Current Electricity Charging Discharging Of Capacitors Question 87
Question: 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\times .8) $ volt sources, then the actual power would be
[CPMT 1989]
Options:
A) $ 100\times 0.8,watt $
B) $ 100\times {{(0.8)}^{2}}watt $
C) Between $ 100\times 0.8 $ watt and 100 watt
D) Between $ 100\times {{(0.8)}^{2}}watt $ and $ 100\times 0.8 $ watt
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Answer:
Correct Answer: D
Solution:
$ P _{1}=\frac{{{(220)}^{2}}}{R _{1}} $ and $ P _{2}=\frac{{{(220\times 0.8)}^{2}}}{R _{2}} $
$ \frac{P _{2}}{P _{1}}=\frac{{{(220\times 0.8)}^{2}}}{{{(220)}^{2}}}\times \frac{R _{1}}{R _{2}} $
Therefore $ \frac{P _{2}}{P _{1}}={{(0.8)}^{2}}\times \frac{R _{1}}{R _{2}} $ Here R2 < R1 (because voltage decreases from 220 V ® 220 ´ 0.8 V It means heat produced ® decreases) So $ \frac{R _{1}}{R _{2}}>1 $
Therefore $ P _{2}>{{(0.8)}^{2}}P _{1} $
Therefore $ P _{2}>{{(0.8)}^{2}}\times 100,W $ Also $ \frac{P _{2}}{P _{1}}=\frac{(220\times 0.8)i _{2}}{220,i _{1}}, $ Since $ i _{2}<i _{1} $ (we expect) So $ \frac{P _{2}}{P _{1}}<0.8 $
Therefore $ P _{2}<(100\times 0.8) $ Hence the actual power would be between $ 100\times {{(0.8)}^{2}}W $ and (100 ´ 0.8) W
BETA
