### Discussion :: RL Circuits - General Questions (Q.No.2)

Mani said: (Jun 23, 2011) | |

For this the inductor resistance need to reduce and will done when frequency decreases. |

Abhinav Kumar said: (Nov 13, 2011) | |

In RL circuit both R & L are connected in series. Voltage drop is more at high impedance as V proportional to R whil I constant. So impedance of inductor should be high. Hence frequency should be increases as Zl=wL. |

Kishore said: (Jan 23, 2012) | |

XL= 2*3.14*f*L Here inductive reactance increases by increasing frequency, so the voltage across the inductor increases. |

Danish Baidar said: (Apr 16, 2013) | |

To make the resistor voltage less than the inductor voltage. Inductor voltage has to be increased. VL=I(2*pi*f*L) So f is Increased to increase Inductor voltage. |

Ganesh Shind said: (Jul 20, 2013) | |

For resistive circuit voltage and current are in phase but for. R-L circuit current lags the voltage by 90 degree so there is phase difference phi because of that in R-L circuit current is out of phase. |

Dhiva said: (May 28, 2014) | |

Generally is formula Xl=2piFL. So. frequency is increased. |

Brundha said: (Jan 19, 2015) | |

To make the resistor voltage less than the inductor voltage? Normally we know the formula for frequency, xl = 2*3.14*l*f. So, the voltage across the inductor is increased. |

Shardul said: (Mar 8, 2015) | |

Xl=2*3.14*f*l and Xc=1/2*3.14*f*c. So for in case of indicator its proportional form. |

Sanu Singh said: (Dec 7, 2015) | |

But we have to inductive voltage more than resistance voltage so we have to decrease Xl. |

Surendra said: (Sep 9, 2016) | |

XL = 2 * &phi * f * L. The frequency increase so the voltage across the inductor also increased. |

Saubhagya said: (Jan 9, 2017) | |

I can't get this question. Anyone explain me clearly. |

Prasad Hajare said: (Sep 7, 2017) | |

The frequency increase so the voltage across the inductor also increased. This one right conclusion. |

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