Find the heat developed in each of the three resistors shown in figure in 1 minute.

Explanation

The equivalent resistance of `6Omega and 3Omega` resistors is
`((6 Omega) xx (3 Omega))/(6 Omega + 3 Omega) = 2 Omega`.
This is connected in series with the `1 Omega` resistor. The equivalent resistance of the circuit is
`R = 2 Omega + 1 Omega = 3 Omega`.
The current through the battery is
`i = 9V/3Omega = 3A`.
The current through the `1 Omega` resistor is, therefore, `3 A`. The heat developed in this resistor is
`H = i^2 Rt`.
`= (3A)^2 xx (1 Omega) xx (60 s) = 540 J`.
The current thrugh the `6 Omega` resistor is
`H = i^2 Rt`
`= (3 A)^2 xx ( 1Omega) xx (60 s) = 540 J`.
The current through the `6 Omega` resistor is
`(3A) xx (3Omega)/(6Omega + 3Omega) = 1A`.
The heat developed in it
`= (1 A)^2 xx (6 Omega) xx (60 s) = 360 J`.