# 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

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

### Preparation Products

##### Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-