Norton’s equivalent current (IN): Is it defined as the open-circuit current between two terminals, or the short-circuit current?

Introduction / Context:Norton and Thevenin equivalents are dual representations of a linear two-terminal network. Correctly identifying Norton’s current prevents mistakes when transforming sources or characterizing loads.

Given Data / Assumptions:

• Linear, bilateral network with independent/dependent sources and resistors.
• Two terminals under test for equivalent replacement.
• Standard definitions of open-circuit and short-circuit measurements.

Concept / Approach:

Norton’s current IN is the short-circuit current between the two terminals (with the terminals shorted). Thevenin’s voltage VTH is the open-circuit voltage between the two terminals. The Norton/Thevenin resistance is the same: RN = RTH (under appropriate deactivation rules).

Step-by-Step Solution:

Verification / Alternative check:

By transformation: if you know VTH and RTH, the equivalent Norton current is IN = VTH / RTH, which numerically equals the short-circuit current you would measure directly.

Why Other Options Are Wrong:

• Calling IN an open-circuit current confuses it with VTH definition.
• No averaging is involved; it is strictly the short-circuit current.
• The statement “divided by the shorted resistance” is unclear; the correct relation is IN = VTH / RTH.
• Rated power of the load is irrelevant to the definition.

Common Pitfalls:

Using open-circuit current (which is zero in resistive networks) instead of short-circuit current for Norton calculations, leading to erroneous equivalences.

Final Answer:

False