In an LR circuit, R $=$ 10 $Ω$ and L $=$ 2H. If an alternating voltage of 120V and 60Hz is connected in this circuit, then the value of current flowing in it will be _______ A (nearly)

0.32

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0.16

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0.48

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0.8

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Solution

The correct option is D 0.16
Impedance(z) = $\sqrt{R^{2} + X_{L}^{2}}$

$= \sqrt{(10)^{2} + (2 π f L)^{2}}$

$= \sqrt{(10)^{2} + (2 π \times 60 \times 2)^{2}}$
$= \sqrt{100 + π^{2} \times 57600}$
$= \sqrt{100 + 568480}$
$= 754.04$
$i = \frac{V}{z}$
$= \frac{120}{754.04}$
$= 0.16 A$

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