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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Aicharya · Accepted Answer

Given $\phi_1=\frac{\pi}{6}$ and $\phi_2=\frac{\pi}{3}$ $	herefore \quad \Delta \phi=\frac{\pi}{3}-\left(\frac{-\pi}{6}ight)=\frac{\pi}{2}$

In an a.c. circuit the instantaneous current and emf are represented as $I=I_0, \sin [\omega t-\pi / 6]$ and…

Solution