### Discussion :: Inductors - General Questions (Q.No.11)

Msr said: (Aug 24, 2011) | |

R =PL/A |

Vijay said: (Mar 27, 2013) | |

What about turns? |

Cm Prasad said: (Nov 6, 2013) | |

Answer D is right if "impedance" has to increase. But here answer should be B. Because here we have to increase only "resistance" which can be done by using thin wire. Increasing in number of turns will increase inductive reactance. |

Hopeso said: (May 21, 2014) | |

R = PL/A. R = Resistance. L= length of wire (wire is wound in, to form coil and number of coils connected in series gives winding)- directly proportional to resistance. A= Area of cross section - inversely proportional to resistance. As number of turns increases the length of coil also increases and using a thinner wire gives lesser area of cross section. Thus answer is D. |

Hubha93 said: (May 27, 2014) | |

Here overall the length of wire is increasing by increasing the no.of turns and in case of using thin wire cross section area is decreasing so resistance is increasing in both the cases. |

Lasya said: (Dec 8, 2015) | |

L = (no. of turns)2*permeability*area/length. Winding resistance of coil means inductance will come to the picture. Hence it can be increased by increasing number of turns & using thinner wire. Option D is correct. |

Nilesh said: (Mar 16, 2017) | |

Is it correct or wrong that large no of coils have less resistance than less no of the same coil? If yes/no please explain. |

Sharan said: (Oct 9, 2017) | |

Well explained @Lasya. |

Ramu said: (Dec 13, 2018) | |

Answer is A. |

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