AC Generator – IIT‑JEE Study Notes
Date: 2026‑10‑15
Target: Students preparing for IIT‑JEE (Main + Advanced) – Physics
1. Simple Explanation
An AC generator (alternating‑current generator) is a device that converts mechanical energy into electrical energy by rotating a conductor (usually a coil) in a magnetic field.
According to Faraday’s law of electromagnetic induction the induced electromotive force (emf) in a coil is proportional to the rate of change of magnetic flux through the coil.
For a coil of N turns rotating with angular velocity ω in a uniform magnetic field B, the instantaneous emf is
[ \boxed{, e(t)=N,B,A,\omega ,\sin(\omega t) ,} ]
where A is the area of the coil.
The output voltage is sinusoidal with frequency
[ f = \frac{\omega}{2\pi} ]
This sinusoidal emf drives an alternating current through the external circuit.
2. Key Points
| Topic | Details |
|---|---|
| Faraday’s Law | ( e = -\dfrac{d\Phi}{dt}) (negative sign → Lenz’s law) |
| Magnetic Flux | ( \Phi = B\cdot A \cdot \cos\theta ) – for a rotating coil, (\theta = \omega t) |
| Induced emf in a rotating coil | ( e = NAB\omega \sin(\omega t) ) |
| Armature Reaction | Interaction of armature field with main field → flux distortion → affects voltage regulation |
| Regulation & Excitation | Excitation current through field winding sets B. Regulation is the change in terminal voltage with load. |
| Synchronous Machines | AC generator operates at synchronous speed ( N_s = \dfrac{120f}{P}) where P = number of poles |
| Power Output | ( P_{\text{out}} = V_{\text{t}} I_{\text{t}} \cos\phi) (real power) |
| Commutation | AC generators do not need commutators; instead they use slip rings to transfer current to external circuit. |
| Voltage Regulation | ( % \text{Regulation} = \dfrac{V_{\text{no load}} - V_{\text{full load}}}{V_{\text{full load}}}\times100% ) |
| Key Equations | – ( e = \omega \Phi_m ) – ( V_{\text{t}} = E - I_a (R_a + X_s) ) – ( f = \dfrac{N_sP}{120}) |
| Common JEE Focus | 1. Deriving emf expression for rotating coil. 2. Understanding voltage regulation & excitation. 3. Significance of synchronous speed & poles. 4. Power factor & efficiency. |
3. Worked Example
Problem (IIT‑JEE style)
A rectangular copper coil of dimensions ( l = 0.5;\text{m} ) (length) and ( w = 0.2;\text{m} ) (width) with ( N = 100 ) turns is rotated in a uniform magnetic field ( B = 0.3;\text{T} ) at a constant angular speed ( \omega = 150;\text{rad/s} ).
(a) Find the maximum induced emf.
(b) If the coil is connected to a 50‑Ω resistor, compute the average power delivered.
(c) Suppose the generator is part of a 4‑pole synchronous machine operating at 50 Hz. What is the synchronous speed in rpm?
Solution
(a) Maximum Induced emf
Area of coil: ( A = l \times w = 0.5 \times 0.2 = 0.10;\text{m}^2 )
Using ( e_{\max} = N B A \omega )
[ e_{\max} = 100 \times 0.3 \times 0.10 \times 150 = 450;\text{V} ]
So the peak emf is 450 V.
(b) Average Power Delivered
The instantaneous emf is ( e(t) = e_{\max}\sin(\omega t) ).
With a resistive load ( R = 50;\Omega ), the instantaneous current is
[ i(t) = \frac{e(t)}{R} = \frac{e_{\max}}{R}\sin(\omega t) ]
The instantaneous power is ( p(t) = e(t)i(t) = \frac{e_{\max}^2}{R}\sin^2(\omega t) ).
Average over a complete cycle:
[ P_{\text{avg}} = \frac{1}{T}\int_0^T p(t),dt = \frac{e_{\max}^2}{2R} ]
Plugging the numbers:
[ P_{\text{avg}} = \frac{(450)^2}{2 \times 50} = \frac{202{,}500}{100} = 2{,}025;\text{W} ]
So about 2.0 kW is delivered on average.
(c) Synchronous Speed
Number of poles ( P = 4 ).
Frequency ( f = 50;\text{Hz} ).
Using ( N_s = \dfrac{120f}{P} ):
[ N_s = \frac{120 \times 50}{4} = 1500;\text{rpm} ]
The synchronous speed is 1500 rpm.
4. Quick Revision Summary
- Faraday’s Law → ( e