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A $12 \mathrm{ohm}$ resistor and a 0.21 henry inductor are connected in series to an ac source operating at 20 volts, 50 cycle/second. The phase angle between the current and the source voltage is
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The correct answer is:
$80^{\circ}$
$\tan \phi=\frac{\omega L}{R}=\frac{2 \pi \times 50 \times 0.21}{12}=5.5 \Rightarrow \phi=80^{\circ}$
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