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$220 \mathrm{~V}, 50 \mathrm{~Hz}, \mathrm{AC}$ source is connected to an inductance of 0.2 H and a resistance of $20 \Omega$ in series. What is the current in the circuit?
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The correct answer is:
3.33 A
From the formula
$\begin{aligned} I & =\frac{V}{\sqrt{R^2+(\omega L)^2}} \\ I & =\frac{220}{\sqrt{(20)^2+(2 \pi \times 50 \times 0.2)^2}} \\ & =\frac{220}{66}=3.33 \mathrm{~A}\end{aligned}$
$\begin{aligned} I & =\frac{V}{\sqrt{R^2+(\omega L)^2}} \\ I & =\frac{220}{\sqrt{(20)^2+(2 \pi \times 50 \times 0.2)^2}} \\ & =\frac{220}{66}=3.33 \mathrm{~A}\end{aligned}$
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