A 100 W bulb B_1 and two 60W bulbs ...
Question

# ( mathrm{A} 100 mathrm{Wbulb} mathrm{B}_{1} ) and two ( 60 mathrm{W} ) bulbs ( mathrm{B}_{2} )and ( B_{3}, ) are connected to a ( 250 mathrm{V} ) source, as shown in the figure. Now ( W_{1}, W, ) and ( W_{2} ) arethe output powers of the bulbs ( B_{1}^{2}, B_{2} ) and ( B_{3} ) respectively. Then

SSC CGL
Physics
Solution
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A bulb is essentially a resistance ( R=frac{V^{2}}{P} ) where denotes the power of the bulb.
( therefore ) Resistance of ( mathrm{B}_{1}left(mathrm{R}_{1}right)=mathrm{V}^{2} / 100 )
Resistance of ( mathrm{B}_{2}left(mathrm{R}_{2}right)=mathrm{V}^{2} / 60 )
Resistance of ( mathrm{B}_{3}^{2}left(mathrm{R}_{3}right)=mathrm{V}^{2} / 60 )
( therefore I_{1}= ) Current in ( B_{1}=frac{250}{left(R_{1}+R_{2}right)}=frac{250 times 300}{8 V^{2}} )
( mathrm{I}_{2}= ) Current in ( mathrm{B}_{2}=frac{250}{left(mathrm{R}_{1}+mathrm{R}_{2}right)}=frac{250 times 300}{8 mathrm{V}^{2}} )
( mathrm{I}_{3}= ) Current in ( mathrm{B}_{3}=mathrm{I}_{1} ) as ( mathrm{B}_{1}, mathrm{B}_{2} ) are in series
( therefore W_{1} ) output power of ( B_{1}=I_{1}^{2} R_{1} )
( therefore W_{1}=left(frac{250 times 300}{8 V^{2}}right)^{2} times frac{V^{2}}{100} )
( mathrm{W}_{2}=mathrm{I}_{2}^{2} mathrm{R}_{2} ) or ( mathrm{W}_{2}=left(frac{250 times 300}{8 mathrm{V}^{2}}right)^{2} times frac{mathrm{V}^{2}}{60} )
( W_{3}=I_{2}^{2} R_{3} ) or ( W_{3}=left(frac{250 times 300}{8 V^{2}}right)^{2} times frac{V^{2}}{60} )
( therefore mathrm{W}_{1}: mathrm{W}_{2}: mathrm{W}_{3}=15: 25: 64 ) or ( mathrm{W}_{1}<mathrm{W}_{2}<mathrm{W}_{3} )