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 (220×.8) volt sources, then the actual power would be

  • Option 1)

    100×0.8 watt

  • Option 2)

    100× (0.8)2 watt

  • Option 3)

    Between 100×0.8 watt and 100 watt

  • Option 4)

    Between 100× (0.8)2 watt and 100×0.8 watt

 

Answers (1)

As we learnt in 

Power dissipiated in external resistance -

P=(\frac{E}{R+r})^{2}R

-

 

 P=100W \Rightarrow R=\frac{V^{2}}{P}=484\Omega

V=220\ Volt

Applied Voltage= 220 \times (0.8)

P=\frac{\left [ 220\times 0.8 \right ]^{2}}{484}=100\times (0.8)^{2}

maximum power = = 100\times 0.8W


Option 1)

100×0.8 watt

This solution is incorrect

Option 2)

100× (0.8)2 watt

This solution is incorrect

Option 3)

Between 100×0.8 watt and 100 watt

This solution is incorrect

Option 4)

Between 100× (0.8)2 watt and 100×0.8 watt

This solution is correct

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