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An AC voltage is applied to a resistance $R$ and an inductor $L$ in series. If $R$ and the inductive reactance are both equal to $3 \Omega$, the phase difference between the applied voltage and the current in the circuit is
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The correct answer is:
$\pi / 4$
$$
\begin{aligned}
\tan \phi=\frac{X_L}{R} & =\frac{L \omega}{R} \\
\tan \phi & =\frac{3 \Omega}{3 \Omega} \\
\tan \phi & =1 \\
\phi & =\tan ^{-1}(1) \\
\phi & =45^{\circ} \\
\phi & =\frac{\pi}{4} \mathrm{rad}
\end{aligned}
$$
\begin{aligned}
\tan \phi=\frac{X_L}{R} & =\frac{L \omega}{R} \\
\tan \phi & =\frac{3 \Omega}{3 \Omega} \\
\tan \phi & =1 \\
\phi & =\tan ^{-1}(1) \\
\phi & =45^{\circ} \\
\phi & =\frac{\pi}{4} \mathrm{rad}
\end{aligned}
$$
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