Norton's Theorem
Norton's theorem is the dual of Thevenin's theorem. Where Thevenin represents any linear network as a voltage source in series with a resistance, Norton represents the same network as a current source in parallel with a resistance. Both representations describe the network's behaviour at its terminals with equal accuracy — they are simply different ways of writing the same electrical reality. The conversion between the two forms takes two lines of arithmetic, and choosing which to use is a matter of convenience for the problem at hand. This lesson explains Norton's theorem from first principles, shows how to find the Norton equivalent, demonstrates the conversion, and introduces source transformation — one of the most useful circuit-simplification techniques in electronics.
Norton equivalent: current source In in parallel with Rn. Thevenin equivalent: voltage source Vth in series with Rth. Both describe the same source at the same terminals.
View LargerThe Theorem and Its Meaning
In = the short-circuit current at the terminals (current through a wire connecting the terminals)
Rn = Rth = the resistance seen at the terminals with all independent sources deactivated
The Norton resistance Rn is identical to the Thevenin resistance Rth — it is found by exactly the same deactivation-and-reduction procedure. Only the source representation changes: voltage-source-in-series versus current-source-in-parallel.
The Norton current In is the current that flows through a direct short circuit placed across the output terminals. This is physically the maximum current the source can deliver (since the short is the minimum possible load). In the Thevenin equivalent, the short-circuit current is Vth / Rth — the two representations are consistent.
Why "current source in parallel"? Because a current source maintains constant current regardless of terminal voltage, and placing Rn in parallel allows the terminal voltage to be determined by whatever load is connected. The current from In divides between Rn and RL (a current divider), and VL = IL × RL.
Finding the Norton Equivalent
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Find In — the short-circuit current.
Connect a short circuit (a wire) directly across the output terminals. Calculate the current flowing through that short, using KVL, KCL, mesh analysis, or any other method appropriate to the network. This current is In. The direction of In in the Norton equivalent circuit is the same as the direction of this short-circuit current (from + terminal through the external short and back to − terminal). -
Find Rn — the Norton resistance.
Deactivate all independent sources: short every ideal voltage source, open every ideal current source. Then calculate the resistance seen looking into the open terminals using series-parallel reduction. Rn = Rth by definition. -
Assemble the Norton equivalent and analyse any load.
Draw In as a current source (arrow pointing in the direction of the short-circuit current) in parallel with Rn. Connect any desired load RL in parallel with this combination. Use the current divider and Ohm's Law:
VL = In × (Rn ∥ RL) = In × Rn × RL / (Rn + RL)
IL = VL / RL = In × Rn / (Rn + RL)
Converting Between Thevenin and Norton
The conversion between the two forms is always valid and can be performed at any point in an analysis:
In = Vth / Rth Rn = Rth
Norton → Thevenin:
Vth = In × Rn Rth = Rn
These relationships follow directly from Ohm's Law applied to the equivalent circuits:
- Short-circuit current of Thevenin circuit: Isc = Vth/Rth = In.
- Open-circuit voltage of Norton circuit: Voc = In × Rn = Vth.
The three quantities Vth, In, Rth are related by Ohm's Law. Any two determine the third:
The Thevenin–Norton conversion: both representations of the same source. Converting between them requires only division or multiplication by Rth = Rn.
View LargerSource Transformation: A Powerful Simplification Tool
Source transformation is the direct application of the Thevenin–Norton equivalence to individual elements within a circuit (not just at the output terminals). Any voltage source Vs in series with a resistance R can be replaced at any point in a circuit by a current source Is = Vs/R in parallel with the same R — and vice versa.
This is enormously useful for circuit simplification. When a circuit has a mix of voltage and current sources, converting all sources to one type allows all the parallel (or series) resistors to be combined in a single sweep. The procedure:
- Identify a voltage source Vs with a series resistor R. Replace with current source Is = Vs/R in parallel with R.
- Combine any parallel resistors and parallel current sources at the transformed node.
- If needed, convert back to a voltage source (Thevenin form) using V = Is × Rparallel.
- Continue until the circuit is reduced to a manageable form.
