An A.C. source is connected across a pure ind | vedclass

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(B) In a pure inductor,the current lags behind the applied $e.m.f.$ (voltage) by a phase angle of $90^{\circ}$ or $\pi/2$ radians.If the voltage is re

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$B$

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(B) In a pure inductor,the current lags behind the applied $e.m.f.$ (voltage) by a phase angle of $90^{\circ}$ or $\pi/2$ radians.If the voltage is represented as $e_L = E_0 \sin(\omega t)$,then the current is $i_L = I_0 \sin(\omega t - \pi/2)$.In phasor diagrams,this means the current vector $i_L$ is $90^{\circ}$ clockwise from the voltage vector $e_L$.Looking at the given options:Figure $(A)$ shows them in opposite directions $(180^{\circ})$.Figure $(B)$ shows $i_L$ lagging $e_L$ by $90^{\circ}$ (clockwise).Figure $(C)$ shows them in phase $(0^{\circ})$.Figure $(D)$ shows $i_L$ leading $e_L$ by $90^{\circ}$ (counter-clockwise).Therefore,figure $(B)$ correctly represents the phase relationship.

An $A.C.$ source is connected across a pure inductor. Which one of the following figures shows the correct phase relation between the current and $e.m.f.$?

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