Current Angle More Than Ninety Degrees

AI Thread Summary
The discussion revolves around calculating the power factor for a sync generator with a current of 1.59 at an angle of -159.9 degrees. The user is confused about the phase angle exceeding 90 degrees and how it relates to capacitive versus inductive loads. The book states that the current leads the terminal voltage by 23.13 degrees, which the user interprets as derived from 180 - 159.9. There is a clarification that in inductive circuits, the voltage leads the current, resulting in a positive phase angle. The conversation suggests that the question may be more suited for a homework forum for better assistance.
jaus tail
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Hi,
I'm studying Sync Generators. In one problem I got Ia as 1.59 angle of -159.9 degrees.

How do you calculate power factor for such a current?
Book says Ia leads Terminal Voltage by 23.13 degrees.
I guess they got that value by 180 - 159.9
But why?
 
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Capacitive load versus Inductive load.
 
But in capacitive and inductive the phase angle is from 0 to 90 degrees.
I'm not able to understand what does angle more than 90 mean.

Mathematically i used 159 degrees by
I angle 159
= I cos 159 + j I sin 159
and that equals I angle -23
 
I fail to see a reason for 159 degrees. The phase angle has already been given:
jaus tail said:
Book says Ia leads Terminal Voltage by 23.13 degrees.
Pay attention to the book quote above as to which is leading, Voltage or Current, and to the fourth and fifth sentences in this tutorial. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/phase.html

It is customary to use the angle by which the voltage leads the current. This leads to a positive phase for inductive circuits since current lags the voltage in an inductive circuit.[i/]
 
I got the phase angle as 159 degrees. In sync alternator I = (E - V)/X
E is 2.341 pu angle + 90
V = 1 angle 0
X = j1.6
 
Sorry we are not making progress. Your problem can likely be better answered in the Homework forum. Be sure to fill in the information in the template there so people here get a better idea of the question details.

Paging @russ_watters to assist in this redirection.
 
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