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Two electric bulbs marked 40 W, 220 V and 60 W, 220 V when connected in series, across same voltage supply of 220 V, the effective power is P_(1) and when connected in parallel, the effective power is P_(2). Then (P_(1))/(P_(2)) is |
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Answer» `0.5` `R_(B_(1))=((220)^(2))/(40)OmegaandR_(B_(2))=((220)^(2))/(60)Omega` When the BULBS are connected in series, `R_(S)=R_(B_(1))+R_(B_(2))=((220)^(2))/(40)+((220)^(2))/(60)=(220)^(2)[(1)/(40)+(1)/(60)]` `=(220)^(2)[(60+40)/(60xx40)]=(220)^(2)((100)/(2400))=((220)^(2))/(24)` `therefore P_(1)=(V_(S)^(2))/(R_(S))=(220)^(2)xx(24)/((220)^(2))=24W` When the bulbs are connected in PARALLEL `(1)/(R_(P))=(1)/(R_(B_1))+(1)/(R_(B_(2)))or(1)/(R_(P))=(40)/((220)^(2))+(60)/((220)^(2))` `(1)/(R_(P))=(100)/((220)^(2))orR_(P)=((220)^(2))/(100)` `therefore P_(2)=(V_(S)^(2))/(R_(P))=(220)^(2)xx(100)/((220)^(2))=100W therefore (P_(1))/(P_(2))=(24W)/(100W)=0.24` |
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