Norton's Theorem: Equivalent Current Source

Every linear circuit with independent sources, viewed from a pair of terminals, can be reduced to something much simpler than its internal topology suggests. Norton's theorem is the current-source counterpart to Thevenin's theorem: it replaces an arbitrarily complex linear network with a single current source in parallel with a single resistance. This matters whenever you need to analyze how a source drives a variable or nonlinear load — a sensor feeding an ADC input, a photodiode into a transimpedance amplifier, or a power rail feeding a switching regulator — without re-deriving the whole network for every load condition.

Statement of the Theorem

Any linear, bilateral two-terminal network composed of independent/dependent sources and resistors (or, in the AC case, impedances) can be replaced, as seen from its output terminals, by:

• A current source IN in parallel with
• A resistance RN

where:

• IN = the short-circuit current between the two terminals (short the terminals and measure/calculate the current that flows)
• RN = the equivalent resistance looking into the terminals with all independent sources deactivated (voltage sources shorted, current sources opened)

This is exactly the dual of the Thevenin equivalent (VTH in series with RTH). In fact RN = RTH always, and the two equivalents are related by a simple source transformation: IN = VTH / RTH.

Original network          Norton equivalent
   ┌────────┐                ┌────┐
   │ Linear │  A              A
   │ network│──o     ⇒     IN(↑) ── RN ──o
   │(sources│  B              │    │
   │+ R's)  │──o              └────┘──o
   └────────┘                          B

Why It Works: Linearity and Superposition

Norton's theorem is a direct consequence of the fact that any linear network's terminal behavior is fully described by a straight line in the I–V plane:

V = VTH − I·RTH        (Thevenin form: V as a function of I)
I = IN − V/RN           (Norton form: I as a function of V)

Both equations describe the same line — one solved for V as a function of I, the other for I as a function of V. Since V(I) and I(V) are inverse functions of one another, their slopes are reciprocals, not equal: dV/dI = −RTH in the Thevenin form, while dI/dV = −1/RN in the Norton form. Because RN = RTH, this just says the same line can be read "rise over run" in either direction — the magnitude of one slope is RTH, the magnitude of the other is 1/RTH, and multiplying them together always gives 1. A linear one-port therefore reduces to exactly two independent numbers — a resistance and either an open-circuit voltage or a short-circuit current — with Thevenin picking the voltage-intercept description and Norton picking the current-intercept description of the identical line.

Finding IN and RN — Method

1. Identify the two terminals of interest (the port where the load will connect).
2. Remove the load if present.
3. Compute IN: place a short circuit across the terminals and find the current flowing through that short, using KCL/KVL, mesh, or node analysis as appropriate. If dependent sources are present, you generally must solve the full circuit (superposition alone won't isolate their effect).
4. Compute RN: deactivate all independent sources (short voltage sources, open current sources) and find the equivalent resistance looking into the terminals. If dependent sources are present, apply a test source (1 A or 1 V) at the terminals and compute RN = Vtest/Itest with independent sources deactivated.
5. Reconstruct: draw IN in parallel with RN between the two terminals — this is the Norton equivalent, valid for any load connected afterward, as long as the load itself doesn't interact with something inside the deactivated network (e.g., no mutual inductance to the load).

Worked Example

Consider a source network feeding terminals A–B:

        ┌───────┐          A
12 V ───┤  R1   ├───●──────┤R2├────● A
 (+)    │  6Ω   │   │            
        └───────┘   │
   │                 ├──┤R3├────● B (GND)
  GND                │  4Ω
                      │
                (middle node)

Concretely: a 12 V source in series with R1 = 6 Ω feeds a middle node; from that middle node, R2 = 12 Ω goes to terminal A, and R3 = 4 Ω goes to terminal B (which is grounded). Terminal A and B are where we want the Norton equivalent — they are two distinct terminals, not tied together.

