Applying nodal analysis

\begin{array}{r} -4+i+\frac{V-20}{2}=0 \\ -4+\frac{V}{2}+\frac{V-20}{2}=0 \\ -4-\frac{2 V-20}{2}=0 \end{array}

So, we have V=14  V \Rightarrow i=\frac{V}{2}=7  A

Note: A resistor in series with an ideal current source is always neglected from the nodal analysis. But in the power calculation, it cannot be neglected.