A capacitor of capacitance 3 muF is charged to 120 V and another of capacitance 6 muF is charged to 30 V. They are joined through a resistor so that…

Initial charges are $q_1=3\times120=360\ \mu\mathrm{C}$ and $q_2=6\times30=180\ \mu\mathrm{C}$. Because opposite plates are joined, charge on the connected pair of plates partly cancels and the net charge is $q_1-q_2=180\ \mu\mathrm{C}$ shared by $9\ \mu\mathrm{F}$ in parallel, so the final common voltage is $V_f=180/9=20$ V. Initial energy: $\tfrac12(3)(120)^2+\tfrac12(6)(30)^2=21\,600+2\,700=24\,300\ \mu\mathrm{J}$. Final energy: $\tfrac12(9)(20)^2=1\,800\ \mu\mathrm{J}$. The heat is the difference, $22\,500\ \mu\mathrm{J}=22.50$ mJ. As a check, $\tfrac12\dfrac{C_1C_2}{C_1+C_2}(V_1+V_2)^2=\tfrac12(2)(150)^2\ \mu\mathrm{J}$ gives the same value. The energy lost is independent of the resistance value, since only the initial and final charge distributions matter.