In an a.c. circuit,the instantaneous current | vedclass

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(A) The phase of the current is $\phi_I = \omega t - \pi / 6$. The phase of the voltage is $\phi_E = \omega t + \pi / 3$. The phase difference between

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$\pi / 2$

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(A) The phase of the current is $\phi_I = \omega t - \pi / 6$. The phase of the voltage is $\phi_E = \omega t + \pi / 3$. The phase difference between voltage and current is given by $\Delta \phi = \phi_E - \phi_I$. Substituting the values,we get $\Delta \phi = (\omega t + \pi / 3) - (\omega t - \pi / 6)$. $\Delta \phi = \pi / 3 + \pi / 6 = 2\pi / 6 + \pi / 6 = 3\pi / 6 = \pi / 2$. Since $\Delta \phi$ is positive,the voltage leads the current by $\pi / 2$.

In an $a.c.$ circuit,the instantaneous current and $emf$ are represented as $I = I_0 \sin(\omega t - \pi / 6)$ and $E = E_0 \sin(\omega t + \pi / 3)$ respectively. The voltage leads the current by

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