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A 100 W bulb B_(1) and two 60W bulbs B_(2) and B_(3) , are connected to a 250 V source as shown in the figure . Now W_(1) , W_(2) and W_(3) are the output powers of the bulbs B_(1) , B_(2) and B_(3) respectively, then |
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Answer» `W_(1) gt W_(2) = W_(3)` Resistance `R_(1)` of sphere `B_(1)= (V^(2))/(P_(1)) ` `= ((250)^(2))/(100)` = `625 Omega ` ` = 1042 Omega` Resistance`R_(3) ` of sphere `B_(3) = (V^(2))/(P_(3)) = ((250)^(2))/(60)` = ` 1042 Omega`  Now Output power of `B_(1) , W_(1) = (V^(2))/((R_(1)+ R_(2))^(2)). R_(1)` `= ((250)^(2))/((625 + 1042)^(2)) xx 625 ` = 14.1 W Output power of `B_(2) , W_(2)= (V^(2))/((R_(1) + R_(2))^(2)) xx R_(2)` `= ((250)^(2))/((625 + 1042)^(2)) xx 1042` 23.4 W Output power of `B_(3) , W_(3) = (V^(2))/(R_(3)^(2)) xx R_(3)` ` =((250)^(2))/( 1042) = 60 ` W `therefore W_(1) lt W_(2) lt W_(3)` |
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