COMEDK Exam 2015 Solved Paper

According to given condition, let n number of bulbs are connected in parallel as shown belowResistance of each bulb, R‌=‌

V2

P

‌‌[∵P=‌

V2

R

]
‌=‌

1002

50

=200Ω
Using KVL in the lowest loop,
‌120−10I−‌

I

n

×200=0
⇒‌‌12−I=‌

20I

n

. . . (i)
Power delivered by battery = Sum of power used in bulbs
⇒‌(120−10I)I‌=50n
‌(12−I)I‌=5n
⇒‌‌

20I

n

⋅I‌=5n [from Eq. (i)]
⇒‌4I2‌=n2
⇒I‌=‌

n

2

‌‌‌ . . . (ii)
From Eqs. (i) and (ii), we get
12−‌

n

2

=‌

20

n

×‌

n

2

⇒n=4