Thevenin's Theorem: Steps and Solved Examples

Thevenin’s theorem is one of the most useful tools in circuit analysis. It lets an engineer collapse a large network into a two-element model and then study what happens when only the load changes. This guide explains the theorem, gives the step by step method, covers circuits with dependent sources, links Thevenin and Norton forms, and solves a complete example including maximum power transfer.

What Thevenin’s theorem says

The French engineer Léon Charles Thévenin published the theorem in 1883. It applies to any circuit made of linear elements (resistors, inductors, capacitors, linear dependent sources) and independent sources. Looking into two terminals A and B, the whole circuit behaves exactly like:

• an ideal voltage source V_th, equal to the voltage across A and B when nothing is connected, in series with
• a resistance R_th, equal to the resistance measured across A and B with independent sources deactivated.

“Behaves exactly like” means that for any load connected across A and B, the load voltage and current are the same in the original circuit and in the equivalent. What happens inside the original circuit, such as the power lost in its internal resistors, is not represented by the equivalent.

The load current follows directly from the model:

$$I_L = \frac{V_{th}}{R_{th} + R_L}$$

Download diagram (WebP)

Steps to apply Thevenin’s theorem

1. Identify the load. Choose the element or branch you want to study and remove it. Mark the two terminals A and B where it was connected.
2. Find V_th. Calculate the open-circuit voltage between A and B using any method: voltage divider, nodal analysis, mesh analysis or superposition. The basic tools are covered in Ohm’s law and Kirchhoff’s laws.
3. Find R_th. Turn off every independent source: replace voltage sources with short circuits and current sources with open circuits. Then find the equivalent resistance between A and B by series and parallel reduction. Where resistors form a bridge, a star delta transformation may be needed.
4. Draw the equivalent. Place V_th in series with R_th, with the polarity that gives the open-circuit voltage you found.
5. Reconnect the load. Calculate the load current and voltage from the simple series circuit.

An alternative to step 3 is to find the short-circuit current I_sc between A and B. Then R_th = V_th / I_sc. This is also how an equivalent is found from measurements on real equipment: measure the open-circuit voltage, then the current into a known load, and solve for R_th.

Circuits with dependent sources

Dependent (controlled) sources must not be switched off, because their value depends on a voltage or current elsewhere in the circuit. Two methods are used:

• Test source method: turn off independent sources only, apply a test voltage V_t across A and B, find the current I_t it delivers, and take R_th = V_t / I_t.
• Open circuit and short circuit method: find V_th and I_sc with all sources active, then divide.

With dependent sources R_th can even come out negative, which indicates that the circuit can supply energy, as in an amplifier model.

Thevenin and Norton equivalents

Norton’s theorem gives the dual result: the same terminals can be modelled by a current source I_N in parallel with a resistance R_N. The two forms are related by source transformation:

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

Use whichever form makes the next step easier. A Thevenin model suits loads added in series; a Norton model suits loads added in parallel.

Maximum power transfer

For a source with a fixed Thevenin resistance, the power delivered to a resistive load is greatest when R_L = R_th. The maximum power is then:

$$P_{max} = \frac{V_{th}^2}{4 R_{th}}$$

At this point the efficiency is only 50 percent, because equal power is lost in R_th. Power systems therefore do not operate at matched conditions; they keep the source resistance much smaller than the load to achieve high efficiency. Matching is used in communication and signal circuits, where getting the most power to a load matters more than efficiency. In AC circuits the condition becomes a load impedance equal to the complex conjugate of the source impedance.

To check the result, solve the original circuit directly with the 4 Ω load. R3 plus R_L is 6 Ω, which in parallel with R2 gives 2 Ω. The total resistance is 8 Ω, so the source current is 3 A and the voltage across R2 is 6 V. The load branch current is 6 / 6 = 1 A, matching the Thevenin result.

Where Thevenin’s theorem is used

• Load studies: finding the current in one branch for many values of load without re-solving the whole network.
• Fault calculations: the Thevenin impedance at a bus in a power system sets the short circuit current at that point.
• Electronics: modelling a sensor, signal source or bias network as a source with an output resistance, which shows how much the reading drops when a meter or amplifier input loads it.
• Instrumentation: checking that a transmitter loop or bridge output is not pulled down by the input resistance of the measuring device.

Common mistakes

• Leaving the load connected while finding V_th.
• Opening a voltage source instead of shorting it when finding R_th.
• Switching off dependent sources.
• Using the equivalent to calculate power inside the original network.
• Getting the polarity of V_th wrong, which reverses the load current.

Related reading

• Ohm’s Law and Kirchhoff’s Laws with Worked Examples
• Power Factor Explained: Causes, Effects and Correction
• Cable Sizing: Current Rating and Voltage Drop Explained