Question

The phase angle between the current and voltage in pure inductive circuit is
A) zero
B) $\pi$
C) $\dfrac{\pi }{4}$
D) $\dfrac{\pi }{2}$

Answer 1

In a purely inductive circuit, there is no resistance or capacitance The circuit only contains an inductor and an AC voltage source. We need to know the equation of the circuit first after solving this we can find the required phase difference.

Step by step solution
Let the input voltage is given by
${v_i} = {v_a}\sin \omega t$
Here, va is the amplitude and ωt is the phase angle of the input voltage.
Now, the equation of the circuit given as
⇒ $L\dfrac{{di}}{{dt}} = {v_a}\sin \omega t$
On integrating the above equation, we get
⇒ $i = - \dfrac{{{v_a}}}{{\omega L}}\cos \omega t + k$
⇒ ${v_i} = \dfrac{{{v_a}}}{{\omega L}}\sin \left( {\omega t - \dfrac{\pi }{2}} \right) + k$
Where, L is the inductance of the inductor and I is the AC current and k is constant which we can neglect while considering the phase differences.
Now, from the above equation the new phase is $\left( {\omega t - \dfrac{\pi }{2}} \right)$.
Hence, clearly the phase of the current lagging behind the voltage by $\dfrac{\pi }{2}$
Therefore, option D is correct.

Additional information
In case of purely capacitive circuit current leads the voltage by $\dfrac{\pi }{2}$And in case of purely resistive circuits there is no phase change i.e. both the voltage and current have the same phase angle.

Note
One must be noticed that we take the input source voltage in terms of sine wave. But here we can take the source voltage in terms of cosine wave i.e. ${v}_{i} = {v}_{a} \cos\omega t$.