How does Norton's theorem represent a network as a current source with parallel resistance?

In one line

Norton's theorem says any linear two-terminal network can be replaced by a current source \( I_N \) in parallel with a resistance \( R_N \), where \( I_N \) is the short-circuit current at the terminals.

Explained simply

Norton's theorem is the twin of Thevenin's theorem. Instead of describing the hidden circuit as "a voltage behind a resistor", it describes it as "a current that splits between an internal resistor and the load". Both descriptions give the same load current, the same way a phone charger can be described by its voltage or by the current it can supply.

Technically, \( I_N \) is the current that flows when the load terminals are short-circuited, and \( R_N \) is found exactly like \( R_{th} \) (sources replaced by internal resistances), so \( R_N = R_{th} \) and \( V_{th} = I_N R_N \).

How it works

1. Remove \( R_L \) and short the terminals A-B with a wire.
2. Find the current in this short circuit. This is \( I_N \).
3. Remove the short, deactivate all sources, and find the resistance seen from A-B. This is \( R_N \).
4. Draw \( I_N \) in parallel with \( R_N \), reconnect \( R_L \) in parallel.
5. Use the current divider rule to get \( I_L \).

Key formula

\( I_L = I_N \dfrac{R_N}{R_N + R_L} \), with \( R_N = R_{th} \) and \( I_N = V_{th}/R_{th} \).

\( I_N \) = short-circuit current at the load terminals (A), \( R_N \) = Norton resistance (\( \Omega \)), \( R_L \) = load resistance (\( \Omega \)), \( I_L \) = load current (A).

Quick example

Same circuit as the Thevenin example: 12 V, 6 \( \Omega \) series, 3 \( \Omega \) to ground, \( R_L = 2\,\Omega \).

Shorting A-B also shorts the 3 \( \Omega \), so \( I_N = 12/6 = 2 \) A. \( R_N = 6 \parallel 3 = 2\,\Omega \). \( I_L = 2 \times 2/(2+2) = 1 \) A, matching Thevenin. Also \( V_{th} = I_N R_N = 4 \) V.

Common mistakes

• Forgetting that resistors directly across the shorted terminals carry no current when finding \( I_N \).
• Drawing \( R_N \) in series with the current source.
• Using a different resistance for \( R_N \) than for \( R_{th} \); they are always equal.

How MU asks it

Asked as "Find the current in the 10 \( \Omega \) resistor using Norton's theorem" (7–10 marks) or "State Norton's theorem and give the relation between Thevenin and Norton equivalents" (5 marks). For full marks draw the short-circuit circuit, the \( R_N \) circuit and the final Norton equivalent, show the current-divider step, and give the answer with unit.