A 120, volt ac source is connected across a pure inductor of inductance 0.70 ,henry. If frequency of

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7.Alternating Current

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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

A

$4.55 \,amps$

B

$0.355 \,amps$

C

$0.455\, amps$

D

$3.55 \,amps$

Solution

(c) $Z = {X_L} = 2\pi \times 60 \times 0.7$
$\therefore \,\,i = \frac{{120}}{Z} = \frac{{120}}{{2\pi \times 60 \times 0.7}} = 0.455\,\,ampere$

Standard 12

Physics

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