# How to determine current direction in node voltage analysis?

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 !

• You assume the direction when starting the analysis and use it through your whole calculation consistently. If the value of it in the end turns out to be negative, then the direction is opposite. Jul 21, 2022 at 13:35
• electronics.stackexchange.com/questions/628008/…
– G36
Jul 21, 2022 at 13:36