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

Question

The phase difference between current and voltage in an AC circuit is π /4 radian. If the frequency of AC is 50Hz , then the phase difference is equivalent to the time difference: (A) 0.78ms (B) 15.7ms (C) 0.25ms (D) 2.5ms

Tardigrade Admin · Accepted Answer

AC frequency f=50Hz Phase difference Δ φ=(π /4) The relation between phase difference and time difference is given by (Δ φ/2 π )=(Δ t/T) ⇒ Δ t=(Δ φ/f (2 π ))=(π / 4/50 (2 π )) ⇒ Δ t=(1/400)=2.5ms

Q. The phase difference between current and voltage in an $A C$ circuit is $π /4$ radian. If the frequency of $A C$ is $50 Hz$ , then the phase difference is equivalent to the time difference:

$0.78 m s$

$15.7 m s$

$0.25 m s$

$2.5 m s$

AC frequency $f = 50 Hz$
Phase difference $Δ ϕ = \frac{π}{4}$
The relation between phase difference and time difference is given by
$\frac{Δ ϕ}{2 π} = \frac{Δ t}{T}$
$\Rightarrow Δ t = \frac{Δ ϕ}{f ( 2 π )} = \frac{π /4}{50 ( 2 π )}$
$\Rightarrow Δ t = \frac{1}{400} = 2.5 m s$

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

0.78ms

15.7ms

0.25ms

2.5ms