Principle of the AC Generator
An AC generator converts mechanical energy into electrical energy by virtue of electromagnetic induction. The development of the modern machine is credited to Nikola Tesla, and a present-day commercial unit can have an output capacity of the order of 100 MW. As established in the discussion of magnetic flux, one way to induce an emf in a loop is to change the orientation of the loop — that is, to rotate it so that the angle between its area vector and the field changes with time.
When a coil of area $A$ is held in a uniform magnetic field $B$ and rotated about an axis perpendicular to the field, the effective area presented to the field is $A\cos\theta$, where $\theta$ is the angle between the area vector $\vec{A}$ and the field $\vec{B}$. As the coil turns, $\theta$ varies continuously, the flux linked with the coil changes, and Faraday's law guarantees an induced emf. Because the rotation is uniform, this emf turns out to be sinusoidal, and its direction reverses every half rotation — the hallmark of alternating current.
Construction and Parts
The basic elements of an AC generator are a coil, a field magnet, and a commutating arrangement of slip rings and brushes. The coil — called the armature — is mounted on a rotor shaft, and the axis of rotation is kept perpendicular to the direction of the magnetic field. The coil is mechanically rotated in the uniform field by some external means, so its flux linkage changes and an emf is induced. The two ends of the coil are connected to the external circuit through slip rings and brushes.
| Part | Description | Function |
|---|---|---|
| Armature coil | Coil of N turns, area A, on a rotor shaft | Site where the emf is induced as flux changes |
| Field magnet | Permanent magnet or electromagnet poles | Supplies the uniform magnetic field B |
| Rotor shaft | Axle carrying the coil, axis ⊥ to B | Driven by an external agency to rotate the coil at ω |
| Slip rings | Two rings rotating with the coil | Keep each coil end on its own contact |
| Brushes | Stationary carbon contacts on the rings | Carry the induced current to the load |
Because each end of the coil stays connected to its own slip ring, the alternating polarity developed across the coil is delivered unchanged to the load — that is precisely why the output is alternating rather than rectified. In commercial machines the mechanical energy needed to spin the armature is supplied by falling water (hydro-electric generators), by high-pressure steam from coal or other fuels (thermal generators), or from nuclear fuel (nuclear power generators).
Derivation of the Induced EMF
Let the coil rotate with constant angular speed $\omega$, and take the angle between $\vec{B}$ and the area vector $\vec{A}$ to be $\theta = 0$ at $t = 0$. At a later instant $t$ the angle is $\theta = \omega t$. From the definition of magnetic flux, the flux linked with the coil at time $t$ is
$$\Phi_B = BA\cos\theta = BA\cos\omega t$$
For a coil of $N$ turns each linking the same flux, Faraday's law gives the induced emf as the negative time rate of change of the total flux linkage:
$$\varepsilon = -N\frac{d\Phi_B}{dt} = -NBA\frac{d}{dt}(\cos\omega t) = NBA\,\omega\sin\omega t$$
The instantaneous emf is therefore $\varepsilon = NBA\omega\sin\omega t$. The largest value the emf can take occurs when $\sin\omega t = \pm 1$; denoting this peak by $\varepsilon_0$,
$$\varepsilon_0 = NBA\omega \qquad\Rightarrow\qquad \varepsilon = \varepsilon_0\sin\omega t$$
Since $\omega = 2\pi\nu$, where $\nu$ is the frequency of revolution of the coil, the emf can equivalently be written as $\varepsilon = \varepsilon_0\sin 2\pi\nu t$, with $\varepsilon_0 = NBA(2\pi\nu)$. The emf varies between $+\varepsilon_0$ and $-\varepsilon_0$ periodically; its polarity reverses with each half-cycle, so the current it drives is alternating.
Sinusoidal Output and Phase
The most NEET-relevant feature of the output is its phase. The flux varies as $\cos\omega t$, while the emf varies as $\sin\omega t$ — the two are a quarter cycle out of step. The emf attains its extremum when $\omega t = 90^\circ$ or $270^\circ$, because the change of flux is greatest at these instants; at the same instants the plane of the coil is parallel to the field and the flux through it is zero. Conversely, when $\omega t = 0^\circ$ or $180^\circ$ the flux is maximum but its rate of change is zero, so the emf is zero.
