A 9 Ω and a 36 Ω resistor are joined in parallel to a 27 V battery…

Step 1. (a) $R_p = \dfrac{9\ \Omega \times 36\ \Omega}{9\ \Omega + 36\ \Omega} = 7.2\ \Omega$.

Step 2. (b) $\dfrac{27\ \text{V}}{9\ \Omega} = 3\ \text{A}$ and $\dfrac{27\ \text{V}}{36\ \Omega} = 0.75\ \text{A}$.

Step 3. (c) $3\ \text{A} + 0.75\ \text{A} = 3.75\ \text{A}$.