Worked Examples
Example 1 — Norton equivalent of a voltage divider
Step 1 — Find In (short-circuit current):
With the output terminals shorted, R2 is bypassed — it connects from node A to ground, but the short also connects A to ground. So R2 is in parallel with a zero-resistance short, making its effective resistance zero. All current from the source flows through R1 and into the short:
In = Vs / R1 = 12 / 8000 = 1.50 mA
Step 2 — Find Rn (deactivate the 12 V source → short it):
With Vs shorted, R1 connects from node A to ground (its left end is now grounded). R2 also connects from A to ground. They are in parallel:
Rn = R1 ∥ R2 = (8000 × 4000)/(8000 + 4000) = 2667 Ω
Verification — convert Norton back to Thevenin:
Vth = In × Rn = 1.50 mA × 2667 Ω = 4.00 V ✓ (matches Thevenin lesson)
Rth = Rn = 2667 Ω ✓
With RL = 2 kΩ from the Norton equivalent:
VL = In × (Rn ∥ RL) = 1.50 mA × (2667 ∥ 2000)
Rn ∥ RL = (2667 × 2000)/(2667 + 2000) = 5,334,000/4667 = 1143 Ω
VL = 1.50 × 10-3 × 1143 = 1.714 V ✓ (same as Thevenin)
Example 2 — Source transformation to simplify a multi-source circuit
Step 1 — Transform both voltage sources to Norton (current source) form:
Source 1: I1 = V1/R1 = 10/5000 = 2.00 mA (in parallel with R1 = 5 kΩ)
Source 2: I2 = V2/R2 = 6/3000 = 2.00 mA (in parallel with R2 = 3 kΩ)
Step 2 — Both Norton current sources point into node A. Combine parallel resistors and parallel current sources:
Both I1 and I2 push current into node A (check the polarity from each voltage source: V1 pushes + toward A, V2 pushes + toward A, so both current arrows point into node A).
Total Norton current: In,total = I1 + I2 = 2.00 + 2.00 = 4.00 mA
Parallel resistors: R1 ∥ R2 ∥ R3 = 1/(1/5000 + 1/3000 + 1/10000)
= 1/(0.2000 + 0.3333 + 0.1000) mS
= 1/0.6333 mS = 1.579 kΩ
Step 3 — Node voltage:
VA = In,total × Rparallel = 4.00 mA × 1579 Ω = 6.316 V
Verification using KCL at node A:
(VA − V1)/R1 + (VA − V2)/R2 + VA/R3 = 0
(6.316 − 10)/5000 + (6.316 − 6)/3000 + 6.316/10000
= −0.7368 + 0.1053 + 0.6316 = 0.0001 mA ≈ 0 ✓ (rounding)
Source transformation converted a problem that would normally require simultaneous equations into a two-step calculation: combine currents, multiply by parallel resistance.
Example 3 — Norton equivalent for the Wheatstone bridge output
Step 1 — Find In (short the galvanometer terminals):
With the galvanometer shorted: top node connects to Vs+; bottom to ground. The bridge becomes two parallel branches:
Left branch (R1 + R3): 1000 + 1000 = 2000 Ω; Ileft = 5/2000 = 2.50 mA
Right branch (R2 + R4): 1000 + 1100 = 2100 Ω; Iright = 5/2100 = 2.381 mA
V at left-mid node (junction R1/R3): 5 − 2.50 mA × 1000 = 2.500 V
V at right-mid node (junction R2/R4): 5 − 2.381 mA × 1000 = 2.619 V
Short-circuit current through galvanometer (from right-mid to left-mid since right is higher):
In = (2.619 − 2.500) / 0 → cannot compute directly; use KCL instead.
In = Ileft,bottom − Iright,bottom: after the short, recalculate mesh currents. (Alternatively, this is the standard bridge sensitivity calculation, showing that even a small imbalance creates a detectable current through a sensitive galvanometer.)
Simpler approach for small imbalance — use Thevenin Voc and Rth:
Voc = Vs × [R4/(R2+R4) − R3/(R1+R3)] = 5 × [1100/2100 − 1000/2000]
= 5 × [0.5238 − 0.5000] = 5 × 0.0238 = 0.119 V
Rth = (R1 ∥ R3) + (R2 ∥ R4) = 500 + 524 = 1024 Ω
In = Voc/Rth = 119 mV/1024 Ω = 116 µA
With Rg = 50 Ω: Ig = 116 µA × 1024/(1024 + 50) = 111 µA — nearly all In since Rg << Rth.
Norton / Thevenin Conversion Calculator
Norton ↔ Thevenin Conversion Calculator
Enter any two of: Vth, In, Rth/Rn. The calculator finds the third and shows both complete equivalent circuits. Optionally enter RL to find load voltage and current.
Current Source Models in Transistors
Norton's theorem becomes indispensable in transistor circuit analysis because transistors are naturally modelled as current sources in their small-signal equivalent circuits.
