Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.

Question

Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.

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Euphemia 4 years 2021-08-04T22:45:35+00:00 1 Answers 49 views 0

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  1. vankhanh
    0
    2021-08-04T22:46:38+00:00
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    Answer:

    The peak voltage is 473.86 V

    Explanation:

    Given;

    number of turns of generator, N = 200  turn

    diameter of the coil, d = 0.1 m

    radius of the coil, r = 0.05 m

    Magnitude of the magnetic field, B = 0.8 T

    angular frequency, ω = 3600 rpm

    angular frequency, ω (rad/s) = \frac{2\pi}{60} *3600\ rpm = 377.04 \ rad/s

    The peak voltage is given by;

    E = NBAω

    Where;

    A is the area of the coil = πr² =  π (0.05)² = 7.855 x 10⁻³ m²

    E = (200)(0.8)(7.855 x 10⁻³)(377.04)

    E = 473.86 V

    Therefore, the peak voltage is 473.86 V

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