Network Theorems: Thevenin, Norton, Superposition

Network theorems turn a large linear circuit into a smaller problem. They do not replace Kirchhoff's laws; they package the same physics into forms that are faster to reuse.

Learning objectives

After this lesson, you will be able to:

• find Thevenin voltage and resistance seen from a load;
• convert between Thevenin and Norton equivalents;
• apply superposition correctly in linear circuits with multiple independent sources;
• recognize when dependent sources require a test-source method;
• avoid incorrect use of superposition for power calculations.

Thevenin's theorem

Any linear two-terminal network of sources and impedances can be replaced, as seen from its output terminals, by:

• a voltage source V_th;
• in series with a resistance R_th in DC resistive circuits.

BT1: Device:Battery value="Vth" rotate=0
R1: Device:R value="Rth" rotate=270
RL: Device:R value="Load RL" rotate=0

layout direction=LR gap=70
group TH label="Thevenin equivalent seen by the load" direction=LR gap=70 {
  BT1 R1 RL
}

BT1.1 --> R1.1 color=#b91c1c
R1.2 --> RL.1 color=#2563eb
RL.2 --> BT1.2 color=#334155

The Thevenin source is the left branch, Rth is in the upper series path, and the load is the right branch with the lower wire completing the loop.

Thevenin procedure

1. Identify the two terminals where the load connects.
2. Remove the load.
3. Find the open-circuit voltage across those terminals: V_th.
4. Find the resistance seen looking back into the network: R_th.
5. Reattach the load to the Thevenin equivalent.

Source deactivation rules for R_th

• independent voltage source -> replace with a short circuit;
• independent current source -> replace with an open circuit.

If dependent sources are present, do not simply turn them off. Use a test source and compute the resulting V/I seen at the port.

Worked example 1: Thevenin equivalent of a divider

Consider a 12 V source feeding R_1 = 10 Ω in series with R_2 = 20 Ω, with the load connected across R_2.

Step 1: find V_th

With the load removed, the output is just the divider voltage:

$$
V_{th} = 12\frac{20}{10 + 20} = 8\ \text{V}
$$

Step 2: find R_th

Short the 12 V source and look back into the output terminals:

$$
R_{th} = R_1 \parallel R_2 = \frac{10 \times 20}{10 + 20} = 6.67\ \Omega
$$

So the original network is equivalent to an 8 V source in series with 6.67 Ω.

Norton's theorem

The same linear two-terminal network can also be replaced by:

• a current source I_N;
• in parallel with a resistance R_N.

I1: Simulation_SPICE:IDC value="IN" caption="↑ IN current source" rotate=0
R1: Device:R value="Rn" rotate=0
RL: Device:R value="Load RL" rotate=0

layout direction=LR gap=95
group NOR label="Norton equivalent seen by the load" direction=LR gap=95 {
  I1 R1 RL
}

I1.1 --> R1.1 color=#b91c1c
R1.1 --> RL.1 color=#b91c1c
I1.2 --> R1.2 color=#334155
R1.2 --> RL.2 color=#334155

The Norton current source, Rn, and the load all connect across the same top and bottom rails. The vertical branches make the parallel relationship explicit.

For resistive linear networks:

$$
R_N = R_{th}
$$

and

$$
I_N = \frac{V_{th}}{R_{th}}
$$

Worked example 2: convert Thevenin to Norton

From the previous result:

$$
I_N = \frac{8}{6.67} \approx 1.20\ \text{A}
$$

So the Norton equivalent is approximately:

• 1.20 A current source;
• in parallel with 6.67 Ω.

Interactive Thevenin/Norton converter

Enter any two values to compute the third.

Superposition theorem

In a linear circuit with multiple independent sources, the total voltage or current in an element equals the algebraic sum of the contributions from each source acting alone.

How to turn off other sources

• independent voltage source -> short circuit;
• independent current source -> open circuit.

Dependent sources remain active because they are part of the circuit behavior.

Worked example 3: superposition current

Suppose two sources drive the same load branch.

• Source V_1 alone produces +3 mA through the load from left to right.
• Source V_2 alone produces -1 mA through the load from left to right, meaning 1 mA right to left.

Then the total load current is:

$$
I_{load} = 3 + (-1) = 2\ \text{mA}
$$

The negative contribution is not a problem; it simply opposes the reference direction.

Which theorem to choose

Practical engineering uses

Common mistakes

• Forgetting to remove the load before finding V_th.
• Turning off a voltage source by opening it instead of shorting it.
• Turning off a current source by shorting it instead of opening it.
• Turning off dependent sources during equivalent-resistance calculation.
• Adding powers from separate superposition runs as though power were linear.
• Using these theorems on nonlinear elements without checking the model assumptions.

Summary

• Thevenin replaces a linear two-terminal network with V_th in series with R_th.
• Norton replaces the same network with I_N in parallel with R_N.
• The two forms are equivalent and convertible.
• Superposition lets you add voltage or current contributions from independent sources one at a time.
• These methods simplify analysis, but only when source deactivation and linearity assumptions are handled correctly.

Next: Measuring Circuits: Multimeter Basics and Safety.

Further reading

• OpenStax: DC Circuit Analysis and Network Theorems
• All About Circuits: Thevenin's Theorem
• All About Circuits: Superposition Theorem