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?

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