The phase difference between current and voltage in an AC circuit is $\frac{\pi }{4}$ radian. If the frequency of AC is 50Hz, then the phase difference is equivalent to the time difference of

We must first determine the time period for one revolution before determining the time period for $\frac{\mathrm{\pi }}{4}$ revolutions in order to solve this problem. Your issue will be resolved by doing this.

Formula used - Time period $=\frac{1}{\text{ frequency }}$
The circuit's frequency is specified as $50 \mathrm{Hz}$.
Consequently, the wave's period is $\frac{1}{f}=\frac{1}{50}=0.02 \mathrm{s}$
As we are aware, a revolution occurs in one time period $2\pi$.
Then $\frac{\pi }{4}$ has the time period of $\frac{0.02}{2\pi }\mathrm{x}\frac{\pi }{4}=0.0025 \mathrm{s}=2.5 \mathrm{ms}$.

Hence, the correct answer is option C.