Alternating Current: Quick Revision

1. RMS Values

RMS from peak

I_{rms} = \dfrac{I_0}{\sqrt{2}}, \qquad V_{rms} = \dfrac{V_0}{\sqrt{2}}

Rated AC values (e.g. “230 V”) are always RMS, not peak.

Inductor (V leads I by 90°)

X_L = \omega L

Capacitor (I leads V by 90°)

X_C = \dfrac{1}{\omega C}

XL grows with frequency; XC falls with frequency.

Impedance

Z = \sqrt{R^2+(X_L-X_C)^2}

Current amplitude

I = \dfrac{V}{Z}

Phase angle

\tan\phi = \dfrac{X_L-X_C}{R}

Resonant frequency

\omega_0 = \dfrac{1}{\sqrt{LC}}

Quality factor (at resonance)

Q = \dfrac{X_L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}

At ω = ω₀: XL = XC exactly, so Z = R (minimum) and I = V/R (maximum).
VL and VC at resonance: each equals Q×V — can be far larger than the source (example: L = 0.5 H, C = 2 µF, R = 10 Ω, V = 10 V gives Q = 50, so VL = VC = 500 V).
Not a KVL violation: VL and VC are 180° apart in phase, so they cancel each other in the loop sum (V_L − V_C = 0), leaving VR = V. They magnify because they fight each other, not because they add onto VR.

Average power

P = V_{rms}I_{rms}\cos\phi

Pure R: φ = 0, power factor = 1, all power dissipated.
Pure L or C: φ = 90°, power factor = 0, zero net power despite nonzero V and I.

Turns ratio

\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}

Full derivations and worked examples: detailed notes →