# Find the Norton equivalent of circuit using superposition

Exercise 3.23a of Kaufman, 2005: Find the Norton equivalent of the circuit in Figure 3.125:

I've run into some trouble when answering this question. The Norton equivalent resistance is clearly $$\R_N=R_6+R_7\$$. The Norton equivalent current can be found using superposition.

In the first sub-circuit, $$\I\$$ is changed to an open circuit, and $$\I_N=v/(R_6+R_7)\$$. According to the answer I was provided, this is the only term. This implies that the current from the other sub-circuit, in which $$\v\$$ is changed to a short circuit, is $$\0\$$.

I don't see how this can be the case. My thinking is that from $$\I\$$, I need to use the current divider relation to determine what portion of the current passes through $$\R_4\$$, then again at the junction of the short circuit when $$\v\$$ used to be, and once again to see what portion passes through $$\R_6\$$.

If $$\R_1\$$ or $$\R_2\$$ were not present, the given answer would make sense to me, because all no current would flow through the resistors to the right of where $$\v\$$ was, but since they do exist, there is some resistance in all paths back to $$\I\$$, so the current divider formula should apply.

Can someone correct/affirm my thinking on this?