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An inductance of 0.2H and resistance $100 Ω$ are connected to an AC 180V – 50Hz. The current in the circuit is

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0.52 A

15.2 A

5.2 A

1.52 A

answer is C.

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Detailed Solution

$ω = 2 π n = 2 π \times 50 = 314 r a d / \sec$ V rms =180 volt Resistance =R=100 ohm, Inductance=L=0.2 H

$i = \frac{V}{\sqrt{R^{2} + {(L ω)}^{2}}}$ =. $\frac{180}{\sqrt{100^{2} + {(0.2 x 314)}^{2}}} = 1.52 A$

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