I have this circuit in which I'm required to find the Power produced by the Current source \$0.2A\$ :
So I Add ground to \$V2\$ and did node voltage analysis such as :
$$ \begin{aligned}V_{0}:\dfrac{V_{0}-30}{10}+\dfrac{V_{0}-60-V_{1}}{45}-0.2+\dfrac{V_{0}}{20}=0\\ V_{1}:\dfrac{V_{1}}{12}+\dfrac{V_{1}+60-V_{0}}{45}+0.2+\dfrac{V_{1}-45}{15}=0\end{aligned} $$
and eventually got :
$$ \begin{pmatrix} 145 & -20 \bigm|& 3760\\ -12 & 102 \bigm|& 1197\\ \end{pmatrix} \longrightarrow \begin{pmatrix} V_0 = 28.004 \quad [V] \\ V_1=15.029 \quad [V] \end{pmatrix} $$
Which according to the final answer is wrong ! as I should get \$ V_0-V_1=15.77 \quad [V]\$
I do get the final answer if I change the direction of \$ 0.2 A \$ in the second equation :
$$ \begin{aligned}V_{0}:\dfrac{V_{0}-30}{10}+\dfrac{V_{0}-60-V_{1}}{45}-0.2+\dfrac{V_{0}}{20}=0\\ V_{1}:\dfrac{V_{1}}{12}+\dfrac{V_{1}+60-V_{0}}{45}-0.2+\dfrac{V_{1}-45}{15}=0\end{aligned} $$
My question is as it follows , If I'm assuming \$ 0.2 A \$ goes from \$ V1 \$ to \$ V0 \$ , why I should assume that in the node \$ V1 \$ it goes from \$ V0 \$ to \$ V1 \$ ?
I remember solving similar questions where I had the current source direction not changed and got right final answers , but this is confusing me .. I'd appreciate some help !
