A) 3.33 A
B) 33.3 A
C) 5 A
D) 10 A
Correct Answer: A
Solution :
From the formula \[I=\frac{V}{\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}}\] \[=\frac{220}{\sqrt{{{(20)}^{2}}+{{(2\pi \times 50\times 0.2)}^{2}}}}\]
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