Electrical Engineering - Passive Filters - Discussion

Discussion :: Passive Filters - General Questions (Q.No.6)

6.

An RL high-pass filter consists of a 470 resistor and a 600 mH coil. The output is taken across the coil. The circuit's critical frequency is

 [A]. 125 Hz [B]. 1,250 Hz [C]. 564 Hz [D]. 5,644 Hz

Explanation:

No answer description available for this question.

 Nil said: (Dec 30, 2011) How? please explain.

 Raju said: (Jan 1, 2012) Use this formula critical frequency is = 1*R/2*pi*L.

 Sushil Gaind said: (Mar 23, 2012) fc =R/2*pi*L

 Kiran Kumar said: (Apr 1, 2012) I'm not getting the answer can any one explain me.

 Rekha said: (Apr 30, 2012) F=R/(2*PI*L) =470/(2*3.14*600*10-3) =125HZ

 Havish said: (Oct 26, 2012) The transfer function for this ckt will be: |Vo/Vi(jw)| = w/[w^2 + (r/l)^2]^(1/2) This Magnitude is Zero for w--> 0 Unity for w--> inf Critical freq will be that freq for which Magnitude is 0.707(i.e 1/sqrt(2)) Hence Wc= r/l or Fc= r/2*pi*l

 Shalin said: (Jan 18, 2015) Z = R+jwL. For critical frequency, R = wL. i.e. R = 2*pi*f*L. So f = R/(2*pi*L) -> critical frequency.

 Umesh Patil said: (Oct 28, 2015) Which formula you are using, I can't understand. Just I know about. F = 1/[2*pie*{sqr root(LC)}]. After solving we can't get given answer.

 Aamir said: (Dec 29, 2015) Well, of course the cut-off frequency is when XL = R. XL = ωL = 2*F*L. And the time constant for LR circuit is L/R. Therefore, the cut-off frequency, no matter if High-Pass/ Low-Pass: fc = R/(2*L).

 Uday Pratap said: (Sep 17, 2019) Critical frequency= R/2πL.