Finding Norton current (I_N): to determine the Norton equivalent current at a pair of terminals, what should you do with the external load connected to those terminals?

Introduction / Context:
Norton’s theorem represents a linear network by a current source I_N in parallel with a resistance R_N as seen from two terminals. Determining I_N correctly is fundamental for simplifying networks and predicting how any load will behave when attached to those terminals.

Given Data / Assumptions:

• We consider a linear circuit with independent/dependent sources and resistors.
• The load initially connects to the terminals of interest.
• The goal is to measure or calculate Norton current I_N.

Concept / Approach:
By definition, I_N equals the short-circuit current at the output terminals (the current that would flow if those terminals were shorted together). Therefore, to find I_N, you remove the actual external load and replace it with a short across the output terminals, then compute the resulting current through that short. Afterward, R_N can be found by deactivating sources or via R_N = V_th / I_sc if you also know V_th (Thevenin voltage).

Step-by-Step Solution:
Disconnect the load so it does not influence the measurement.Short the output terminals together.Compute the current through the short; this is I_sc.Set I_N = I_sc as the Norton current.

Verification / Alternative check:
Use duality with Thevenin: if V_th is known, then R_N = V_th / I_sc. Simulated or measured results using a circuit tool confirm that the short-circuit current equals the current source value in the Norton model.

Why Other Options Are Wrong:
Opening the load gives open-circuit voltage V_th, not I_N. Shorting or opening sources (C, D) applies to finding R_N by deactivation, not to finding I_N directly. “None” is wrong because shorting the terminals is exactly correct.

Common Pitfalls:
Forgetting dependent sources remain active during R_N determination via test sources; confusing open-circuit voltage with short-circuit current.

Final Answer:
Short the load resistor