Electronics - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 76)
76.
What is the phase angle of a 24 Vac parallel RL circuit when R = 45 ohms and XL = 1100 ohms?
Discussion:
6 comments Page 1 of 1.
Historia said:
5 years ago
Since, it is a parallel RL circuit the formula for getting the phase angle is:
Arctan(B/G).
Susceptance (B) = (1/1100).
Conductance (G) = (1/45),
Angle = arctan((1/1100)/(1/45)).
Angle = 2.342612006 degrees.
Arctan(B/G).
Susceptance (B) = (1/1100).
Conductance (G) = (1/45),
Angle = arctan((1/1100)/(1/45)).
Angle = 2.342612006 degrees.
Pritam said:
5 years ago
For parallel R || L.
z=product/sum=(45*1100j)/(45+1100j).
By simplifying using calculator in cmplx form we get;
44.92+1.83j.
angle=tan-1(1.83/44.92)=2.3°.
z=product/sum=(45*1100j)/(45+1100j).
By simplifying using calculator in cmplx form we get;
44.92+1.83j.
angle=tan-1(1.83/44.92)=2.3°.
Guest said:
1 decade ago
I=V/R=V/z=24/(45+j1100)=(24/45)+(24/j1100)
tan(angle)=(Il/Ir)
Where Il=V/Xl
And Ir=V/R
So by putting values we get
Angle=2.344
tan(angle)=(Il/Ir)
Where Il=V/Xl
And Ir=V/R
So by putting values we get
Angle=2.344
Attitude King said:
1 decade ago
z=R+jXl=45+j1100.
V=IR
I=V/R=V/z=24/(45+j1100)=(24/45)+(24/j1100)
I=0.53-j0.0218=-0.53<2.355
So,2.355 degree is the phase angle.
V=IR
I=V/R=V/z=24/(45+j1100)=(24/45)+(24/j1100)
I=0.53-j0.0218=-0.53<2.355
So,2.355 degree is the phase angle.
Titi Malaki said:
6 years ago
The problem says parallel RL circuit.
So, R||X_L=45<2.3°.
So, R||X_L=45<2.3°.
Raja said:
1 decade ago
45+1100i =1100 angle 87.600
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