Alternating current fundamentals

Most electrical power in Nepal and the world is generated, transmitted and used as alternating current (AC), because AC voltages can be stepped up and down easily with transformers. This topic explains how a sinusoidal voltage is produced by rotating a coil in a magnetic field, how it is described mathematically, the meaning of peak, average and RMS values, and how three-phase systems are connected. Nepal's power system operates at 50 Hz.

Key concepts

• Generation of AC: When a coil of $N$ turns and area $A$ rotates with angular speed $\omega$ in a uniform magnetic field $B$, the flux linkage changes sinusoidally. By Faraday's law the induced emf is $e = NBA\omega\sin\omega t = E_m\sin\omega t$. EMF is maximum when the coil plane is parallel to the field (flux linkage zero, rate of change maximum) and zero when the plane is perpendicular.
• Terms: Cycle (one complete set of positive and negative values), time period $T$ (seconds per cycle), frequency $f = 1/T$ (Hz), amplitude (peak value), phase (position in the cycle) and phase difference (angle between two waves of the same frequency).
• Frequency of an alternator: $f = PN/120$, where $P$ is the number of poles and $N$ is speed in rpm. A 4-pole machine needs 1500 rpm for 50 Hz.
• Equation of a waveform: $v = V_m\sin(\omega t \pm \phi)$, with $\omega = 2\pi f$ rad/s. A "+φ" wave leads; "−φ" lags.
• Peak value: the maximum instantaneous value $V_m$. Peak-to-peak = $2V_m$ for a sine.
• Average value: For a symmetrical wave the full-cycle average is zero, so the average is taken over a half cycle. For a sine wave $V_{avg} = 2V_m/\pi = 0.637V_m$.
• RMS (effective) value: The DC value that produces the same heating in a resistor. For a sine wave $V_{rms} = V_m/\sqrt2 = 0.707V_m$. Meters read RMS; "230 V" is an RMS value, so the peak is about 325 V.
• Form factor and peak factor:

Waveform	RMS	Average (half cycle)	Form factor	Peak (crest) factor
Sine	$V_m/\sqrt2$	$2V_m/\pi$	1.11	1.414
Square	$V_m$	$V_m$	1	1
Triangular	$V_m/\sqrt3$	$V_m/2$	1.155	1.732
Half-wave rectified sine	$V_m/2$	$V_m/\pi$ (full cycle)	1.57	2

• Phasors: A sinusoid can be represented by a rotating vector of length equal to the RMS (or peak) value. In a pure resistor V and I are in phase; in a pure inductor I lags V by 90°; in a pure capacitor I leads V by 90° (memory aid "CIVIL").
• Three-phase system: Three equal voltages 120° apart, produced by three windings spaced 120° electrical in the alternator. Phase sequence is R-Y-B. Advantages: constant instantaneous power, about 25% less conductor material for the same power, self-starting induction motors, and smaller machines for the same rating.
• Star (Y) connection: $V_L = \sqrt3 V_{ph}$, $I_L = I_{ph}$; provides a neutral, so both phase and line voltages are available (e.g. 230 V and 400 V).
• Delta (Δ) connection: $V_L = V_{ph}$, $I_L = \sqrt3 I_{ph}$; no neutral.
• Three-phase power (balanced): $P = \sqrt3 V_L I_L\cos\phi$ for both star and delta. In a balanced star, neutral current is zero.

High-yield points

• Sine wave: $V_{rms} = 0.707V_m$, $V_{avg} = 0.637V_m$ (half cycle).
• Sine wave form factor = 1.11, peak factor = 1.414.
• $f = PN/120$: 4-pole → 1500 rpm, 2-pole → 3000 rpm at 50 Hz.
• Star: $V_L = \sqrt3V_{ph}$, $I_L = I_{ph}$. Delta: $V_L = V_{ph}$, $I_L = \sqrt3I_{ph}$.
• $P = \sqrt3V_LI_L\cos\phi$ for balanced star and delta.
• Three phases are 120° apart; neutral current of a balanced star load is zero.

Formulas

$e = E_m \sin\omega t, \quad E_m = NBA\omega$ $N$ = turns, $B$ = flux density (T), $A$ = coil area (m²), $\omega$ = angular speed (rad/s).

$V_{rms} = \frac{V_m}{\sqrt2}, \qquad V_{avg} = \frac{2V_m}{\pi}$

$\text{Form factor} = \frac{V_{rms}}{V_{avg}}, \qquad \text{Peak factor} = \frac{V_m}{V_{rms}}$

$P_{3\phi} = \sqrt3\, V_L I_L \cos\phi$ $V_L$ = line voltage (V), $I_L$ = line current (A), $\cos\phi$ = power factor.

Worked example

$v = 325\sin(314t)$ V. Find RMS value, frequency and average value.

1. $V_{rms} = 325/\sqrt2 = 325/1.414 \approx 229.8$ V ≈ 230 V.
2. $\omega = 314 = 2\pi f \Rightarrow f = 314/6.283 \approx 50$ Hz.
3. $V_{avg} = 0.637\times325 \approx 207$ V.

Three-phase check: a balanced star load draws 10 A at 400 V line, pf 0.8. $P = 1.732\times400\times10\times0.8 \approx 5542$ W ≈ 5.54 kW.

Exam tips

• Form factor of a sine = 1.11; peak factor = 1.414. These are asked very often.
• Average over a full cycle of a pure sine is zero; MCQs usually mean the half-cycle average.
• In star, line voltage leads phase voltage by 30°; in delta, line current lags phase current by 30°.
• $f = PN/120$: a 2-pole 50 Hz machine runs at 3000 rpm.
• Do not confuse $\sqrt3 V_LI_L\cos\phi$ with $3V_{ph}I_{ph}\cos\phi$ — they are equal; use whichever matches the given data.

Exam-style questions

From Calmon's question bank, which has 23 questions on this topic.

1. In a circuit, $v = 100\sin(\omega t + 30^\circ)$ V and $i = 10\sin(\omega t - 15^\circ)$ A. The current:

• A. Lags the voltage by 45°
• B. Leads the voltage by 45°
• C. Lags the voltage by 15°
• D. Is in phase with the voltage

Answer: A. Phase difference $= 30^\circ - (-15^\circ) = 45^\circ$, with the current behind the voltage, so it lags by 45°.

2. A balanced star-connected supply has a line voltage of 400 V. Its phase voltage is about:

• A. 231 V
• B. 400 V
• C. 133 V
• D. 693 V

Answer: A. In star, $V_L = \sqrt{3}V_{ph}$, so $V_{ph} = 400/\sqrt{3} \approx 231$ V.

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