← All printables Chapter 06 · Electromagnetism · 2 sheets · read the chapter →
Vedatom Physics
FORMULA CARD · PHYSICS.VEDATOM.COM
Chapter 06 · Part V · Electromagnetism
Revised on ____________________
Chapter 06 · Alternating Current

Phasors, impedance, and the power that is really delivered

2 sheets
2 diagrams
26 results

Formulas and conditions only — no derivations, no solved numbers. The diagrams are the reader's own.

I

Peak, mean and rms

“230 V” IS AN RMS VALUE
I.Supply  v = V0 sin ω t ω = 2π f — 314 rad s−1 at 50 Hz
II.RMS, sinusoid  Vrms = V0√2 230 V rms ⇒ 325 V peak
III.Mean over a cycle  0 over a half cycle, 2V0/π
IV.Other waveforms  square V0 , triangular V0√3
V.Form factor  VrmsVmean = 1.11 sinusoid only
VI.A meter reads  rms, always
II

Reactance and impedance

PYTHAGORAS, NOT ADDITION
I.Resistor  V and I in phase
II.Inductor  XL = ω L current lags by 90°
III.Capacitor  XC = 1ω C current leads by 90°
IV.Impedance  Z = √( R² + (XL − XC)² )
V.Phase angle  tan φ = XL − XCR
VI.Current  Irms = VrmsZ
VII.Voltages  V = √( VR² + (VL − VC)² ) they do not add arithmetically
VIII.Mnemonic  ELI the ICE man E leads I in L; I leads E in C
V_R ∥ IV_LV_CVV_L−V_Cφvoltage phasorsRX_L−X_CZφimpedance triangle
III

Resonance

Z = R, AND THE CURRENT PEAKS
I.Condition  XL = XC
II.Frequency  ω0 = 1√(L C) , f0 = 12π√(LC)
III.At resonance  Z = R, φ = 0, I is maximum
IV.Quality factor  Q = 1R √( LC ) = ω0 LR
V.Bandwidth  Δω = RL = ω0Q
VI.Voltage magnification  VL = VC = Q V each can exceed the supply
ω / ω₀I_rms1high Q (small R)low Q (large R)
IV

Power and transformers

cos φ IS WHERE THE MONEY GOES
I.Average power  P = Vrms Irms cos φ
II.Power factor  cos φ = RZ
III.Wattless current  the component in quadrature with V pure L or C: P = 0
IV.Transformer  VsVp = NsNp = IpIs ideal, no power gain
V.Efficiency  η = PoutPin
VI.Transmission  high V, low I, low I²R loss
V

Where the marks go

FOUR ERRORS, EVERY YEAR
Where marks are lost in this chapter
1Writing XL = f L. It is ω L — at 50 Hz that is 314, not 50, and the answer is out by 2π.
2Adding R, XL and XC arithmetically. Z = √(R² + (XL − XC)²).
3Reading “zero mean” as “zero power”. Power follows v², whose average is V0²/2.
4Thinking L and C vanish at resonance. Their reactances cancel; VL and VC are each Q times the supply.