simulate this circuit – Schematic created using CircuitLab
By calculating the equivalent Norton circuit between a and b, I first calculated \$R_{eq}\$ and after \$i_{eq}\$.
When I calculated \$i_{eq}\$, I short-circuited the terminals a and b, and so the circuit becomes:
So \$i_{eq}=-\frac{E}{R_1}\$, but the solution gives \$\frac{E}{R_1}\$. Is it a typo?
Best Answer
The difference in the signs of the solutions lies in the directions chosen for the current. So, the answers are equivalent, the current is just flowing in the opposite direction to which you have selected.