Electronics - RL Circuits - Discussion

Discussion Forum : RL Circuits - General Questions (Q.No. 7)
7.
What is the magnitude of the phase angle of a 24 Vac parallel RL circuit when R = 45 omega.gif and XL = 1100 omega.gif?
0.001°
2.3°
87.6°
89.9°
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Historia said:   5 years ago
The formula for phase angle in series and parallel is basically the same the only difference it the magnitude.

Arctan θ = -(XL/R) = -87.65739 which is also the same as +2.342612 it just depends on where you measure the angle from it's phase diagram. Since in the choices the given value of 87.6 is positive it would be the incorrect answer since it should be negative. That is why the correct answer is B.

Len said:   7 years ago
XL/R is for Series, here the question is parallel.

Atul said:   8 years ago
Answer is 87.6.

As the formula for phase angle is tan(θ)=XL/R.
θ=tan-1(1100/45)=87.6.
or we can use cos(θ)=R/Z=same answer.
Correct me if I am wrong.

KIRAN V said:   9 years ago
Phase angle:

Tan(θ) = XL/R.

Samir said:   9 years ago
Phase angle = image part/real part.
=XL/R.

Safi said:   1 decade ago
No need to convert it into current.

Phase angle = (inv)tan(R/XL) (in degrees).

Phase angle = (inv)tan(45/1100) (in degrees).

= 2.3°

@Ritu, Titos: conversion to current will lead to same answer. The current will again be divided, result=1. The only thing remaining will be resistance and impedance.

Search google for phase angle explanation and formula.

Titos said:   1 decade ago
Phase angle = -tan-1(IL/ IR).

Phase angle = -tan(.02181/.533) (in degrees).

= 2.3°

Ritu said:   1 decade ago
Here, Ir= 24/45= .533
Il= 24/1100= .02181

phase angle for parallel RL ckt= atan (Il/Ir)

So, ans= atan (.02181/.533) (in degrees)

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