As `P = I^2 R, I = sqrt((P)/(R ))`

Thus, maximum current that can flow through `2 Omega` resistor (rating 18 W), `I = sqrt((18 W)/(2 Omega)) = 3 A`.

Since resistor `B (= 2 Omega)` and resistor `C(= 2 Omega)` are in parallel, and the current through their combination, which is in series with resistor `A(= 2 Omega)` is also `3 A`, current through `B and C` (having equal resistance and in parallel) `= (3 A)/(2) = 1.5 A`.