1. The problem statement, all variables and given/known data 2. Relevant equations HEAT PRODUCED=WORK DONE BY BATTERIES - ENERGY ABSORBED BY CAPACITORS 3. The attempt at a solution INITIALLY BEFORE SWITCH CLOSED :Initially a current flows only in the left most loop involving 4 microfarad capacitor.The charged on the capacitor during steady state is (20*4)=80 microcoulumbs. FINALLY AFTER SWITCH IS CLOSED:Now a decaying current current flows in all the three loops -left,right, and bottom.However we are concerned only with steady state.So suppose the potential at the topmost junction (between the two capacitors and 2 ohm resistor) is x volts.Assuming that the node on the left (connected to 20 V cell,resistance R and resistance 4 ohm) is 0 volts.So we can write the nodal equation at x. (x-20)4+(x-0)5=0.Hence,x=80/9 volts. Now work done by cell= (Charge Flown )* Cell Voltage = Final Charge On Left Plate Of Capacitor 4 microfarads - Initial Potential On Left Plate Of 4 microfarad= 4(20-80/9) - 4*20=-35.56 microjoule(negative). Energy Absorbed By Capacitor = 1/2(4)(20-80/9)^2 + 1/2(5)(80/9)^2 - 1/2(4)(20)^2= -355.56 microjoule HEAT PRODUCED=WORK DONE BY BATTERIES - ENERGY ABSORBED BY CAPACITORS=-35.56 + 355.56=320 microjoules But this does not match with any of the options.Please help.