Learn › Topic 5

Thévenin, Norton and Superposition with Worked Examples

8 min read · bitGateLab Team

Often you are not interested in the whole circuit, only in what happens to one part of it: a load, a sensor, the input of another stage. The Thévenin and Norton theorems let you replace everything else with a model made of just two elements, and the superposition principle lets you analyze one source at a time. Here you'll learn all three, how to convert from one equivalent to the other, and which load receives the most power from a source.

Thévenin's theorem

Any linear network of resistors and sources, seen from two terminals a and b, behaves exactly like a voltage source V_th in series with a resistance R_th. "Exactly like" means that any load connected between a and b receives the same voltage and the same current in both cases.

V_th is the open-circuit voltage between a and b (with no load).
R_th is the resistance seen from a and b with the independent sources turned off: voltage sources are replaced by a wire and current sources by an open circuit.

Left: the original network seen from terminals a and b. Right: its Thévenin equivalent.

Example 1. Find the Thévenin equivalent of the circuit on the left.

V_th: with no load, no current flows through R₃, so no voltage drops across it and V_ab is the voltage across R₂. It is a divider: V_th = 12 × 3 / (6 + 3) = 4 V.
R_th: turn off the 12 V source (replace it with a wire). From a you see R₃ in series with the parallel combination of R₁ and R₂: R_th = 2 + (6 ∥ 3) = 2 + 2 = 4 kΩ.
With a 4 kΩ load, for example, the current is 4 V / (4 + 4) kΩ = 0.5 mA and the voltage across the load is 2 V. Solving the full network gives the same result, but with far more work.

Norton's theorem

The same network is also equivalent to a current source I_N in parallel with a resistance R_N. I_N is the current that flows if you short terminals a and b together, and R_N is the same as R_th. The two models are converted into each other with Ohm's law:

R_N = R_th        I_N = V_th / R_th        V_th = I_N · R_N

Example 2. Find the Norton equivalent of the same circuit by measuring the short-circuit current, and verify that it agrees with the conversion.

With a and b shorted, R₃ is in parallel with R₂: 2 ∥ 3 = 1.2 kΩ.
The source sees 6 + 1.2 = 7.2 kΩ, so it delivers 12 / 7.2 = 1.667 mA.
The voltage across the parallel combination is 1.667 × 1.2 = 2 V, and the current through R₃ (the short-circuit current) is 2 V / 2 kΩ = 1 mA.
By conversion: I_N = 4 V / 4 kΩ = 1 mA. It matches. The Norton equivalent is 1 mA in parallel with 4 kΩ.

Key point: if you can measure the open-circuit voltage and the short-circuit current of a real source, you already know its complete model: R_th = V_oc / I_sc. When a short circuit would be dangerous, measure the voltage with two different known loads and solve for the model.

Superposition principle

In a linear circuit with several independent sources, any voltage or current is the sum of the contributions of each source acting alone. To find the contribution of one source, you turn off all the others (voltage → wire, current → open), solve, and at the end add the results with their signs.

Example 3. A source V₁ = 12 V is connected to node X through R₁ = 6 kΩ, another source V₂ = 6 V reaches the same node through R₃ = 2 kΩ, and from X to ground there is R₂ = 3 kΩ. Find V_X by superposition.

V₁ alone (V₂ shorted): R₃ is in parallel with R₂, 2 ∥ 3 = 1.2 kΩ. Divider: V′ = 12 × 1.2 / (6 + 1.2) = 2 V.
V₂ alone (V₁ shorted): R₁ is in parallel with R₂, 6 ∥ 3 = 2 kΩ. Divider: V″ = 6 × 2 / (2 + 2) = 3 V.
Total: V_X = 2 + 3 = 5 V.
Check with nodal analysis: (12 − V)/6 + (6 − V)/2 = V/3. Multiplying by 6: 12 − V + 18 − 3V = 2V, so V = 5 V.

Superposition works for voltages and currents, but not for power, because power depends on the square. In the example, R₂ dissipates 5² / 3 = 8.33 mW; adding the partial powers would give 2²/3 + 3²/3 = 4.33 mW, which is wrong.

Maximum power transfer

If you connect a load R_L to a Thévenin equivalent, the power in the load is P = V_th² · R_L / (R_th + R_L)². This power is greatest when R_L = R_th, and its value is:

P_max = V_th² / (4 R_th)

For the example (4 V, 4 kΩ):

R_L	Current	Power in R_L
2 kΩ	0.667 mA	0.889 mW
4 kΩ	0.500 mA	1.000 mW
8 kΩ	0.333 mA	0.889 mW

Careful: at the maximum-power point, half the energy is lost in R_th, so the efficiency is only 50 %. That is why this kind of matching is sought for weak signals (antennas, sensors) but not in power distribution, where you want R_L to be much larger than R_th.

Common mistakes

When calculating R_th, removing the current sources by replacing them with a wire instead of leaving them open.
Calculating V_th with the load still connected; it must be the open-circuit voltage.
Adding powers in superposition.
Forgetting the signs when adding contributions: a source can subtract voltage at a node.
Believing that maximum power transfer also means maximum efficiency.

Try it in the simulator

You can verify that the original circuit and its equivalent are indistinguishable from the load's point of view. Build the two side by side and connect the same load to each.

Build the original circuit: a 12 V source, R₁ = 6 kΩ, R₂ = 3 kΩ and R₃ = 2 kΩ.
With no load, measure the voltage between a and b with the multimeter: it should read 4 V.
Connect an ammeter between a and b (a short circuit) and observe 1 mA.
Separately, build a 4 V source with a 4 kΩ resistor in series.
Connect the same load (2 kΩ, 4 kΩ, 8 kΩ) to both circuits and compare voltage and current.
Calculate the power in the load for each value and notice that it peaks at 4 kΩ.

Exercises

A divider with two 1 kΩ resistors is powered from 10 V. Find its Thévenin equivalent seen from the output and the maximum power it can deliver to a load.

Show answer

V_th = 5 V, R_th = 1 ∥ 1 = 500 Ω. P_max = 5² / (4 × 500) = 12.5 mW, with a 500 Ω load.
A network has a Norton equivalent of I_N = 2 mA and R_N = 5 kΩ. What is its Thévenin equivalent?

Show answer

R_th = 5 kΩ and V_th = 2 mA × 5 kΩ = 10 V.
A 9 V source is connected to node X through 3 kΩ; from X to ground there is 6 kΩ, and a 3 mA current source injects current into node X. Find V_X by superposition.

Show answer

The 9 V source alone (current source open): 9 × 6 / 9 = 6 V. The 3 mA source alone (voltage source shorted): 3 mA × (3 ∥ 6 = 2 kΩ) = 6 V. Total: 12 V. Nodal check: (9 − 12)/3 + 3 = 2 mA = 12/6.
An output stage has V_th = 8 V and R_th = 8 Ω. What is the maximum power it delivers, and into what load?

Show answer

With R_L = 8 Ω: P_max = 64 / 32 = 2 W.

Review how to set up systems of equations in nodal and mesh analysis and continue with capacitors, inductors and RC and RL circuits.