COMEDK Exam 2019 Solved Paper

Given, power of bulb A,PA=60W
Power of bulb B,PB=100W
Since, both bulbs are connected to the same power source in parallel combination. Thus,
‌PA=‌

V2

RA

=60W‌ and ‌PB=‌

V2

RB

=100W
‌RA=‌

V2

60

Ω‌ and ‌RB=‌

V2

100

Ω
‌‌ Current in bulb ‌A,IA=‌

‌ Potential ‌

‌ Resistance ‌

‌⇒‌‌IA=‌

V

V2 60

⇒IA=‌

60

V

‌ Similarly, current in bulb B,IB=‌

V

V2 100

=‌

100

V

It is clear that IB>IA.