In the circuit shown below, the power developed in `6Omega` resistor is 6 watt. The power in watts developed in the `4Omega` resistor is

Options

A. `y = x + 2`
B. `y = 4x + 2`
C. `y = 2x - 4`
D. `y = 2x + 4`

Answer

B

Explanation

`vec(r )_(cm) = (m_(1)vec(r )_(10) + m_(2) vec(r )_(2))/(m_(1) + m_(2)) = (1 (3 hat(i)) + 2 (9 hat(j)))/(1 + 2) = (hat(i) + 6hat(j))m`
`vec(v)_(cm) = (m_(1) vec(v)_(1) + m_(2)vec(v)_(2))/(m_(1) + m_(2)) = (1(3hat(i)) + 2(6hat(j)))/(1 + 2) (hat(i) + 4hat(j))m//s`
Now `Deltavec(r )_(cm) = vec(v)_(cm) t rArr vec(r )_(cm) - vec(r)_(cm_(0)) = vec(v)_(cm) t`
`rArr (x - 1)hat(i) + (y - 6) hat(j) = (hat(i) + 4hat(j)) t`
`rArr x = 1 + t` and `y = 6 + 4t rArr y = 4x + 2`