Source transformation to Norton form: A 120 V ideal voltage source has source resistance RS = 60 Ω. What is the magnitude of the equivalent Norton current source?

Introduction / Context:Transforming between Thevenin (voltage source in series with resistance) and Norton (current source in parallel with resistance) models is a routine skill for simplifying circuits and analyzing loading effects. Here we convert a given voltage source to its Norton equivalent.

Given Data / Assumptions:

• Thevenin voltage source: VS = 120 V.
• Series (source) resistance: RS = 60 Ω.
• Linear, time-invariant, resistive network; no dependent sources are involved.

Concept / Approach:The Norton current is the short-circuit current delivered by the Thevenin source: IN = VS / RS. The Norton resistance equals the Thevenin resistance: RN = RS. Only the current magnitude is asked in the options, but the resistance is useful context.

Step-by-Step Solution:

Verification / Alternative check:Reverse check to Thevenin: VTH = IN * RN = 2 A * 60 Ω = 120 V, which is the original source voltage. This confirms a correct, lossless transformation between equivalent models.

Why Other Options Are Wrong:

• 4 A: Would require RS = 30 Ω for a 120 V source.
• 400 mA or 200 mA: These are 0.4 A and 0.2 A, far below the correct short-circuit current for 60 Ω at 120 V.

Common Pitfalls:

• Misapplying the relation by multiplying instead of dividing (VS * RS instead of VS / RS).
• Confusing the resistance placement (series vs. parallel) when visualizing the Norton model.

Final Answer:2 A