Physics - Capacitor Question with Solution | TestHub

With S at 1 for a long time, the capacitor is charged to ε₁, with the polarity set by ε₁ as shown.

After shifting to 2, the battery ε₂ finally charges C to ε₂, but with the opposite polarity on the plates.

Charge driven through the battery: $\Delta q=C{\epsilon }_1+C{\epsilon }_2=C({\epsilon }_1+{\epsilon }_2)$

Work done by the battery: $W={\epsilon }_{2}C({\epsilon }_1+{\epsilon }_2)$

Change in stored energy: $\Delta U=\frac{1}{2}C{\epsilon }_2^2-\frac{1}{2}C{\epsilon }_1^2$

Heat generated = W − ΔU:$H=C{\epsilon }_2({\epsilon }_1+{\epsilon }_2)-\frac{1}{2}C({\epsilon }_2^2-{\epsilon }_1^2)=\frac{C}{2}({\epsilon }_1+{\epsilon }_2{)}^2$

Hence (C).