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  4. A 120 volt AC source is connected across a pure inductor of inductance 0.70 henry. If the frequency of the source is 60 Hz, the current passing through the inductor is

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Q. A $120$ volt AC source is connected across a pure inductor of inductance $0.70$ henry. If the frequency of the source is $60 \,Hz$, the current passing through the inductor is

Alternating Current

Solution:

$Z=X_{L}=2 \pi \times 60 \times 0.7$
$\therefore i=\frac{120}{Z}$
$=\frac{120}{2 \pi \times 60 \times 0.7}=0.455 \,A$.