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In a given series $L C R$ circuit $R=4 \Omega, X_L=5 \Omega$ and $X_C=8 \Omega$, the current
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Verified Answer
The correct answer is:
Leads the voltage by $\tan ^{-1}(3 / 4)$.
The phase angle
$\theta=\tan ^{-1}\left(\frac{X_C-X_L}{R}\right)$
$=\tan ^{-1}\left(\frac{8-5}{4}\right)$
$\theta=\tan ^{-1}\left(\frac{3}{4}\right)$
Therefore, the current leads the voltage by $\tan ^{-1}\left(\frac{3}{4}\right)$
$\theta=\tan ^{-1}\left(\frac{X_C-X_L}{R}\right)$
$=\tan ^{-1}\left(\frac{8-5}{4}\right)$
$\theta=\tan ^{-1}\left(\frac{3}{4}\right)$
Therefore, the current leads the voltage by $\tan ^{-1}\left(\frac{3}{4}\right)$
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