AC Fundamentals: RMS, Peak Value, Phase Angle and Power Factor Explained
The socket in an Indian home supplies 230 V at 50 Hz. Both numbers hide a lot of detail. The voltage is not fixed at 230 V; it swings between about +325 V and -325 V fifty times every second. The 230 V figure is an RMS value, a way of comparing AC with DC by the heating effect it produces. Before you can analyse motors, transformers or power factor correction, you need to be comfortable with these basic descriptions of a sine wave.
This article covers the instantaneous equation of a sinusoid, peak, average and RMS values, frequency and angular frequency, phase difference, reactance and impedance, and ends with a fully worked series RL circuit calculation.
Describing a Sinusoidal Voltage
An alternator produces a voltage that varies sinusoidally with time. The instantaneous value at any moment is written as:
v(t) = Vm sin(ωt + φ) Vm = peak (maximum) value, volts ω = angular frequency = 2πf, rad/s f = frequency, Hz T = time period = 1/f, s φ = phase angle at t = 0, degrees or radians
For the Indian supply, f = 50 Hz, so T = 1/50 = 20 ms and ω = 2 × π × 50 = 314.16 rad/s. A 230 V supply can therefore be written as v(t) = 325.3 sin(314.16t) volts, taking the reference phase angle as zero.
Peak, average and RMS values
The average value of a full sine cycle is zero because the positive and negative halves cancel, so the average is defined over a half cycle. The RMS (root mean square) value is the DC value that would produce the same heat in a resistor. Meters, nameplates and tariffs all use RMS values unless stated otherwise.
Vavg (half cycle) = 2Vm/π = 0.637 Vm Vrms = Vm/√2 = 0.707 Vm Form factor = Vrms / Vavg = 1.11 Crest factor = Vm / Vrms = 1.414 For the 230 V mains: Vm = 230 × 1.414 = 325.3 V Vavg = 0.637 × 325.3 = 207.2 V
Phase Difference, Reactance and Impedance
Two sine waves of the same frequency can be shifted in time. The shift, expressed as an angle, is the phase difference. If the current reaches its peak after the voltage, the current is said to lag; if before, it leads. In a pure resistor the current is in phase with the voltage. In a pure inductor it lags by 90 degrees, and in a pure capacitor it leads by 90 degrees.
Inductors and capacitors oppose AC through reactance, measured in ohms. Inductive reactance rises with frequency, while capacitive reactance falls with frequency.
XL = 2πfL (ohms, L in henry) XC = 1/(2πfC) (ohms, C in farad) Series RLC impedance: Z = √(R² + (XL − XC)²) φ = tan⁻¹((XL − XC)/R) I = V/Z
| Element | Opposition | Phase of current vs voltage | Power factor |
|---|---|---|---|
| Pure resistor R | R | In phase (0°) | 1 (unity) |
| Pure inductor L | XL = 2πfL | Lags by 90° | 0 lagging |
| Pure capacitor C | XC = 1/(2πfC) | Leads by 90° | 0 leading |
| R and L in series | √(R² + XL²) | Lags by 0° to 90° | R/Z lagging |
| R and C in series | √(R² + XC²) | Leads by 0° to 90° | R/Z leading |
Power in an AC circuit
Because voltage and current are not always in step, the product V × I does not give the useful power. Three quantities are defined. Real power P in watts does useful work and produces heat. Reactive power Q in VAR flows back and forth between source and the magnetic or electric fields. Apparent power S in VA is simply V × I and decides the size of cables, switchgear and transformers.
P = V I cos φ (W) Q = V I sin φ (VAR) S = V I (VA) S² = P² + Q² Power factor = cos φ = P/S = R/Z
Worked Example: Series RL Circuit on 230 V, 50 Hz
A coil with resistance R = 30 Ω and inductance L = 127.3 mH is connected to the 230 V, 50 Hz supply. Find the reactance, impedance, current, phase angle, power factor and all three powers.
- Reactance: XL = 2π × 50 × 0.1273 = 40.0 Ω.
- Impedance: Z = √(30² + 40²) = √2500 = 50 Ω.
- Current: I = 230/50 = 4.6 A (RMS).
- Phase angle: φ = tan⁻¹(40/30) = 53.13°, current lagging.
- Power factor: cos φ = 30/50 = 0.6 lagging.
- Real power: P = 230 × 4.6 × 0.6 = 634.8 W. Check: I²R = 4.6² × 30 = 634.8 W.
- Reactive power: Q = 230 × 4.6 × 0.8 = 846.4 VAR. Check: I²XL = 4.6² × 40 = 846.4 VAR.
- Apparent power: S = 230 × 4.6 = 1058 VA. Check: √(634.8² + 846.4²) = 1058 VA.
Notice that only 634.8 W is consumed, yet the supply must deliver 4.6 A. A purely resistive load taking the same 634.8 W would draw only 2.76 A. That extra current, caused by low power factor, heats cables and transformers without doing useful work. This is the reason utilities penalise industrial consumers with poor power factor.
Checking These Values on the Bench
You can verify the theory with a step-down transformer instead of working directly on mains. A 230/12 V transformer, a 10 Ω power resistor and a small inductor give a safe low-voltage circuit. Measure the supply voltage, the voltage across R and the current with a true RMS multimeter, then calculate cos φ = VR/Vsupply. An oscilloscope with both channels referenced to the same ground point shows the time shift directly.
Common Mistakes
- Using the peak value 325 V instead of 230 V RMS in power formulas. All standard power formulas assume RMS values.
- Forgetting to convert mH to H or µF to F when computing reactance.
- Adding R and XL arithmetically instead of as a right-angled triangle. 30 Ω plus 40 Ω gives 50 Ω, not 70 Ω.
- Calling power factor leading or lagging without checking which element dominates. Inductive loads such as motors give lagging power factor.
- Assuming a meter reading is correct on a non-sinusoidal waveform when the meter is not true RMS.
Summary
A sinusoidal voltage is fully described by its peak value, frequency and phase. The RMS value, equal to 0.707 times the peak for a sine wave, is what nameplates and meters report. Inductors and capacitors add reactance that shifts the current in time relative to voltage, and the cosine of that shift is the power factor. With Z, I, cos φ, P, Q and S you have every tool needed for the three-phase, motor and power factor correction topics that follow.