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core30 min studyLesson 283 in Physics

Complex impedance and power factor

Use an explicit RMS phasor convention to calculate impedance, phase, average power, and reactive power without confusing storage with dissipation.

Before you begin
On this page

The question for today

What you’ll understand

  • Translate linear element laws into RMS phasor impedances.
  • Calculate real and reactive power using complex conjugation.
  • Interpret power factor and ideal reactive compensation at one frequency.

A sinusoidal circuit contains two kinds of information: how large each oscillation is and when its peaks occur. Complex phasors encode both while replacing differentiation with multiplication by angular frequency. They describe a steady state at one frequency, after transients have been handled separately. The convention for amplitude matters just as much as the algebra: using peak amplitudes in a formula intended for RMS values creates a factor-of-two error in power.

Declare the time and amplitude conventions

Use the real part of an RMS phasor multiplied by a positive-frequency complex exponential and by square root of two to reconstruct a physical voltage. Current follows the same convention. A derivative then becomes multiplication by positive j omega. The inductor impedance has positive imaginary part, while the capacitor has negative imaginary part. Reversing the time convention reverses these signs consistently; it does not change measurable power or which response leads in physical time.

Impedance is voltage divided by current at the same port, using current entering the positive voltage terminal. Series impedances add because currents are shared, whereas parallel admittances add because voltages are shared. Adding magnitudes instead of complex numbers destroys phase cancellation. Phasors also cannot directly represent the full waveform of a nonlinear rectifier: each harmonic needs its own frequency component, and nonlinear coupling between components requires a broader model than a single impedance.

v(t)=2Re⁡{Vejωt},ZR=R,ZL=jωL,ZC=1jωCv(t)=\sqrt2\operatorname{Re}\{Ve^{j\omega t}\},\qquad Z_R=R,\quad Z_L=j\omega L,\quad Z_C=\frac1{j\omega C}
V is an RMS voltage phasor in volts, j² = −1, and Z is in ohms. R, L, and C are constant linear element values. These expressions assume sinusoidal steady state and the e^{jωt} convention.↔ Scroll wide equations horizontally.

Average power selects the in-phase part

Multiplying real voltage and current produces a constant term and an oscillating term at twice the drive frequency. Averaging over a full period removes the latter. The remaining power depends on the cosine of the voltage-current phase difference. Complex power packages that real average with a signed reactive term. Conjugating the current phasor is essential: it forms a phase difference rather than a phase sum, making the result independent of the arbitrary common time origin.

A positive reactive power indicates net inductive behavior with current lagging voltage under the chosen convention. A negative value indicates capacitive behavior. Reactive power measures cyclic exchange rather than net energy converted to heat each period. Its unit is conventionally var, while apparent power is in volt-amperes and real power in watts. Their numerical relationship is useful, but calling every volt-ampere a dissipated watt would misstate both heating and source requirements.

S=VI∗=P+jQ,P=VrmsIrmscos⁡ϕ,Q=VrmsIrmssin⁡ϕS=VI^*=P+jQ,\qquad P=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi,\quad Q=V_{\mathrm{rms}}I_{\mathrm{rms}}\sin\phi
φ = arg V − arg I is in radians, P is absorbed average power in watts, Q is reactive power in var, and S is complex power in VA. RMS phasors give no factor 1/2; peak phasors would require it.↔ Scroll wide equations horizontally.

Power factor describes a source burden

For sinusoidal voltage and current, power factor is the cosine of their phase difference, equivalently real power divided by apparent power. At fixed delivered real power and voltage, a lower magnitude power factor requires more RMS source current and therefore more resistive heating in supply wiring. The device's own real power need not change when reactive compensation reduces that extra current. A capacitor can supply part of an inductive load's reactive exchange locally.

Compensation is a frequency-specific design decision. An ideal shunt capacitor adds a current leading the applied voltage by a quarter cycle, so it can cancel an inductive load's lagging reactive current without canceling its in-phase current. Its physical capacitance must be chosen from voltage and frequency, not simply from a desired phase angle. Harmonics, resonances, component ratings, and varying loads complicate actual installations. For distorted waveforms, total power factor also includes waveform effects, so the single cosine formula no longer captures everything.

pf=P∣S∣=cos⁡ϕ,QC=−ωCVrms2\mathrm{pf}=\frac P{|S|}=\cos\phi,\qquad Q_C=-\omega C V_{\mathrm{rms}}^2
The capacitor is connected in parallel across an RMS voltage V at angular frequency ω. C is in farads and QC in var. This compensation formula assumes an ideal capacitor and a purely sinusoidal steady state.↔ Scroll wide equations horizontally.

Work it through

A load with a 3–4–5 impedance

A 10 V RMS source drives a series load Z = 3 + j4 Ω. Find the RMS current, real power, reactive power, and power factor.

  1. The impedance magnitude is ∣Z∣=32+42=5 Ω|Z|=\sqrt{3^2+4^2}=5\,\Omega. Therefore current magnitude is 10/5 = 2 A RMS, with current lagging the voltage.

  2. Complex division gives I=10/(3+j4)=1.2−j1.6 AI=10/(3+j4)=1.2-j1.6\,\mathrm A. Conjugating current yields S=10(1.2+j1.6)=12+j16 VAS=10(1.2+j1.6)=12+j16\,\mathrm{VA}.

  3. The apparent power magnitude is 10 × 2 = 20 VA. Real power is also Irms2R=4×3=12 WI_{\mathrm{rms}}^2R=4\times3=12\,\mathrm W, so power factor is 12/20 = 0.6 lagging.

The result

The load takes 2 A RMS, absorbs 12 W, exchanges +16 var, and has power factor 0.6 lagging.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Project an apparent-power vector

Reset and set BOTH Angle to 0 degrees and Magnitude to 2; predict zero y projection. For the second fresh comparison Reset and set BOTH Angle to 60 degrees and Magnitude to 2; predict x = 1 and positive y. Regard x and y as normalized real and reactive components. This one-vector model does not solve impedances, calculate RMS averages, or synthesize capacitor compensation.

Project an apparent-power vectorPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA phase angle reduces the in-phase projection even when the vector magnitude stays fixed.

Pause & reason

Check your understanding

With RMS phasors V and I under the passive sign convention, which expression gives absorbed average power?

Your turn

Try it yourself

A sinusoidal load at 100 V RMS absorbs 600 W and +800 var. An ideal shunt capacitor supplies −800 var. Find source current magnitudes before and after compensation, and C if ω = 100 rad/s.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute apparent power before compensation; afterward only the in-phase current remains. Use the capacitor reactive-power formula.

Compare with the solution

Before compensation, ∣S∣=6002+8002=1000 VA|S|=\sqrt{600^2+800^2}=1000\,\mathrm{VA}, giving 10 A RMS. After cancellation the source supplies 600 W and zero reactive power, so its current is 6 A RMS. The capacitor satisfies 100C(100)2=800100C(100)^2=800, giving C=0.0008 F=800 μFC=0.0008\,\mathrm F=800\,\mu\mathrm F. The original load still absorbs 600 W; the ideal capacitor redistributes reactive exchange and absorbs zero average power.

Keep the idea

Take it with you

  • An RMS phasor convention fixes both impedance signs and power normalization.
  • Real power is the real part of voltage times conjugate current.
  • Reactive compensation can reduce source current without reducing a load’s useful real power.

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