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A $220 \mathrm{~V}, 50 \mathrm{~Hz}$ ac source is connected to an inductance of $0.2 \mathrm{H}$ and a resistance of $20 \mathrm{ohm}$ in series. What is the current in the circuit
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The correct answer is:
$3.33 \mathrm{~A}$
$i=\frac{220}{\sqrt{(20)^2+(2 \times \pi \times 50 \times 0.2)^2}}=\frac{220}{66}=3.33 \mathrm{~A}$
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