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

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?


1 Answer 1


v is a fixed voltage source, nothing connected to it changes the voltage, so you can ignore the entire circuit left of v as well as R5.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.