Resistors of 4 Ω, 6 Ω and 10 Ω are joined in series with an 8 V battery…

Step 1. (a) $R_s = 4\ \Omega + 6\ \Omega + 10\ \Omega = 20\ \Omega$.

Step 2. (b) $I = \dfrac{8\ \text{V}}{20\ \Omega} = 0.4\ \text{A}$.

Step 3. (c) $1.6\ \text{V}$, $2.4\ \text{V}$ and $4\ \text{V}$.