\sin(\omega t)\). Then:

\[ V_\text{RMS} = \sqrt{\frac{1}{T}\int_0^T V_m^2\sin^2(\omega t)\,dt} = V_m\sqrt{\frac{1}{T}\int_0^T \sin^2(\omega t)\,dt} \]

Over a full period, the average of \(\sin^2\) is \(1/2\), so:

\[ V_\text{RMS} = \frac{V_m}{\sqrt{2}} \]

7) Power in phasor form (where things finally get clean)

Using RMS phasors, complex power is:

\[ \underline{S} = \underline{V}\,\underline{I}^* \]

From this:

• \(P = \Re\{\underline{S}\}\) is real power (watts)
• \(Q = \Im\{\underline{S}\}\) is reactive power (var)
• \(|S|\) is apparent power (VA)

Power factor follows the triangle:

\[ \cos\varphi = \frac{P}{|S|} \]

Example: cos φ = 0.7

If \(\cos\varphi=0.7\), then \(\varphi \approx \cos^{-1}(0.7) \approx 45.57^\circ\). If we choose \(|S| = 10\text{ kVA}\) for round numbers:

\[ P = |S|\cos\varphi = 10\cdot 0.7 = 7\text{ kW} \]

\[ Q = |S|\sin\varphi \approx 10\cdot \sqrt{1-0.7^2} \approx 7.14\text{ kVAr} \]

8) Power factor correction: the practical problem

Utilities charge for poor power factor because reactive power increases current (and thus losses) without delivering useful work. The solution: add capacitors to cancel inductive reactive power.

Problem setup

Consider an industrial load:

• Real power: \(P = 50\text{ kW}\) (constant)
• Voltage: \(V = 480\text{ V (line-to-line, RMS)}\)
• Frequency: \(f = 60\text{ Hz}\), so \(\omega = 2\pi f \approx 377\text{ rad/s}\)
• Original power factor: \(\cos\varphi_1 = 0.70\) (lagging)
• Target power factor: \(\cos\varphi_2 = 0.95\) (lagging)

Step 1: Find original reactive power

From \(\cos\varphi_1 = 0.70\):

\[ \varphi_1 = \cos^{-1}(0.70) \approx 45.57^\circ \]

\[ \tan\varphi_1 = \frac{Q_1}{P} \quad\Longrightarrow\quad Q_1 = P\tan\varphi_1 \]

\[ Q_1 = 50\text{ kW} \cdot \tan(45.57^\circ) \approx 50 \cdot 1.020 = 51.0\text{ kVAr} \]

Step 2: Find target reactive power

From \(\cos\varphi_2 = 0.95\):

\[ \varphi_2 = \cos^{-1}(0.95) \approx 18.19^\circ \]

\[ Q_2 = P\tan\varphi_2 = 50 \cdot \tan(18.19^\circ) \approx 50 \cdot 0.329 = 16.4\text{ kVAr} \]

Step 3: Required capacitor reactive power

The capacitor must supply the difference:

\[ Q_C = Q_1 - Q_2 = 51.0 - 16.4 = 34.6\text{ kVAr} \]

Step 4: Calculate capacitance

For a capacitor bank on a three-phase system (delta connection):

\[ Q_C = 3\omega C V^2 \]

Solving for \(C\):

\[ C = \frac{Q_C}{3\omega V^2} \]

Plugging in values (convert kVAr to VAr):

\[ C = \frac{34{,}600}{3 \cdot 377 \cdot 480^2} = \frac{34{,}600}{260{,}198{,}400} \approx 133\text{ µF} \]

Wiring diagram (simplified)

9) Three-phase: the real payoff

Three-phase systems are where phasors stop being “nice” and become “mandatory”. Balanced systems collapse to one phase, and the 120° separation is naturally represented as angles.

Teaser: Δ and Y connections

In a balanced system:

• Line-to-line voltages in Δ are separated by 120°.
• Relationships like \(\sqrt{3}\) and 30° shifts appear as simple phasor geometry.

10) What phasors are not

• They do not handle transients well (switching, startup, non-sinusoidal waveforms).
• They assume a single frequency steady state.
• They hide time dependence; they don’t deny it.

11) Why this page exists

Because the smooth wave on the front page looks simple. It isn’t.

Phasors are the reason engineers can design AC systems without drowning in trig identities. They are not a trick. They’re a mercy.

SevIQ • phasors chapter • shareable notes • back to home