Generation of Alternating Voltage/Current

Basic Electrical & Electronics Engineering — Alternating Current Fundamentals, NEC licence examination syllabus (Nepal Engineering Council).

Generation of Alternating Voltage: where the sine wave comes from

Nobody chose the sine shape. It falls out of the geometry of a spinning coil.

Start here: a shadow on the wall

Stick a bright dot of paint on the edge of a bicycle wheel, and spin it slowly in front of a lamp so it casts a shadow on the wall.

The dot goes round in a circle. But its shadow only moves up and down — and it does not move at a steady speed. It is fastest passing the middle, and it slows to a stop at the top and bottom before coming back.

Plot that shadow's height against time and you get a sine wave. Not because anyone designed it that way — it is simply what a circular motion looks like from the side.

Why that gives you electricity

A generator is a coil of wire spinning between two magnets. And the rule — Faraday's law — is that voltage appears when the coil cuts through magnetic field lines, with more voltage the faster it cuts.

Now the same geometry applies. As the coil spins, the rate at which it cuts field lines rises and falls exactly like the shadow's speed:

🔄 One full turn

Coil flat to the fieldIt is slicing across the field lines as fast as possible → maximum voltage.

Coil edge-on to the fieldIt is sliding along the lines rather than cutting them → zero voltage.

Half a turn laterIt cuts the lines the other way → voltage reverses.

💡 That third row is why it is called alternating current. Nothing switches it — the coil simply comes round to the other side, and cutting the field in the opposite direction produces the opposite voltage. Alternation is a consequence of going round in a circle.

The counter-intuitive bit

Most people guess the voltage is largest when the coil is nearest the magnets. It is not.

Voltage depends on how fast the field through the coil is changing, not on how strong it is. When the coil faces the magnet squarely, the field through it is at its maximum — and for that instant it is not changing at all, so the voltage is zero.

Think of the swing again: at the top of its arc it is highest, and momentarily not moving. Height and speed peak at different moments. Field strength and voltage do the same.

Words for parts of the wave

📏 The vocabulary

CycleOne complete up-and-down — one full turn of the coil.

FrequencyCycles per second, in hertz. Nepal uses 50 Hz, so the coil turns 50 times a second and the voltage reverses 100 times a second.

PeriodHow long one cycle takes — just 1/frequency. At 50 Hz, one cycle takes 0.02 s.

PeakThe highest the wave reaches.

Here's a question worth sitting with: why is mains electricity a sine wave, and not a square or triangular one? It would be easy to assume engineers picked it for convenience. They didn't. Spin a loop of wire in a uniform magnetic field, and the sine wave appears on its own. Why a sine? Because the voltage depends on how much field the loop cuts through as it rotates, and that amount varies sinusoidally. Understanding that derivation makes every later AC formula feel inevitable rather than memorised.

Faraday's law is the engine

e = −N dΦ/dt e = induced EMF (volts) N = number of turns in the coil Φ = magnetic flux linking the coil (webers) The minus sign is Lenz's law: the induced EMF opposes the change that created it. For magnitude work we drop it.

The key phrase is rate of change. A coil sitting still in a field, however strong, induces nothing. Only changing flux produces EMF — which is why the coil must move.

Why rotation gives a sine

Take a rectangular coil of area A, with N turns, spinning at angular velocity ω inside a uniform field B. At any instant the coil's plane makes angle θ = ωt with its starting position. The flux actually threading the coil is not BA — it's the component perpendicular to the coil:

Φ = B A cos(ωt) ← flux linking the coil e = −N dΦ/dt = −N d/dt [B A cos(ωt)] = −N B A × (−ω sin(ωt)) = N B A ω sin(ωt) So: e = E_m sin(ωt) where E_m = N B A ω

The sine appears because differentiating a cosine gives a sine. The flux varies as cos, so the EMF — its rate of change — varies as sin. That's the whole reason mains is sinusoidal: it is the derivative of a rotating geometry, nothing more.

Where the EMF is largest — the counter-intuitive bit

Most students guess the EMF peaks when the coil is perpendicular to the field, because that's when flux is maximum. It's the opposite. EMF depends on the rate of change of flux, not its value.

🔄 The two extremes, and why

θ = 0°Coil plane parallel to the field lines. Flux linkage is maximum (BA), but momentarily it isn't changing — the curve is at its flat top. So e = 0.

θ = 90°Coil plane perpendicular to the field. Flux linkage is zero, but it is sweeping through zero at the steepest possible rate. So e is maximum.

The lessonMaximum flux ⇒ zero EMF. Zero flux ⇒ maximum EMF. The EMF wave leads the flux wave by exactly 90°.

The quantities that describe the wave

e = E_m sin(ωt ± φ) E_m = peak (maximum) value ω = angular frequency = 2πf (rad/s) f = frequency = 1/T (Hz) T = time period (s) φ = phase angle (shift from the reference) For a machine with P poles running at N rpm: f = P·N / 120 (Hz)

Worked numerical 1 — from coil geometry to EMF

A 200-turn coil of area 0.05 m² rotates at 1500 rpm in a uniform field of 0.8 T. Find the peak EMF, the RMS EMF, the frequency, and the instantaneous EMF at t = 4 ms.

Angular velocity: N = 1500 rpm = 1500/60 = 25 rev/s ω = 2πN = 2π × 25 = 157.08 rad/s f = 25 Hz, T = 1/25 = 0.04 s = 40 ms Peak EMF: E_m = N B A ω = 200 × 0.8 × 0.05 × 157.08 = 1256.6 V RMS value: E_rms = E_m/√2 = 1256.6/1.414 = 888.6 V Instantaneous value at t = 4 ms: ωt = 157.08 × 0.004 = 0.6283 rad = 36° e = 1256.6 × sin(36°) = 1256.6 × 0.5878 = 738.7 V Sanity check: 4 ms out of a 40 ms period is one-tenth of a cycle = 36°. ✔ Consistent.

Worked numerical 2 — reading a waveform equation

An alternating voltage is given by v = 325 sin(314t). Find the peak value, RMS value, frequency, period, and the time at which v first reaches 230 V.

Comparing with v = V_m sin(ωt): V_m = 325 V ω = 314 rad/s f = ω/2π = 314/(2π) = 49.97 ≈ 50 Hz T = 1/50 = 0.02 s = 20 ms V_rms = 325/√2 = 229.8 ≈ 230 V ← this is standard Nepali/Indian mains: 230 V RMS, which peaks at 325 V Time to first reach 230 V: 230 = 325 sin(314t) sin(314t) = 0.7077 314t = 0.7857 rad t = 0.7857/314 = 2.502 ms Note 0.7077 ≈ 1/√2, so this happens at 45° — exactly one-eighth of the way through the cycle. ✔

Worked numerical 3 — a generator's pole count

An alternator must produce 50 Hz while running at 375 rpm. How many poles does it need?

f = P·N/120 50 = P × 375/120 P = 50 × 120/375 = 16 poles Cross-check: 16 poles at 375 rpm f = 16 × 375/120 = 50 Hz ✔ Slower machines need more poles. This is why large hydro alternators (low rpm, like Nepal's plants) are physically huge with many poles, while high-speed turbo-alternators use just 2 or 4.

💡 Exam angle: the derivation e = NBAω·sin(ωt) is a standard 5-mark question — start from Faraday's law, write Φ = BA·cos(ωt), differentiate, and state E_m = NBAω. Add the physical insight that EMF is maximum when flux is zero; examiners specifically look for whether you understand that, because it separates memorisation from comprehension. Also memorise f = PN/120 — it appears in one-line questions.

Syllabus points

Principle of generation
Equations and waveforms

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