The phase difference between current and voltage in an AC circuit is $\dfrac{\pi }{4}$ radian. If the frequency of AC is 50Hz, then the phase difference is equivalent to the time difference of
A. 0.75s
B. 10.5s
C. 2.5ms
D. 0.25ms

Hint – In order to solve this problem we need to find the time period of one revolution and then find the time period of $\dfrac{\pi }{4}$ revolution. Doing this will solve your problem.
Formula used - ${\text{Time}}\,{\text{period = }}\dfrac{{\text{1}}}{{{\text{frequency}}}}$.

Complete Step-by-Step solution:
Frequency of the circuit is given as 50Hz.
Therefore the time period of the wave is $\dfrac{1}{f} = \dfrac{1}{{50}} = 0.02s$
As we know that in one time period the revolution is $2\pi $.
Then $\dfrac{\pi }{4}$ has the time period of $\dfrac{{0.02}}{{2\pi }}{\text{x}}\dfrac{\pi }{4} = 0.0025s = 2.5ms$.
Hence, the correct option is C.

Note – To solve such problems we need to know that in AC circuit the equations are sinusoidal and the frequency is inverse of time period and when there is the difference between the phases of current and voltage then the circuit is not purely resistive and when there is no phase difference then the circuit is purely resistive. To solve this problem you only need to know that the inverse of frequency is the time period and vice-versa.