Two identical bulbs A and B are connected in series across a source of emf E. now, bulb B is removed from the circuit. Compare the brightness of bulb A in both arrangements .

Explanation

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 .