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.
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.
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.
D gomathi said:
10 years ago
I don't understand please explain.
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.
B Sharath Kumar said:
9 years ago
Please, anyone explain how to calculate tan inverse of 120/40 without calculator?
BUDDHADEB SINGHA said:
8 years ago
How to calculate?
47^2 + 120^2= 128.87.
47^2 + 120^2= 128.87.
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.
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