The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected to a (220  x .8) volt source, then the actual power would be

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected to a (220  x .8) volt source, then the actual power would be

  1. A

    100×  0.8  watt

  2. B

    100×  0.82  watt

  3. C

    between 100 x 0.8 watt and 100 watt

  4. D

    between 100 x (0.8)2 watt and 100 x 0.8 watt

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    Solution:

    P1=2202R1  andP1=220×0.82R2

    P2P1=  220×0.822202  ×R1R2  P2P1=0.82  ×  R1R2

    Here R2 < R1

    (because voltage decreases from 220 V  220 x 0.8 V. It means heat produced  decreases)

    SoR1R2>1    P20.82P1P2>  0.82×  100  W

    AlsoP2P1=  220×0.8i2220i1,  Since  i2<i1  we  expect

    So  P2P1<  0.8  P2  <  100  ×  0.8

    Hence, the actual power would be between 100 x (0.8)2 W and (100 x 0.8) W

     

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