All bulbs consume same power. The resistance of bulb `1` is ` 36 Omega` . Answer the following questions:

What is the resistance of bulb `3`?

Options

A. `4 Omega`
B. `9 Omega`
C. `12 Omega`
D. `18 Omega`

Answer

B

Explanation

`i_(2)=i_(3)=I_(b)`
`V_(1)=(V_(2)+V_(3))=V`
and `P_(2)=P_(3)`
`RightarrowR_(2)=R_(3)`
and `V_(2)=V_(3)=V//2`
`P_(1)=P_(2)=P_(3)=P_(4)`(given)
`P_(1)=V^(2)/36,P_(3)=(V//2)^(2)/R_(3)`
As `P_(1)=P_(3) RightarrowR_(3)=9Omega`
Also `R_(3)=R_(2)=9 Omega`
`I_(a)=1/3` and `I_(4)=I`
`P_(1)=P_(4)=4W`
`(1/3)^(2)R_(1)=(I)^(2)R_(4)=4W`
Rightarrow `R_(1)/9=R_(4)`
Rightarrow `4Omega=R_(4)`
Also `I= IA`
`R_(eq)=16 Omega` and `I = 1A`
`varepsilon =16 V`