Find the heat developed per minute in each of the resistors.

Answer

`540 J, 480J, 960J`

Explanation

Equivalent resistance of the circuit is
`(8 xx 4)/(8 + 4) xx 1 = (8)/(3) xx 1 = (11)/(3) Omega`
current drawn from the battery is
`1 = (11V)/((11//3) Omega) = 3A`
Since the current through `1 Omega` tresistor is `3A` so heat produced in this resistor in `1` minut `(= 60` second ) is
`H = l^(2)Rt = (3)^(2) xx 1 xx 60 = 540 J`
current through `8 Omega` resistor
`l_(1) = (3)/(8 + 4) xx 4 = 1A`
Head produced in `8 Omega` resistor in `1` minut
`H_(1) = l_(1)^(2) R_(1)t = (1)^(2) xx 8 xx 60 = 480 J`
Current through `4 Omega` resistor
`l_(2) = 1 - l_(1) = 3 - 1 = 2A`
Current through `4 Omega` resistor in `1` minute
`H_(2) = l_(2)^(2) R_(2) t = (2)^(2) xx 4xx 60 = 960 J`