In the hybrid-π model of a BJT, the collector appears as a controlled current source gmVbe in parallel with the output resistance ro. This is a Norton equivalent of the transistor's output port. The transconductance gm = IC/VT (where VT ≈ 26 mV at room temperature) plays the role of In, and ro plays the role of Rn. When a load resistor RC is connected to the collector, the Norton current In = gmVbe divides between ro and RC:
Vout = −gmVbe × (ro ∥ RC)
The voltage gain is −gm × (ro ∥ RC). Norton's theorem makes this calculation transparent: the current source drives a parallel combination, giving a voltage proportional to the parallel resistance.
FETs are modelled similarly: a controlled drain current Id = gm × Vgs in parallel with drain-source resistance rds. Both models are Norton equivalents. The load at the drain is in parallel with rds — exactly the setting where Norton analysis is most natural.
Norton's Theorem in Ham Radio
Current source modelling of a receiving antenna
At resonance, a receiving antenna can be modelled as a Norton equivalent: a short-circuit current In proportional to the incident electric field, in parallel with the radiation resistance Rrad. The received power into a matched load (RL = Rn = Rrad) is P = In2 × Rrad / 4. This Norton representation is the basis for the relationship between antenna effective aperture, noise temperature, and received signal power. Connecting the feeder (50 Ω) to an antenna that has a different Rrad is exactly a Norton source driving a mismatched load — the mismatch loss can be calculated directly from the Norton parameters.
Transistor amplifier output impedance
The output stage of a transistor amplifier has a Norton equivalent: the controlled current source (gmVbe) in parallel with the output resistance ro. When a load (the next stage, a speaker, or an antenna) is connected, the current divides between ro and RL. Low-output-impedance stages (Rn much larger than RL) deliver most of the current to the load — a desirable property for voltage amplifiers driving low-impedance loads.
RF power amplifiers
Load-line analysis of RF power amplifiers starts from the Norton equivalent of the active device. The optimum load for maximum output power is derived from the short-circuit current and saturation voltage of the device. Knowing In and Rn directly gives the load line and matching network specifications. This is the starting point for designing output matching networks in HF and VHF PA stages. The output matching network transforms 50 Ω to Ropt for the transistor, while presenting 50 Ω to the antenna connector.
Source transformation in filter design
In ladder filter design, it is mathematically convenient to convert a series voltage source to a parallel current source to maintain the preferred element topology (series or shunt) for the filter synthesis procedure. Source transformation — Norton ↔ Thevenin — allows this conversion at any node without changing the circuit's terminal behaviour, simplifying the design of filter networks from impedance specifications.
Frequently Asked Questions
When should I use Norton rather than Thevenin?
Use Norton when: (1) the load connects in parallel with the source (as in transistor output stages where the load parallels ro); (2) the circuit already contains current sources, making the Norton representation more natural; (3) the short-circuit current is easier to find directly than the open-circuit voltage. Use Thevenin when: (1) the load is in series with the source resistance; (2) you need the open-circuit voltage directly; or (3) a voltage divider or series-path analysis is more natural. Both give identical results for any load — the choice is purely computational convenience.
What is an ideal current source?
An ideal current source maintains a constant current regardless of the voltage across its terminals. Its terminal voltage adjusts automatically to whatever the external circuit demands. The Norton equivalent uses an ideal current source In in parallel with Rn — Rn accounts for the finite output resistance of the real source. A transistor biased in its active region approximates a current source: the collector current is set by the base current and changes relatively little with VCE over a wide range (the flat part of the output characteristic curves).
Is the Norton resistance always equal to the Thevenin resistance?
Yes, always — for linear networks. Rn = Rth by definition. Both are found by the same procedure: deactivate all independent sources and find the resistance at the output terminals. The fact that Vth = In × Rn is consistent with this. If you compute Rth by the open-circuit/short-circuit method (Rth = Voc/Isc), the same result follows from the definitions Voc = Vth and Isc = In.
Can Norton's theorem handle non-linear components?
No — like Thevenin's theorem, Norton's theorem applies only to linear networks. Diodes, transistors in saturation or cut-off, and other non-linear components invalidate the theorem when treated as part of the network being reduced. However, transistors operating in their active region can be replaced by their small-signal linear equivalent circuits, and then Norton's theorem applies to that linear model. This is the standard approach in small-signal RF amplifier analysis.
What is source transformation and when is it most useful?
Source transformation converts a voltage source Vs with a series resistor R into an equivalent current source Is = Vs/R with the same R in parallel (or vice versa). It is most useful when a circuit has multiple sources of mixed type (some voltage, some current) — converting all to one type lets you combine parallel current sources and parallel resistors in one step, then convert back to a Thevenin voltage source. The technique is especially powerful for node analysis: it eliminates voltage sources that would otherwise require constraint equations or supernode treatment.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.