Norton theorem with example

Statement

Norton's theorem: any linear two-terminal network containing sources and resistors can be replaced, as seen from its two terminals, by a current source I_N in parallel with a resistance R_N.

• I_N = the short-circuit current between the terminals.
• R_N = the resistance seen from the terminals with all independent sources set to zero (voltage sources shorted, current sources opened). R_N equals the Thevenin resistance.

 Linear network          Norton equivalent
 o--------+-----o a       o----+--------+---o a
          |                   (I_N)    [R_N]   load
 (with    |                    |        |
 sources) |                 o--+--------+---o b
 o--------+-----o b

Steps

1. Remove the load and short the two terminals.
2. Find the short-circuit current I_N.
3. Set all sources to zero and find R_N across the terminals.
4. Connect the load across the Norton source: I_L = I_N × R_N/(R_N + R_L).

Example

A 12 V source with 4 Ω in series feeds a 6 Ω load (terminals a, b across the 6 Ω).

• Short a-b: I_N = 12/4 = 3 A.
• Source off (shorted): R_N = 4 Ω.
• Load current: I_L = 3 × 4/(4 + 6) = 1.2 A.
• Load voltage: V_L = 1.2 × 6 = 7.2 V.

Check by the original circuit: I = 12/(4 + 6) = 1.2 A. The results are the same.