The maximum current if this inductor is connected to a 50.0 hertz source that produces a 105 v rms voltage is 3.87A
X= ωl
where,
X is inductive reactance
ω is thee angular frequency
L is the inductance
therefore,
46= 2π60.L eq-1
and
X° is inductive reactance when 50 hz source is used
X°=2π50.L eq-2
dividing eq 2 by eq 1 we get
X°=38.33 Ω
maximum current (I°) = maximum voltage divided by inductive reactance
I°=√2Vrms/38.33
I°=3.87A
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