All bulbs consume same power. The resistance of bulb 1 is `36 Omega`

What is the resistance of bulb 4?

Options

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

Answer

A

Explanation

`i_(2) = i_(3) = I_(b) , V_(1) = (V_(2) + V_(3)) = V , and P_(2) = P_(3)`
or ` R_(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) , so R_(3) = 9 Omega`
Also `R_(3) = R_(2) = 9 Omega`
`I_(a) = (I)/(3) and I_(4) = I`,
`P_(1) = P_(4) = 4 W`
`((I)/(3))^(2) R_(1) = (I)^(2) R_(4) = 4 W or (R_(1))/(9) = R_(4) or R_(4) = 4 Omega`
Also `I = 1 A , R_(eq) = 16 Omega and I = 1 A , epsilon = 16 V`