### Discussion :: Circuit Theorems in AC Analysis - General Questions (Q.No.2)

Biswajit Patra said: (Feb 11, 2014) | |

I want the process to solve this? |

Nikhil Jain said: (Jun 29, 2014) | |

The value of C is j12 kohm and l is j12 kohm. First remove RL and consider vth at the place of Rl then apply nodal analysis. You'll get the answer |

Chinna said: (Aug 25, 2014) | |

Can you please elaborate the explanation by substituting values @Nikhil Jain. |

Sky_Net said: (Dec 5, 2014) | |

Can anybody tell true procedure? |

Annnonymus said: (Sep 24, 2015) | |

Use kcl to find current in 15 kohm resistance. Then find voltage and that will answer. |

Nhuss said: (Jan 30, 2016) | |

XL = 12 K OHMS. XC = 12 K OHMS. RL = 49 K OHMS. R = 15 K OHMS. VS = 6 < 0 V. Z = SQRT ((R+RL)^2 + (XL-XC)^2). Z = R + RL = 49 + 15 = 64 K OHMS. I = V/Z = 6/64K = 0.09375 mA. VTH = 0.09375 mA x 49 k OHMS = 4.59 V. |

Abdi said: (Apr 21, 2016) | |

If you write equation for loop 1(branch with the inductor) and loop 2(the branch with the capacitor and R after removing RL): Loop 1: I1Xl = 6V Loop 2: I2(15000-j12000) = 6 (since we are interested Vth across R, we can find I2, then find the voltage across R. Because Rl is in parallel with R, the voltage across R is the same as Vth) I2 = 6V/(15000-j120000)= 3.1235*10^-4<theta = 38.66> Vr=Vth = 15000 * 3.1235*10^-4<theta = 38.66> = 4.685<theta = 38.66> Hope this helps! |

Hasan said: (May 12, 2016) | |

Eth is the voltage across R resistance. So, Eth = 6*15/(15 - 12j) = 4.68 < 38.7. |

Siddhant Mukherjee said: (Dec 12, 2016) | |

The Vth across RL is equal to the voltage across the resistor R. So for determining the voltage across R we need to find out the current flowing through R. The Current across R would be equal to I (across R) = VS/(R+XC) So the Voltage across R would be equal to IR. Vth = I (across R) * R. = (VS*R)/(R + XC). Put the values of the respective parameters and the answers would be option b = 4.69<38.7° V. |

Kishorvarma said: (Dec 12, 2016) | |

It's very good explanation @Nhuss. |

Azeem said: (Jul 24, 2020) | |

It is very simple. 1) As we know voltage across parallel is same so XL is useless to calculate; 2) To find vth RL is open circuit so remove RL. 3) Now we have VS= 6 V and Xc= 12 K is in series with R = 15 K. 4) Z = (R^2+Xc^2)^1/2 = 19.2 K. 5) Vth = R/Z * Vs = .78*6 = 4.69. 6) θ = tan-1 ( Xc/R) = .6747 rad * 57 = 38.5°. |

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