Electrical Engineering - Passive Filters - Discussion

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

Answer: Option A


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)  

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))
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.

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