(A) The phase of the current is $\phi_i = -\frac{\pi}{6}$.The phase of the voltage is $\phi_E = +\frac{\pi}{3}$.The phase difference $\phi$ by which t

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(A) The phase of the current is $\phi_i = -\frac{\pi}{6}$.The phase of the voltage is $\phi_E = +\frac{\pi}{3}$.The phase difference $\phi$ by which the voltage leads the current is given by $\phi = \phi_E - \phi_i$.Substituting the values,we get $\phi = \frac{\pi}{3} - (-\frac{\pi}{6})$.$\phi = \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi + \pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.Therefore,the voltage leads the current by $\frac{\pi}{2}$.

Class 12 Physics Alternating Current Alternating Current, Voltage (rms and Average)

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In an $A.C.$ circuit,the instantaneous current and $E.M.F.$ are represented as $i = i_0 \sin(\omega t - \frac{\pi}{6})$ and $E = E_0 \sin(\omega t + \frac{\pi}{3})$ respectively. The voltage leads the current by

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A
$\frac{\pi}{2}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{6}$

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Alternating Current, Voltage (rms and Average)

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In an $A.C.$ circuit,the instantaneous current and $E.M.F.$ are represented as $i = i_0 \sin(\omega t - \frac{\pi}{6})$ and $E = E_0 \sin(\omega t + \frac{\pi}{3})$ respectively. The voltage leads the current by

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