Find the heat developed per minutes in each of the three resistors `R_(1),R_(2) and R_(3)`

Answer

`600J,1200J 960 J`

Explanation

Equivalent resistance of `6 Omega and 3 Omega` in parallel
`R_(p) = (6 xx 3)/(6+ 3) = 2 Omega`
Total resistance of circuit `R = 2 + 1 = 3 = 3 Omega`
Total current `I = (12V)/(3 Omega) = 4 A`
current through `6 Omega` resistor
`l_(1) = I xx (R_(2))/(R_(1) + R_(2)) = 4 xx (3)/(6 + 3) = (4 )/(3) A`
Heat developed in it per minute
`= l_(1)^(2)k_(1)t = ((4)/(3))^(2) xx 6 xx 60 = 640 J`
current through `3 Omega` resistor
`l_(2) = I xx (R_(1))/(R_(1) + R_(2)) = 4 xx (6)/(6 + 3) = (8)/(3) A`
Heat developed in it minut
`l_(2)^(2) R_(2) t = ((8)/(3))^(2) xx 3 xx 60 = 1200J`
Head produced per minut in `R_(3)`
`= l^(2)R_(3) t = (4)^(2) xx 1 xx 60 = 960 J`