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Question

In an AC generator, what happens to the induced emf when the rotational speed of the coil doubles?

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Explanation

**Back emf** in motor opposes applied voltage, e_b = N B A ω sin ωt, reduces net current, at start ω=0 e_b=0 current large, as speed increases e_b increases limiting current, power conversion mechanical, principle of motor and generator reciprocity. The emf is proportional to the angular speed ( ε = N B A ω ), so doubling the rotational speed doubles the emf. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L I², result Doubles follows, reflecting Faraday's law and Lenz's opposition.

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