In the circuit shown in figure switch S is thrown to position 1 at  t=0. When the current in the current in the resistor is 1A it is then shifted to position 2. The heat generated in the circuit after shifting to position 2 is :

When current through the resistor is 1A, the p.d. across capacitor plates is 5V. when switch is shifted to position 2 and steady state is reached. The p.d. across capacitor plates is still 5V but with opposite polarity. Thus the energy of the capacitor does not change.
Therefore heat generated=work done by the battery.
 =charge flown through battery  EMF
 = $2\times 25\times 5\times 5\mu J=1250\mu J$