Question

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

Answer 1

Step 1: Given

Phase difference, $\mathrm{\varphi }=\frac{\mathrm{\pi }}{4}$.

Frequency, $\mathrm{f}=50\mathrm{Hz}$

Step 2: To Find

Time difference.

Step 3: Calculate the time difference ($∆\mathrm{t}$)

$$\begin{gathered} ∆\mathrm{t}=\frac{\mathrm{\varphi }}{2\mathrm{\pi f}} \\ =\frac{\mathrm{\pi }}{4\left(2\mathrm{\pi }\times 50\right)} \\ =0.0025\mathrm{s} \end{gathered}$$

Hence, the time difference is $0.0025\mathrm{s}$.