### Discussion :: Passive Filters - General Questions (Q.No.5)

Alaa said: (Dec 31, 2014) | |

Explain it please. |

Kailash said: (Feb 9, 2015) | |

At resonance: xl = xc. sub fof l $ c ,to find f. |

Nyce said: (Mar 27, 2015) | |

Just solve for resonant freq given the formula Fr = 1/(2pi.sqrt(LC)). |

Rashid said: (Jun 13, 2015) | |

XL=Xc. 2*pi*f*L = 1/(2*pi*f*C). f^2 = 1/(4*pi^2*L*C). f^2 = 21129.99 hz. f = sqrt (21129.99). f = 4596.7 Hz. |

Ali said: (Aug 31, 2018) | |

How 21129.99hz came? |

Ahmadnour said: (May 29, 2019) | |

Center frequency = Resonance Frequency. At Resonance Frequency XL = Xc . Fr = 1/2*p* √(L*C). |

Harshal said: (Jul 29, 2019) | |

XL = XC is for the series RLC circuit and in this question the mentioned circuit is parallel circuit. |

Udaypratapsingh said: (Sep 17, 2019) | |

Frequency = 1/2π√Lc. |

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