Q.If the instantaneous emf and the instantaneous current equations of an A.C. circuit are respectively $e = E_0 \sin(\omega t - \pi/6)$, $i = I_0 \sin(\omega t + \pi/6)$, then the phase difference between voltage and current is :

Definition. A sinusoidal voltage or current can be represented by a phasor: a vector of fixed length, rotating anticlockwise about the origin at constant angular velocity $\omega$ equal to the quantity's own angular frequency. A phasor's length equals the peak value ($V_m$ or $I_m$); its angular velocity equals $\omega$; and its projection onto a fixed vertical axis at any instant equals the INSTANTANEOUS value of the quantity at that instant. As the phasor rotates, one full rotation traces out exactly one complete cycle of the corresponding waveform.

Phasor diagrams and phase angle. A phasor diagram draws several such phasors together to show the phase relations between them. For $v=V_m\sin\omega t$ (phasor $\vec{OA}$, length $V_m$, angle $\omega t$ from the reference axis) and a current $i=I_m\sin(\omega t+\phi)$ in the same circuit (phasor $\vec{OB}$), the angle $\phi$ between $\vec{OA}$ and $\vec{OB}$, measured in the rotation direction, is the phase angle. If $\vec{OB}$ sits AHEAD of $\vec{OA}$ in the rotation sense, current LEADS voltage by $\phi$; if current instead lags, $i=I_m\sin(\omega t-\phi)$. …