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A 100 W bulb B_(1) and two 60 W 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» SOLUTION :A bulb is essentially a resistance `R=(V^(2))/(P)` where P denotes the POWER of the bulb. `therefore"Resistance of "B_(1)(R_(1))=V^(2)//100` `"Resistance of "B_(2)(R_(2))=V^(2)//60` `"Resistance of "B_(3)(R_(3))=V^(2)//60` `therefore I_(1)="CURRENT in "B_(1)=(250)/((R_(1)+R_(2)))=(250xx300)/(8V^(2))` `I_(2)="Current in "B_(2)=(250)/((R_(1)+R_(2)))=(250xx300)/(8V^(2))` `I_(3)="Current in "B_(3)=(250)/(R_(3))=(250xx60)/(V^(2))` `therefore""W_(1)" OUTPUT power of "B_(1)=I_(1)^(2)R_(1)` `therefore W_(1)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(100)` `W_(2)=I_(2)^(2)R_(2)"or"W_(2)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(60)` `W_(3)=I_(2)^(2)R_(3)"or"W_(3)=((250xx60)/(V^(2)))^(2)xx(V^(2))/(60)` `therefore W_(1):W_(2):W_(3)=15:25:64"or"W_(1)LT W_(2) lt W_(3)` |
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