A 12.0 V battery with an internal resistance of 0.25 Ω is charged from a 15.0 V supply with an…

Step 1. The current. $I = \dfrac{15.0\ \text{V} - 12.0\ \text{V}}{1.9\ \Omega + 0.35\ \Omega + 0.25\ \Omega} = \dfrac{3.0\ \text{V}}{2.5\ \Omega} = 1.2\ \text{A}$.

Step 2. The terminal voltage. The battery is charging, so $V = E + Ir = 12.0\ \text{V} + 1.2\ \text{A} \times 0.25\ \Omega = 12.3\ \text{V}$.

Step 3. The heat. $I^2r = (1.2\ \text{A})^2 \times 0.25\ \Omega = 0.36\ \text{W}$.

Where the rest goes. $EI = 12.0\ \text{V} \times 1.2\ \text{A} = 14.4\ \text{W}$ is stored in the battery as chemical energy.