A bulb of 7 Ω and a bulb of 11 Ω are joined in series to a 9 V battery…

Step 1. Total. $R_s = 7\ \Omega + 11\ \Omega = 18\ \Omega$.

Step 2. Current. $I = \dfrac{9\ \text{V}}{18\ \Omega} = 0.5\ \text{A}$.

Step 3. Shares. $0.5\ \text{A} \times 7\ \Omega = 3.5\ \text{V}$ and $0.5\ \text{A} \times 11\ \Omega = 5.5\ \text{V}$.