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

Options

A. `100 xx 0.8 watt`
B. `100 xx (0.8)^(2) watt`
C. Between `100 xx 0.8 `watt and `100 watt`
D. Between `100 xx (0.8)^(2) `watt `100 xx 0.8` watt`

Answer

D

Explanation

(d) `P_(1) = ((220)^(2))/(R_(1))` and `P_(2) =((220 xx 0.8)^(2))/(R_(2))`
`(P_(2))/(P_(1)) = ((220 xx 0.8)^(2))/((220)^(2)) xx (R_(1))/(R_(2)) implies (P_(2))/(P_(1)) = (0.8)^(2) xx (R_(1))/(R_(2))`
`R_(2) lt R_(1)`
because voltage decreases from `220 V rarr 220 xx 0.8 V`. It means heat produced rarr decrease)
So `(R_(1))/(R_(2)) gt 1 implies P_(2) gt (0.8)^(2) P_(1) implies P_(2) gt (0.8)^(2) xx 100 W`
Also `(P_(2))/(P_(1)) = ((220 xx 0.8)i_(2))/(220 i_(1))`, Since `i_(2) lt i_(1)` (we expect)
So `(P_(2))/(P_(1)) lt 0.8 implies P_(2) lt (100 xx 0.8)`
Hence, the actual power would be between `100 xx (0.8)^(2) W` and `(100 xx 0.8) W`