Question

A coil of inductive reactance $\dfrac{1}{{\sqrt 3 }}\Omega$ and resistance $1\Omega$ is connected to $200V$, $50Hz$ supply A.C. supply. The time lag between maximum voltage and current is
A) $\dfrac{1}{{600}}\sec$
B) $\dfrac{1}{{200}}\sec$
C) $\dfrac{1}{{300}}\sec$
D) $\dfrac{1}{{500}}\sec$

Answer 1

From the given value of frequency, first calculate the angular frequency using its mathematical expression. And then find the phase difference between the current and the voltage. The phase difference must be the product of the angular frequency and the time difference.

Formula used:
\eqalign{ & \omega = 2\pi f \cr & \tan \phi = \dfrac{R}{L} \cr & \phi = \omega t \cr}

Complete solution:
Here, in the AC source along with the resistor an inductor has been connected.
In a Direct current, the inductance would be similar to a conductor, and thus the only opposition to the current is provided by the resistor. But here as we know the power source is alternating and thus the potential difference across the resistor and inductor system would be continuously varying with time. And when the emf changes continuously the current too changes with time.
Electric current produces a magnetic field. And here, the current is continuously changing thus, the magnetic field would change continuously too. This creates a change in magnetic flux around the coil. And this in turn creates an opposing current to the initial current. This whole process happening around a coil is called self-induction and the coil is called an inductor.
Now let us have a proper look at the question.
We were provided with the frequency of A.C source which is $50Hz$
And we can find the value of angular frequency from the value of frequency
i.e.
$\omega = 2\pi f = 2\pi \times 50 = 314rad/\sec$
The inductance connected in the circuit by the process of self-induction creates a lag in the current. i.e. unlike a circuit with only a resistor the voltage and current are not in the same phase. They will have a phase difference $\phi$.
Let us find the value of $\phi$
$\tan \phi = \dfrac{R}{L}$
So, $\tan \phi = \dfrac{1}{{\dfrac{1}{{\sqrt 3 }}}} = \sqrt 3$
$\Rightarrow \phi = \dfrac{\pi }{6}$
We can now express phase difference as a product of angular frequency and time
i.e. $\omega t = \dfrac{\pi }{6}$
substituting value of angular frequency, we obtain
$\therefore t = \dfrac{1}{{600}}\sec$
Hence, the time difference is $\dfrac{1}{{600}}\sec$

Thus, option A is the right answer.

Note:
Remember to convert frequency to angular frequency. A capacitor has the opposite effect on an AC circuit. Do not make mistakes while taking inverse to find phase angle, else might land in the wrong answer.