### Discussion :: Resistance and Power - General Questions (Q.No.5)

Manikanta said: (Apr 19, 2012) | |

P=I^2R, So P = 1*10^-6*470 = 0.00047W. How will come 1/4. Please explain? |

Rathi said: (Apr 19, 2012) | |

@Manikanta. Yes you are right. But we should take the wattage of the resistance greater than calculated value. 0.25(i.e 1/4) is greater than 0.00047W, So 1/4 is enough. That means 1/4 watt is the right answer. Note: If 1/8watt is given in option then 1/8 will be the correct answer. |

Domy Dimal said: (Oct 22, 2014) | |

0.25 w is greater than 0.00047w. Then how? |

Theophillus said: (May 7, 2015) | |

I don't understand how did you get 0.25 W there seriously. |

Vasava said: (Aug 31, 2015) | |

I can't understand how it come 0.25 watt? |

Priyank Sharma said: (Jun 2, 2016) | |

Why its answer is 0.25watt? |

A.Sathish said: (Jun 29, 2016) | |

0.00047w is correct, which is the nearest value. |

Lewis said: (Aug 19, 2016) | |

On our resistor wattage we have 1/8W, 1/4W, 1/2W, 1W,2W so when we rate we take the closest to 0.00047W which is the 0.25W. If there was the 0.125 W(1/8) resistor would be more suitable. |

Fred Simz Jr said: (Feb 9, 2017) | |

I can't understand how 0.25w came? Please explain. |

Vignesh said: (Apr 11, 2017) | |

Here, the m value is 10^-3 then how to get the answer 1/4? |

Soni said: (Sep 16, 2017) | |

P= IV, P=I*I*R. 1*10^-3*10^-3*470, =0.00047w. |

Engr Faizan Rais said: (Sep 28, 2018) | |

p=i^2*R. So putting the values: p = (40mA)^2*470 p = 0.751W So, 0.751 w is nearer value to 1 w, the ans is 1 watt. |

Gopal said: (Sep 17, 2019) | |

I am not getting the solution. Please explain it. |

Gopala said: (Sep 17, 2019) | |

Explain how it come 1/4? |

Rajesh said: (Oct 4, 2019) | |

0.00047 less than 1/4 watt so least one is 1/4 watt is right answer. 1000mW = 1W, 0.250W is correct. |

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