A 15.4 V battery is joined in one loop with a 3 Ω and a 4 Ω resistor…

Step 1. The current. $I = \dfrac{15.4\ \text{V}}{3\ \Omega + 4\ \Omega} = 2.2\ \text{A}$.

Step 2. The shares. $2.2\ \text{A} \times 3\ \Omega = 6.6\ \text{V}$ and $2.2\ \text{A} \times 4\ \Omega = 8.8\ \text{V}$.