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

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