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I need help with 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

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

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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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