(i) With the help of a labelled diagram, describe the principle and working of an AC generator. Hence obtain an expression for the instantaneous value of the emf generated. (ii) The coil of an AC generator consists of $100$ turns of wire, each of area $0.5\,\mathrm{m}^{2}$. The resistance of the wire is $100\,\Omega$. The coil rotates in a magnetic field of $0.8\,\mathrm{T}$ perpendicular to its axis of rotation at a constant angular speed of $60\,\mathrm{rad\,s^{-1}}$. Calculate the maximum emf generated and the power dissipated in the coil.

$\varepsilon = N B A \omega \sin\omega t$; $\varepsilon_{\max} = 2400\,\mathrm{V}$; power dissipated (mean) $= 28.8\,\mathrm{kW}$ (peak $57.6\,\mathrm{kW}$).

CBSE Class 12 · 2023 · Electromagnetic InductionLong Answer

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(i) With the help of a labelled diagram, describe the principle and working of an AC generator. Hence obtain an expression for the instantaneous value of the emf generated.
(ii) The coil of an AC generator consists of

100

turns of wire, each of area

0.5 m^{2}

. The resistance of the wire is

100 Ω

. The coil rotates in a magnetic field of

0.8 T

perpendicular to its axis of rotation at a constant angular speed of

60 rad s^{- 1}

. Calculate the maximum emf generated and the power dissipated in the coil.

Solution

(i) Principle: an AC generator works on electromagnetic induction — when a coil is rotated in a uniform magnetic field, the flux linked with it changes continuously and an emf is induced.
A rectangular coil of

N

turns and area

A

rotates at angular speed

ω

in field

B

. When its normal makes angle

θ = ω t

with

B

, the flux linkage is

N ϕ = N B A cos ω t

The induced emf is

ε = - \frac{d ( N ϕ )}{d t} = N B A ω sin ω t = ε_{0} sin ω t, ε_{0} = N B A ω

Slip rings and brushes carry this alternating emf to the external circuit.
(ii) Maximum emf:

ε_{0} = N B A ω = 100 \times 0.8 \times 0.5 \times 60 = 2400 V

ε_{0} = 2400 V

Mean power dissipated over a cycle:

P = \frac{ε _{rms}^{2}}{R} = \frac{( ε _{0} / 2 ) ^{2}}{R} = \frac{ε _{0}^{2}}{2 R} = \frac{( 2400 ) ^{2}}{2 \times 100} = 28800 W

P = 28.8 kW

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CBSE Class 12 · 2023 · Electromagnetic InductionLong Answer

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