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Two identical bulbs A and B are connected in series acrossa source of emf E. now, bulb B is removed from the circuit. Compare the brightness of bulb A in both arrangements . |
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Answer» <P> Solution :The POWER dissipated across bulb is `P=I^2R`Current `I=E/(R+R)=E/(2R)`, when both BULBS are CONNECTED in series . `therefore` Power across bulb A, `P=(E/(2R))^2=R=(E^2)/(4R)` Now, when bulb B is removed , current `I=E/R` `therefore` Power across bulb A, `P^.=(E/R)^2R=(E^2)/R` SINCE, power dissipated across bulb A is more when it is connected alone therefore it will glow more brightly in this arrangement . |
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