Alternating current fundamentals

Almost all electrical energy in Nepal and the world is generated, transmitted and used as alternating current (AC), because AC voltage can be stepped up or down easily with transformers. This topic explains how a sinusoidal EMF is produced by a rotating coil, how it is written mathematically, how its average, peak and RMS values are defined, and how three-phase systems work. Nepal's supply is 50 Hz, with 230 V single-phase and 400 V three-phase at the consumer end.

Key concepts

Generation of AC

• When a coil of N turns, area A, rotates with angular velocity $\omega$ in a uniform magnetic field B, the flux linkage is $N B A \cos\omega t$. 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 (conductors cut flux at the fastest rate) and zero when the coil plane is perpendicular to the field (maximum flux linkage).
• In practical alternators the field rotates and the armature is stationary. Frequency $f = PN/120$ (P = poles, N = rpm).

Terminology

• Cycle, time period $T = 1/f$, frequency $f$ (Hz), angular frequency $\omega = 2\pi f$ (rad/s).
• Instantaneous value: $v = V_m\sin(\omega t + \phi)$, where $\phi$ is the phase angle. A quantity with larger phase angle "leads".
• Phasor: a rotating vector whose length equals the RMS (or peak) value; used to add sinusoids of the same frequency.

Average value

• The mean over a half cycle (the full-cycle average of a symmetrical sine wave is zero). For a sine: $V_{av} = 2V_m/\pi = 0.637V_m$.

RMS (effective) value

• The DC value that produces the same heating in a resistor. For a sine: $V_{rms} = V_m/\sqrt2 = 0.707V_m$. Meters read RMS; 230 V supply has peak $230\sqrt2 \approx 325$ V.

Form and peak factors

Waveform	RMS	Average	Form factor	Peak factor
Sine	$0.707V_m$	$0.637V_m$	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	$V_m/2$	$V_m/\pi$	1.57	2

Three-phase system

• Three EMFs of equal magnitude and frequency, displaced by 120° electrical. Phase sequence R-Y-B (also written a-b-c).
• Advantages: more power per conductor weight, constant instantaneous power in a balanced load, self-starting rotating magnetic field for induction motors, smoother rectified output.
• Star (Y): $V_L = \sqrt3V_{ph}$, $I_L = I_{ph}$; neutral available (4-wire). Line voltage leads phase voltage by 30°.
• Delta (Δ): $V_L = V_{ph}$, $I_L = \sqrt3 I_{ph}$; no neutral.
• In a balanced star system, neutral current is zero; phasor sum of three balanced quantities is zero.

High-yield points

• Sine wave: $V_{rms} = 0.707V_m$, $V_{av} = 0.637V_m$, form factor 1.11, peak factor 1.414.
• Star: $V_L = \sqrt3V_{ph}$, $I_L = I_{ph}$; delta: $I_L = \sqrt3I_{ph}$, $V_L = V_{ph}$.
• Three-phase power $P = \sqrt3V_LI_L\cos\phi$ for both star and delta.
• $f = PN/120$: a 4-pole machine runs at 1500 rpm for 50 Hz.
• Two-wattmeter method: at pf 0.5 one wattmeter reads zero.

Formulas

$e = E_m\sin\omega t,\quad E_m = NBA\omega$

$V_{rms} = \sqrt{\frac{1}{T}\int_0^T v^2\,dt},\qquad V_{av} = \frac{1}{T/2}\int_0^{T/2} v\,dt$

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

$P_{3\phi} = \sqrt3 V_L I_L\cos\phi\ \text{(W)},\quad Q_{3\phi} = \sqrt3V_LI_L\sin\phi\ \text{(VAR)}$

Two-wattmeter method: $P = W_1 + W_2$, $\tan\phi = \sqrt3\frac{W_1 - W_2}{W_1+W_2}$.

Worked example

A balanced star load draws 10 A per phase at 0.8 pf lagging from a 400 V, 50 Hz three-phase supply. Find phase voltage and total power.

1. $V_{ph} = 400/\sqrt3 = 230.9$ V.
2. In star, $I_L = I_{ph} = 10$ A.
3. $P = \sqrt3 \times 400 \times 10 \times 0.8 = 1.732\times3200 = 5543$ W ≈ 5.54 kW.
4. Check: $3V_{ph}I_{ph}\cos\phi = 3\times230.9\times10\times0.8 = 5542$ W. ✓

Exam tips

• Sine wave: form factor 1.11, peak factor 1.414 — both asked frequently.
• Average over a full cycle of a pure sine is zero; "average value" in MCQs means half-cycle average.
• In two-wattmeter method, at pf = 0.5 one wattmeter reads zero; below 0.5 one reading goes negative.
• $f = PN/120$: a 4-pole alternator must run at 1500 rpm for 50 Hz.
• Star: line voltage = √3 phase voltage; Delta: line current = √3 phase current.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. For $v = 100\sin(314t + \pi/6)$ V, what is the instantaneous voltage at $t = 0$?

• A. 86.6 V
• B. 0 V
• C. 100 V
• D. 50 V

Answer: D. $v(0) = 100\sin(30°) = 50$ V.

2. A coil of 100 turns, each of area 0.02 m$^2$, rotates at $\omega = 100\pi$ rad/s in a uniform field of 0.5 T. What is the maximum induced emf?

• A. 31.4 V
• B. 314 V
• C. 157 V
• D. 100 V

Answer: B. $E_m = NBA\omega = 100 \times 0.5 \times 0.02 \times 100\pi = 100\pi \approx 314$ V.

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