Alternating current fundamentals

Almost all electrical power is generated, transmitted and used as alternating current because AC voltages can be changed easily with transformers and AC machines are simple and robust. Nepal's supply is 230 V single-phase / 400 V three-phase at 50 Hz. This topic explains how a sinusoidal voltage is generated, how it is written as an equation and drawn as a waveform, and the key measures of an AC quantity — peak, average and RMS values, with form and peak factors.

Key concepts

• Generation of AC: a coil of N turns and area A rotating with angular velocity ω in a uniform magnetic field B cuts flux at a varying rate. Flux linkage $N\phi = NBA\cos\omega t$, so by Faraday's law $e = NBA\omega\sin\omega t = E_m\sin\omega t$. EMF is maximum when the coil plane is parallel to the field (cutting flux fastest) and zero when perpendicular to it. Direction by Fleming's right-hand rule.
• Practical alternators use a rotating field and stationary armature; frequency $f = PN/120$.
• Equation and terms: $v = V_m\sin(\omega t + \phi)$.
  • Amplitude/peak value $V_m$; instantaneous value v.
  • Cycle, time period T (s), frequency f = 1/T (Hz), angular frequency ω = 2πf (rad/s). For 50 Hz, T = 20 ms and ω = 314.16 rad/s.
  • Phase φ — position relative to reference; phase difference between two quantities of the same frequency; one leads or lags the other.
• Waveforms: sinusoidal (standard because it gives least distortion in machines, its derivatives and integrals are also sinusoidal), square, triangular, sawtooth. Non-sinusoidal waves can be resolved into harmonics by Fourier series.
• Average value: the arithmetic mean over a half cycle (the full-cycle average of a symmetrical AC wave is zero). Relevant for rectifiers and charging (DC effect).
• RMS (effective) value: the steady DC value that produces the same heat in a resistor in the same time. AC voltmeters and ammeters are calibrated to indicate RMS; ratings like 230 V are RMS values. The peak of 230 V RMS is about 325 V.
• Form factor = RMS/average; peak (crest) factor = peak/RMS.
• Phasor representation: a sinusoid can be represented by a rotating vector of length equal to its RMS (or peak) value at angle φ, allowing addition by vector methods: $V\angle\phi$ in polar or $a + jb$ in rectangular form.

Waveform	$V_{avg}$ (half cycle)	$V_{rms}$	Form factor	Peak factor
Sine	$0.637V_m$	$0.707V_m$	1.11	1.414
Half-wave rectified sine (full cycle)	$0.318V_m$	$0.5V_m$	1.57	2
Full-wave rectified sine	$0.637V_m$	$0.707V_m$	1.11	1.414
Square	$V_m$	$V_m$	1	1
Triangular	$0.5V_m$	$0.577V_m$	1.155	1.732

High-yield points

• Sine wave: $V_{avg} = 2V_m/\pi = 0.637V_m$, $V_{rms} = V_m/\sqrt2 = 0.707V_m$.
• Sine form factor 1.11, peak factor 1.414; square wave both 1.
• Half-wave rectified sine: average $V_m/\pi$, RMS $V_m/2$.
• At 50 Hz: T = 20 ms, ω ≈ 314 rad/s; 230 V RMS ≈ 325 V peak.
• Coil EMF $e = NBA\omega\sin\omega t$; maximum when the coil plane is parallel to the field.

Formulas

$e = E_m\sin\omega t,\quad E_m = NBA\omega,\quad \omega = 2\pi f$

$V_{avg} = \frac{1}{\pi}\int_0^{\pi}V_m\sin\theta\,d\theta = \frac{2V_m}{\pi},\qquad V_{rms} = \sqrt{\frac{1}{T}\int_0^T v^2dt} = \frac{V_m}{\sqrt2}$

$k_f = \frac{V_{rms}}{V_{avg}},\qquad k_p = \frac{V_m}{V_{rms}}$

Worked example

An alternating voltage is $v = 325\sin(314t + 30^\circ)$ V. Find RMS value, average value, frequency and the instantaneous value at t = 0.

1. $V_{rms} = 325/\sqrt2 = 229.8$ V ≈ 230 V.
2. $V_{avg} = 2\times325/\pi = 206.9$ V.
3. $f = 314/(2\pi) = 49.97$ ≈ 50 Hz; T = 20 ms.
4. At t = 0: $v = 325\sin30^\circ = 162.5$ V.

Exam tips

• Form factor of a sine wave = 1.11; peak factor = 1.414 — the most asked values.
• Average value of a full sine cycle is zero; use the half-cycle average.
• RMS is defined by heating effect; AC meter scales are calibrated in RMS.
• For a square wave, peak = RMS = average, so both factors are 1.
• $\omega = 2\pi f$: at 50 Hz, ω ≈ 314 rad/s.

Exam-style questions

From Calmon's question bank, which has 24 questions on this topic.

1. For $v = 100\sin(100\pi t)$ V, the time after $t = 0$ at which the voltage first reaches 86.6 V is about:

• A. 1.67 ms
• B. 3.33 ms
• C. 6.67 ms
• D. 5 ms

Answer: B. $\sin(100\pi t) = 0.866 \Rightarrow 100\pi t = \pi/3 \Rightarrow t = 1/300$ s $\approx 3.33$ ms.

2. The rms value of an alternating current is equal to the steady direct current that:

• A. Produces the same magnetic flux at the peak instant
• B. Gives the same average value over a half cycle
• C. Produces the same heat in a given resistance in the same time
• D. Has the same peak value

Answer: C. The rms (effective) value is defined by equal heating effect: $I_{rms}^2RT$ equals the AC heat over the same time.

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