### Discussion :: RLC Circuits and Resonance - General Questions (Q.No.4)

Jay said: (Oct 17, 2012) | |

Find XL=2*3.14*(5*10^3)*(120*10^-3) =3728 ohm. Xc=1/2*3.14*(5*10^3)*(0.05*10^-6) =636.94 ohm. Z=sqrt[R^2+(XL-Xc)^2] =4,337.3 ohm. |

Bashishth Singh said: (Apr 14, 2014) | |

Give that, L = 120mH = 1.2*10^-1H and R = 3*10^3 ohm. c = 0.05uF = 5*10^-8F. F = 50kHz = 5*10^4Hz. XL = 2*3.14*5*10^4*1.2*10^-1 = 3728 ohm. Xc = 1/2(3.14*5*10^4*5*10^-8) = 636.94 ohm. Z = (R^2+(XL-XC)^2)^1/2. = 4337.3 ohm. |

Girmay Abadi said: (Jul 22, 2015) | |

Given parameters; L = 120 mH, R = 3*10^3 ohm, c = 0.05 uF and F = 50 kHz. XL = 2*3.14*5*10^4*1.2*10^-1 = 3728 ohm. Xc = 1/2 (3.14*5*10^4*5*10^-8) = 636.94 ohm. Magnitude |Z| = (R^2+(XL-XC)^2)^1/2 = 4337.3 ohm. Tan (Q) = (XL-XC)/R = (3728-636.94)/3000 = 3091.06/3000 = 1.03035333. Q = tan^-1 (1.03035333) = 45.85 degree. Therefore the impedance as polar form is Z=|Z|at angle of Q = 4337.3 ohms at angle of 45.85 degree. So as me the answer is not given here; no answer! |

Aleksandr said: (Feb 8, 2016) | |

Also must be angle 46. |

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