Electrical Engineering - RL Circuits - Discussion

Discussion :: 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

[A]. 4768.6°
[B]. 12068.6°
[C]. 12931.4°
[D]. 12968.6°

Answer: Option D

Explanation:

No answer description available for this question.

Arjun said: (Jan 4, 2012)  
Magnitude=( r^2+ x^2)^12
tan(phase angle) = 120/47

So ans is magnitude*phase angle = Ans D.

Priya said: (Mar 11, 2015)  
Z = R+jXL.

= 47+j120.

In polar form: = 128.87 angle 68.61.

D Gomathi said: (Dec 24, 2015)  
I don't understand please explain.

Gnani said: (Feb 5, 2016)  
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.

B Sharath Kumar said: (Jan 31, 2017)  
Please, anyone explain how to calculate tan inverse of 120/40 without calculator?

Buddhadeb Singha said: (Dec 29, 2017)  
How to calculate?

47^2 + 120^2= 128.87.

Raj said: (Jan 30, 2018)  
Z= R+jXL.
= (47^2+120^2)1^1/2.
= 128.87.
= tan^-1(XL/R)=68.8.

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