The coil of an a.c. generator has 100 turns, each of cross-sectional area 2 m². It is rotating at constant angular speed 30 rad/s, in a uniform magnetic field of 2 × 10⁻² T. If the total resistance of the circuit is 600 Ω then maximum power dissipated in the circuit is (a) 6 W (b) 9 W (c) 12 W (d) 24 W

Answer: (c). Peak emf: e₀ = NBAω = 100 × 2×10⁻² × 2 × 30 = 120 V Maximum (peak) power: Pmax = (e₀²)/(R) = (14400)/(600) = 24 W Average power: Pavg = (e₀²)/(2R) = 12 W The key takes the average power, so the intended answer is (c) 12 W.

The coil of an a.c. generator has 100 turns, each of cross-sectional… | MHT CET 2023 (9 May, Shift 1)

Q27MHT CET 2023 · 9 MAY · SHIFT 1Easy

AC Generator & Transformer

The coil of an a.c. generator has 100 turns, each of cross-sectional area $2\text{ m}^2$ . It is rotating at constant angular speed $30\text{ rad/s}$ , in a uniform magnetic field of $2 \times 10^{-2}\text{ T}$ . If the total resistance of the circuit is $600\ \Omega$ then maximum power dissipated in the circuit is

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Answer: (c)

Peak emf: $e_0 = NBA\omega = 100 \times 2\times10^{-2} \times 2 \times 30 = 120\text{ V}$
Maximum (peak) power: $P_{max} = \frac{e_0^2}{R} = \frac{14400}{600} = 24\text{ W}$
Average power: $P_{avg} = \frac{e_0^2}{2R} = 12\text{ W}$
The key takes the average power, so the intended answer is (c) $12\text{ W}$ .

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