VITEEE 2010 Paper

Power, P=

V2

R

,
⇒R=

V2

P

For the first bulb,
R1=(

V2

P1

)=(

(220)2

25

)=1936Ω
For the second bulb,
R2=(

V2

P2

)=(

(220)2

100

)=48Ω
Current in series combination is the same in the two bulbs,
i=

V

R1+R2

=

220

1936+484

=

220

2420

=

1

11

A
If the actual powers in the two bulbs be P1 and P2 then
P′1=i2R1=(

1

11

)2×1936=16W
P′2=i2R2=(

1

11

)2×484=4W
Since P′1>P′2, so, 25 W bulb will glow more brightly.