In a circuit, an ideal battery of emf 12 V is connected in series with a 2 Ω resistor, and a 4 Ω resistor in parallel with an uncharged 6 muF…

In the steady state no current flows through the capacitor branch, so the current is $12/(2+4)=2\ \mathrm{A}$ and the capacitor voltage equals the voltage across the $4\ \Omega$ resistor, $V_C=2\times4=8\ \mathrm{V}$. The stored energy is $U=\tfrac12CV_C^2=\tfrac12(6\ \mu\mathrm{F})(64)=192\ \mu\mathrm{J}$. After the battery is replaced by a wire, the $2\ \Omega$ and $4\ \Omega$ resistors are both directly across the capacitor, i.e. in parallel with each other, and the capacitor discharges completely through them. At every instant they share the same voltage $V$, so their powers are $V^2/R$ and the heat divides in inverse ratio of resistance: $H_2:H_4=4:2$. Hence the $2\ \Omega$ resistor receives $\dfrac{4}{6}$ of the energy: $H=\tfrac23\times192=128\ \mu\mathrm{J}$. A numerical integration of the discharge gives the same value. Assuming the energy is all taken by the $2\ \Omega$ resistor would wrongly give $192$.