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

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. |

#### Post your comments here:

Name *:

Email : (optional)

» Your comments will be displayed only after manual approval.