Step 1 — Short-circuit current IN. Short A to B. Now R2 (12 Ω) is directly in parallel with R3 (4 Ω), since both run from the middle node to the same shorted point (A=B).

R2 ∥ R3 = (12·4)/(12+4) = 48/16 = 3 Ω

Total resistance seen by the 12 V source: R1 + (R2∥R3) = 6 + 3 = 9 Ω

Total current from source: I_total = 12 V / 9 Ω = 1.333 A

This current splits between R2 and R3 (current divider). The current through R2 (which is the short-circuit current IN, since all of R2's current flows into the short at A-B):

I_R2 = I_total × R3/(R2+R3) = 1.333 × 4/16 = 0.333 A

So IN = 0.333 A (333 mA), flowing from A to B through the short.

Step 2 — Norton resistance RN. Deactivate the 12 V source (replace with a short). Looking into terminals A–B: R2 (12 Ω, from middle node to A) is in series with (R1 ∥ R3), since with the source shorted, R1 and R3 both connect the middle node to ground, and B is that same ground.

R1 ∥ R3 = (6·4)/(6+4) = 24/10 = 2.4 Ω

RN = R2 + (R1∥R3) = 12 + 2.4 = 14.4 Ω

Step 3 — Verification via Thevenin cross-check. Compute VTH (open-circuit voltage at A-B, no load) independently and confirm IN = VTH/RTH.

With A-B open, no current flows through R2 (dead end), so no drop across R2. The voltage at the middle node is a simple divider between the 12 V source through R1 into R3 (R2 carries no current since A is open):

V_mid = 12 × R3/(R1+R3) = 12 × 4/10 = 4.8 V

Since no current flows through R2, V_A = V_mid = 4.8 V. So VTH = 4.8 V.

Check: IN = VTH / RN = 4.8 / 14.4 = 0.333 A ✓ — matches Step 1 exactly.

Result: The network is equivalent to a 0.333 A current source in parallel with a 14.4 Ω resistor, as seen from A-B. Any load — resistive, nonlinear, or even a capacitor for transient studies — can now be attached to this simple two-element equivalent instead of the original three-resistor, one-source network.

Practical Design Implications

• Load sweeps without re-solving the network: once IN and RN are known, computing the voltage or current for any load RL is a one-line current divider: I_RL = IN × RN/(RN+RL).
• Source transformation: Norton and Thevenin equivalents convert into each other freely (VTH = IN·RN, IN = VTH/RN), which is useful when a datasheet gives one form (e.g., a current-output sensor) but your analysis is easier in the other (e.g., mesh analysis prefers voltage sources).
• Sensor modeling: photodiodes, phototransistors, and many current-output transducers are naturally modeled as Norton sources — a current source with finite shunt (leakage) resistance — making Norton's theorem the natural framework rather than an afterthought conversion from Thevenin.
• Superposition limitation: with dependent sources, IN and RN must be found by direct analysis (test source method for RN) — you cannot simply "turn off" a dependent source, since its value depends on other circuit variables.
• AC/impedance version: the theorem extends directly to phasor analysis by replacing R with complex impedance Z — IN and ZN become frequency-dependent, useful for analyzing frequency-domain source loading (e.g., an amplifier's output modeled as a Norton equivalent at each frequency).

Key Takeaways

• Norton's theorem reduces any linear two-terminal network to a current source IN in parallel with resistance RN — the dual of Thevenin's VTH in series with RTH.
• IN = short-circuit current at the terminals; RN = resistance looking in with independent sources deactivated; RN always equals RTH.
• The two equivalents are linked by IN = VTH/RTH, letting you convert between them via source transformation whenever convenient.
• With dependent sources present, compute RN using a test source (Vtest/Itest) rather than simple source deactivation.
• Cross-checking IN, RN, and VTH against each other (IN = VTH/RN) is a fast way to catch analysis errors, as shown in the worked example.
• The equivalent simplifies repeated load analysis to a single current-divider calculation and matches naturally with current-output sensors and transducers.