# Calculation of a Norton equivalent current 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: simulate this circuit

So $i_{eq}=-\frac{E}{R_1}$, but the solution gives $\frac{E}{R_1}$. Is it a typo?

• Is the arrow representing I on the original problem? If you should consider the current between a and b its value is E/R1. If you should consider the current at the direction of arrow, then it's -E/R1. To avoid misunderstood, represent the direction of current by an arrow. – Pedro Quadros Jun 9 '15 at 18:57
• This is a common thing that people get hung up on in early courses. If you say that current going into something is -X amps and the answer says current coming out of something is X amps, they're the same thing. PS - Look at the current source. It shows current flowing up and then down from a to b. It would make sense then that current is positive from a to b and negative from b to a (hence why you got negative answer). – I. Wolfe Jun 9 '15 at 21:00