Figure `7.37` shows a network of three resistances. When some potential difference is applied across the network , thermal powers dissipated by `A, B and C` are in the ratio

Options

A. `2:3:4`
B. `2:4:3`
C. `4:2:3`
D. `3:2:4`

Answer

C

Explanation

`i_(1) = (6R)/(3R+6R) I = (2i)/(3)`
`i_(2) = I - i_(1) = i/3`
`P_(A) : P_(B) : P_(C) = (i_(1)^(2).3R) : (i_(2)^(2).6R):(i^2R)`
`= 4/3 : 2/3 : 1 = 4:2:3`.