Question

Three bulb each having resistance of 90 Ω are connected in parallel to an ideal battery of emf 60V. Then the current delivered by battery when all the bulb are ON is
I = 0.5 A
I = 1.0 A
I = 1.5 A
I = 2.0 A​

Answer 1

Answer:

a) R

eff

=

3

180

=60Ω

i=60/60=1A

b) R

eff

=

2

180

=90Ω

i=60/90=0.67A

c) R

eff

=180Ω⇒i=60/180=0.33A

Answer 2

Answer:

Option 4

Explanation:

In parallel connection,

1/Reqivalent = 1/90 + 1/90 + 1/90

or, 1/Requivalent = 3/90 = 1/30

or, Reqivalent = 30

Now, V = IR

or, I = V/R = 60/30 = 2A