Any linear two-terminal network can be replaced, as seen from its output terminals, by one ideal voltage source in series with one resistance. This Thevenin equivalent gives exactly the same terminal behaviour for every load, which makes it a compact way to reason about sensors, dividers and interfaces.
Finding the equivalent
The Thevenin voltage is the open-circuit voltage at the terminals. The Thevenin resistance is the resistance seen from the terminals with independent sources reduced to zero: voltage sources replaced by short circuits and current sources by open circuits.
Divider with a load
Problem. A 12 V source feeds a divider with R1 = 4 Ω and R2 = 8 Ω; the output is taken across R2. Find the Thevenin equivalent and the output voltage with an 8 Ω load.
Solution. Open-circuit output: V_th = 12 × 8 ÷ (4 + 8) = 8 V. With the source shorted, R_th = 4 × 8 ÷ (4 + 8) = 2.67 Ω. With an 8 Ω load: V_load = 8 × 8 ÷ (2.67 + 8) = 6.0 V.
Why loading changes the output
The Thevenin resistance forms a second divider with the load. A high R_th relative to R_load means the output sags heavily when loaded. This is why a measurement instrument should have a high input resistance and why a sensor output must be buffered before it drives a low-resistance load.
- V_th is measured with no load connected.
- R_th is found with sources set to zero.
- A load always reduces the terminal voltage of a real source.
- A low R_th source holds its voltage better under load.
- A linear network is equivalent to V_th in series with R_th.
- V_th is the open-circuit terminal voltage.
- R_th is the terminal resistance with sources zeroed.
- Loading a divider always lowers its output.
Worked example
A sensor has V_th = 2.5 V and R_th = 10 kΩ. Connected to a 10 kΩ load, the output is 2.5 × 10 ÷ 20 = 1.25 V — half the open-circuit value. A buffer amplifier prevents this loss.