Question

In a purely inductive circuit, the current:
A. is in phase with the voltage
B. is out of phase with the voltage
C. leads the voltage by $\dfrac {\pi}2$
D. lags behind the voltage by $\dfrac {\pi}2$

Answer 1

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

Formula used:

$L\dfrac{di}{dt}={{v}_{a}}.\sin \omega t$

Complete step-by-step solution:
The voltage may be assumed to be given by,

${{v}_{i}}={{v}_{a}}.\sin \omega t$

Here, $v_a$ is the amplitude of input voltage and $\omega t$ is called the phase angle of the voltage.

Now, the equation of the circuit can be given as,

$\begin{aligned} & L\dfrac{di}{dt}={{v}_{a}}.\sin \omega t \\\ & \Rightarrow i=-\dfrac{{{v}_{a}}}{\omega L}\cos \omega t+k=\dfrac{{{v}_{a}}}{\omega L}\sin \left( \omega t-\dfrac{\pi}{2} \right)+k \\\ \end{aligned}$

Where, L is the inductance of the inductor, i is the ac current and k is a constant and can be neglected while considering phase differences.

The new phase of the current of the circuit is $\omega t-\dfrac{\pi}{2}$. Hence, clearly the phase of the current is lagging behind the voltage by $\dfrac{\pi}{2}$. So option D is the correct answer.

Additional information:
In case of purely capacitive circuits, the current of the circuit 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: A student may choose to assume the source voltage as ${{v}_{a}}.\cos \omega t$. Again option D will be the correct answer. In higher studies, students take the input ac voltage as $v_a.e^{i\omega t}$. Because ${{e}^{i\omega t}}=\cos \omega t+i.\sin \omega t$. So, there’s a combination of sine and cosine components of the voltage. In all possible methods, the answer is the same.