Let us assume that the phase difference between current and voltage in an AC circuit is $\dfrac{\pi }{4}$ radian. The frequency of AC circuit has been given as $50Hz$, then the phase difference will be equivalent to the time difference of

Hint: The frequency has been defined as the inverse of the time period of a wave. Find out the time period of the wave like this. This time period will be around the one full complete cycle. So find out the time period up to the mentioned angle. This will help you in answering this question.

Complete answer:
It has been mentioned in the question that the frequency of the AC circuit is,
\[f=50Hz\]
The frequency has been defined as the reciprocal of the time period of the wave. Therefore we can write that,
\[T=\dfrac{1}{f}=0.02s\]
The full complete cycle is having a time period of \[0.02s\]. The full complete cycle will take a total of \[2\pi \]radians phase difference. We have to calculate the time period of the situation where the phase difference is given as \[\dfrac{\pi }{4}\]radians. This can be found as,
\[\begin{align}
  & 2\pi rad=0.02s \\
 & \Rightarrow \dfrac{\pi }{4}=0.0025s=2.5ms \\
\end{align}\]

Therefore the time difference which is equivalent to the phase difference of \[\dfrac{\pi }{4}\] has been determined as \[2.5ms\]. Therefore the answer has been calculated.

Note:
The time period has been defined as the time taken for a complete wave to pass through a medium. It is having the units in second. The frequency of a wave has been defined as the number of waves passing through the material medium in a time interval of one second. The unit has been mentioned as hertz. The phase difference between the current and the voltage is occurring because of the presence of a capacitor or inductor in the circuit.