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9 September, 14:03

An inductor with L = 9.80 mH is connected across an ac source that has voltage amplitude 41.5 V. What value for the frequency of the source results in a current amplitude of 4.50 A?

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  1. 9 September, 14:16
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    149.8Hz

    Explanation:

    V=I. (XL) ... {1}

    where (XL) is inductive reactance

    (XL) = 2πfL ... {2}

    V = 41.5, l = 4.5, L = 9.8 * 10^-³

    take π to be 3.14

    (XL) = V/I

    = 41.5/4.5

    = 9.22

    f = (XL) / 2πL

    = 9.22 * 10^³/6.28 * 9.8

    =149.8
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