Electronics - RL Circuits - Discussion

2. 

The phase angle in the circuit in the given circuit is approximately equal to ______.

[A]. 53°
[B]. 90°
[C]. 37°
[D].

Answer: Option C

Explanation:

No answer description available for this question.

Vishnu said: (Aug 22, 2011)  
Phase angle
= cos inverse((1k)^2/((1k^2)+(2*3.142*5k*0.024)^2)^0.5
= 37.015

Manmeeth said: (Apr 9, 2012)  
Phase angle = tan(wL/R)

Monica said: (Apr 25, 2012)  
XL = 2*(3.14)*f*L
=2*3.14*5*10^3*24*10^-3
=240(3.14)
Total phase angle= tan inverse(XL/ R)
= tan inverse(240*(3.14)/10^3)
=37.015

Srijith Raju said: (Jun 9, 2012)  
tan (q) = XL/R
q = Phase Angle (deg)
XL = Inductance Reactance(ohm)
R = Resistance (ohm)

Shahid said: (Sep 26, 2012)  
Phase angle=tan^-1(wL/R)
=tan^-1(2*3.14*24*10^-3/1*10^-3)
=37

Dharani said: (Feb 22, 2014)  
Phase angle = tan^-1(wL/R).

= tan^-1(2*3.14*5*10^3*24*10^-3/1*10^3).

= 37.

Shivany said: (Mar 10, 2015)  
How to calculate inverse values? We do not have calculator in the exam halls right.

S.Bala Kumar said: (May 19, 2015)  
I studied phase shift is tan inverse of (imaginary part/real part). Why that formula applied here?

Shiv said: (Dec 4, 2016)  
This question asked in exam where calculator is allowed.

Samurtza said: (Aug 31, 2017)  
Just keeping in mind the tangent graph and its relation in this circuit: (When Calculator is not allowed).

tan(angle)=2*π*f*L/R; .

For this question we can use the act==> angle is less than 45 if 0<(2*π*f*L/R)<1.

Otherwise greater than that in positive values. Here we have (2*π*f*L/R) <1. Also, note that (2*π*f*L/R) will be positive always.

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