Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 14)
14.
A 47 
resistor is in series with an inductive reactance of 120 across an ac source. The impedance, expressed in polar form, is
resistor is in series with an inductive reactance of 120 across an ac source. The impedance, expressed in polar form, is47
68.6°
68.6° 12068.6°
68.6° 12931.4°
31.4° 12968.6°
68.6° Discussion:
7 comments Page 1 of 1.
Raj said:
8 years ago
Z= R+jXL.
= (47^2+120^2)1^1/2.
= 128.87.
= tan^-1(XL/R)=68.8.
= (47^2+120^2)1^1/2.
= 128.87.
= tan^-1(XL/R)=68.8.
BUDDHADEB SINGHA said:
8 years ago
How to calculate?
47^2 + 120^2= 128.87.
47^2 + 120^2= 128.87.
B Sharath Kumar said:
9 years ago
Please, anyone explain how to calculate tan inverse of 120/40 without calculator?
Gnani said:
10 years ago
R = 47; XL = 120 then R+jXL = R^2+X^2.
= 47^2 + 120^2.
= 128.87.
tan^-1 (XL/R) = tan^-1 (120/47) = 68.8.
= 47^2 + 120^2.
= 128.87.
tan^-1 (XL/R) = tan^-1 (120/47) = 68.8.
D gomathi said:
10 years ago
I don't understand please explain.
Priya said:
1 decade ago
Z = R+jXL.
= 47+j120.
In polar form: = 128.87 angle 68.61.
= 47+j120.
In polar form: = 128.87 angle 68.61.
Arjun said:
1 decade ago
Magnitude=( r^2+ x^2)^12
tan(phase angle) = 120/47
So ans is magnitude*phase angle = Ans D.
tan(phase angle) = 120/47
So ans is magnitude*phase angle = Ans D.
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