Question

The phase difference between current and voltage in an AC circuit is π/4 radian. If the frequency of AC is 50 Hz, then the phase difference is equivalent to the time difference of how many seconds?

Answer 1

Answer. Phase difference ϕ = π/4 radianFrequency f = 50 Hz Time difference Δt = ϕ/(2πf) = π/(4 x 2π x 50) = 1/(400) = 0.0025 s = 2.5 ms

Answer 2

Answer: The phase difference is equivalent to the time difference of 2.5 ms.

Explanation:

Given that,

Phase difference $\Delta\phi = \dfrac{\pi}{4}$

Frequency $f = 50 hz$

We know that,

The time period $T = \dfrac{1}{f}$

$T = \dfrac{1}{50}sec$

$T = 0.02 s$

We know,

The time period is 0.02s in $2\pi$ radian.

So, the time period will be in $\dfrac{\pi}{4}$ radian

$T= \dfrac{0.02}{2\pi}\times\dfrac{\pi}{4}$

$T = 0.0025 s$

$T = 2.5 ms$

Hence, the phase difference is equivalent to the time difference of 2.5 ms.