The voltage at 5 Ω resistor is
\begin{aligned} & V_{5 \Omega}=10 V \\ & I_{5 \Omega}=\frac{10}{5}=2 A \end{aligned}
Then applying KCL, we have
−i – 2 + 2 = 0 ⇒ i = 0
Thus, the power delivered by the 10 V source =10 \cdot(0)=0 Watts
\begin{aligned} & P_{2 A }=2 \times 10=20 \text { Watts (delivered) } \\ & P_{5 \Omega}=2^2 \times 5=20 \text { Watts (absorbed) } \end{aligned}