How it works
Norton equivalent circuit: short the output terminals and find the short-circuit current IN (using superposition, mesh, or node analysis). Find RN by deactivating all independent sources (voltage sources → short, current sources → open) and calculating the resistance seen at the terminals — same procedure as finding Rth in Thevenin's theorem, so RN = Rth. Conversion: Vth = IN·RN. For circuits with dependent sources: do not deactivate them; instead, apply a test voltage Vt at the terminals and find the resulting current It; then RN = Vt/It. Load current IL = IN · RN/(RN+RL) by current divider.
Key points to remember
Norton's theorem is valid only for linear bilateral networks — non-linear devices like diodes or transistors cannot be included unless the theorem is applied to the linearised small-signal model. The relationship Vth = IN·RN holds exactly between the two equivalent forms of the same network, so once you have one form you can derive the other by simple multiplication. When all sources are deactivated in a network containing only dependent sources, the resistance seen at the terminals is not simply calculated by inspection — always use the test source method. A current source in the Norton equivalent facing a purely resistive load gives maximum power when RL = RN, confirming the link with the maximum power transfer theorem.
Exam tip
The examiner always asks you to find the Norton equivalent of a network at specified terminals when a dependent source is present — always state that you cannot deactivate the dependent source, then apply a 1 V test source at the terminals and find the resulting current to get RN = 1/Itest.