Question

Explain briefly, with the help of a labelled diagram, the basic principle of the working of an a.c. generator. In an a.c. generator, a coil of N turns and area A is rotated at an angular velocity ω in a uniform magnetic field B. Derive an expression for the instantaneous value of the emf induced in coil. What is the source of energy generation in this device?

Answer 1

AC Generator: It is a device used to convert mechanical energy into electrical energy.

Principle: It is based on the principle of electromagnetic induction. When a closed coil is rotated rapidly in a strong magnetic field, the magnetic flux linked with the coil changes continuously. Hence an emf is induced in the coil and a current flows in it. In fact, the mechanical energy expended in rotating the coil appears as electrical energy in the coil.

Construction: 

1. Armature: It is a rectangular coil ABCD having a large number of turns of insulated copper wire wound on a soft-iron core. The use of soft-iron core increases the magnetic flux linked with the armature.
2. Field Magnet: It is a strong electromagnet having concave pole pieces N and S. The armature is rotated between these pole pieces about an axis perpendicular to the magnetic field.
3. Slip Rings: The leads from the armature coil ABCD are connected to two copper rings R₁ and R₂ called the ‘slip rings’. These rings are concentric with the axis of the armature coil and rotate with it.
4. Brushes: These are two carbon pieces B₁ and B₂ called brushes, which remain stationary pressing against the slip rings R₁ and R₂, respectively. The brushes are connected to an external circuit.

Working Theory: When the coil ABCD is rotated inside the field, an emf is induced between its two ends. Let the plane of the coil be at right angles to the magnetic field at t = 0 and the angular speed of the rotation of the coil be ω. Then at time t, θ = ωt. The magnetic flux linked with the coil at time t is

Φ = n BA cosωt

Induced emf e = `(-dphi)/(dt) = (-d)/(dt)[nBAcosωt]`

⇒ e = n BA ω sinωt

e = e₀sinωt   

Where e₀ = nBAω is the peak value of emf.

The current in the external load is given by

i = `(e_0sinomegat)/(R_L)`

i = i₀sinωt, here i₀ is the peak value of the current.

In an ac generator, the source of electrical energy is the mechanical energy.