SSC JE Electrical Engineering 2014 Set 1 Paper

Question : 113 of 200

Marks: +1, -0

A 200 W, 200 V bulb and a 100 W, 200 V bulb are connected in series and the voltage of 400 V is applied across the series connected bulbs, Under this condition

100 W bulb will be brighter than 200 W bulb
200 W bulb will be brighter than 100 W bulb
Both the bulbsz will have equal brightness
Both the bulbs will be darker than when they are connected across rated voltage

Validate

Solution:

For 200 W, 200 V,

R_1=200^2/200=200Ω

For 100 W, 200 V,

R_2=200^2/100=400Ω

current

I=400/{200+400}=4/6=2/3A

For Bulb (1), power divided

=200×({2/3})^2={800/9}W

Bulb (2), power divided

=400×(2/3)^2=1600/9W

Hence 100 W bulb will be bighter than 2000 W bulb.

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