Resistance, `R = 2 Omega`

Maximum power, `P_(max) = 18 W`

Maximum current, `I_(max) =` ?

`P = I^(2)R`

`rArr " " I= sqrt((P)/(R)) = sqrt((18)/(2)) = 3A= I_(max)`

Maximum current that can flow through `2 Omega` resistor is `3A`. This current divides along B and C because in parallel combination, voltage across B and C remain same and hence, `I prop (1)/(R)`.

Since, B and C have same resistance `2 Omega` each, therefore, same current i.e., `(3)/(2) = 1.5 A` flows through B and C.