Maximum flux is not maximum emf
Students routinely assume the emf is largest when the flux through the coil is largest. The opposite is true. The emf depends on the rate of change of flux, not on the flux itself. When the coil plane is perpendicular to $\vec{B}$ the flux is maximum but momentarily unchanging, so $\varepsilon = 0$. When the coil plane is parallel to $\vec{B}$ the flux is zero, yet it is changing fastest, so $\varepsilon = \varepsilon_0$. Equally, do not confuse the peak emf $\varepsilon_0 = NBA\omega$ with the instantaneous emf $\varepsilon = \varepsilon_0\sin\omega t$.
Coil plane ⊥ B → flux max, emf zero. | Coil plane ∥ B → flux zero, emf peak ($\varepsilon_0$).
Factors Affecting the Peak EMF
Every quantity in $\varepsilon_0 = NBA\omega$ is a design lever. The peak emf rises in direct proportion to the number of turns $N$, the field strength $B$, the coil area $A$, and the angular speed $\omega$. Because $\omega = 2\pi\nu$, spinning the armature faster raises both the peak emf and the output frequency together.
| Quantity | Symbol | Effect on peak emf $\varepsilon_0$ |
|---|---|---|
| Number of turns | N | $\varepsilon_0 \propto N$ — more turns, larger emf |
| Magnetic field | B | $\varepsilon_0 \propto B$ — stronger field, larger emf |
| Coil area | A | $\varepsilon_0 \propto A$ — larger coil, larger emf |
| Angular speed | ω | $\varepsilon_0 \propto \omega = 2\pi\nu$ — faster spin, larger emf and frequency |
In most large generators the coils are held stationary and it is the electromagnets that are rotated. The frequency of rotation, and hence of the supply, is 50 Hz in India, while in some countries such as the USA it is 60 Hz.
Worked Examples
A 100-turn coil of area $0.10\ \text{m}^2$ rotates at half a revolution per second in a uniform field of $0.01\ \text{T}$ perpendicular to the axis of rotation. Find the maximum voltage generated.
Here $\nu = 0.5\ \text{Hz}$, $N = 100$, $A = 0.10\ \text{m}^2$, $B = 0.01\ \text{T}$. Using $\varepsilon_0 = NBA(2\pi\nu)$:
$\varepsilon_0 = 100 \times 0.01 \times 0.10 \times 2 \times 3.14 \times 0.5 = 0.314\ \text{V}$.
The maximum voltage generated is $0.314\ \text{V}$.
A circular coil of radius 10 cm, 500 turns and resistance $2\ \Omega$ is placed with its plane perpendicular to the horizontal component of the earth's field $(3.0\times 10^{-5}\ \text{T})$. It is rotated about its vertical diameter through $180^\circ$ in $0.25\ \text{s}$. Estimate the emf and current induced.
Initial flux $\Phi_B = BA\cos 0^\circ = 3.0\times10^{-5}\times(\pi\times10^{-2})\times 1 = 3\pi\times10^{-7}\ \text{Wb}$; final flux after $180^\circ$ rotation $= -3\pi\times10^{-7}\ \text{Wb}$, so the change is $6\pi\times10^{-7}\ \text{Wb}$.
$\varepsilon = N\,\dfrac{\Delta\Phi_B}{\Delta t} = 500\times\dfrac{6\pi\times10^{-7}}{0.25} = 3.8\times10^{-3}\ \text{V}$, and $I = \varepsilon/R = 1.9\times10^{-3}\ \text{A}$.
These are estimated (average) values; the instantaneous emf depends on the speed of rotation at that instant.
AC Generator in one screen
- An AC generator converts mechanical energy to electrical energy through electromagnetic induction in a rotating coil.
- Flux: $\Phi_B = BA\cos\omega t$; instantaneous emf: $\varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$.
- Peak emf $\varepsilon_0 = NBA\omega = NBA(2\pi\nu)$; it grows with $N$, $B$, $A$ and $\omega$.
- EMF is peak when the coil plane is parallel to $B$ (flux zero); emf is zero when the coil plane is perpendicular to $B$ (flux maximum).
- Slip rings keep each coil end on its own contact, and brushes pass the alternating output to the load. Indian mains frequency is 50 Hz.