Physics Hypertextbook writes that "The current through the 1 Ω resistor most certainly runs from right to left."
Why is that true?
My approach:
I arbitrarily assigned a counterclockwise direction to the current running through the 1 $\Omega$ resistor. I then applied the junction rule at the node connecting the 2 $\Omega$ Resistor the 20V emf and the wire leading to the 1 $\Omega$ Resistor:
I$_{1\Omega}$ + I$_{2\Omega}$ = I$_{emf}$
where
I$_{2\Omega}$ = 4. If the labeled 3A current loops down and to the right.
Also,
I$_{emf}$ = 3
It can be shown that the current running through the 2 $\Omega$ resistor is 4A:
By the loop rule (for the smaller right-hand circuit): 20V – 2I – 4*3 = 0; I = 4.
If the I$_{emf}$ < 0 (which, by my reasoning, it is), the current actually runs clockwise.
Best Answer
The earlier part of the solution (which can be found here) finds that $I_2$ and $I_3$ both have current from right to left. So the left side must be at lower potential. The passive path on the top loop must likewise carry current from right to left.
Yes, $I_2$ is $4A$, but since you haven't calculated the current through the $20V$ source (it isn't $3A$!), you can't use that node to find $I_1$ yet.