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In an a.c. circuit the instantaneous current and emf are represented as $I=I_0, \sin [\omega t-\pi / 6]$ and $\mathrm{E}=\mathrm{E}_0 \sin [\omega \mathrm{t}+\pi / 3]$ respectively. The voltage leads the current by
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The correct answer is:
$\frac{\pi}{2}$
Given $\phi_1=\frac{\pi}{6}$ and $\phi_2=\frac{\pi}{3}$
$$
\therefore \quad \Delta \phi=\frac{\pi}{3}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{2}
$$
$$
\therefore \quad \Delta \phi=\frac{\pi}{3}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{2}
$$
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