The theorem that a circuit of sources and linear elements, seen from a pair of terminals, can be replaced by a single voltage source Vth in series with a single impedance Zth. Vth is the voltage with the terminals open, and Zth is the impedance seen from the terminals with the independent sources in the circuit set to 0 (voltage sources shorted, current sources opened). It gives the current and power for any load immediately, and works directly for AC with phasors.
Japanese name: テブナンの定理
Thevenin's theorem appears in 7 practice questions across 3 topics, including Circuit Analysis: AC, Linear Circuits and Transients, Circuit and electromagnetism basics, Analog and electronic circuits.
Thevenin equivalent circuit with a load impedance · View in diagrams
First remove the load and find the open-circuit voltage between the terminals; this is Vth. Next set the independent sources in the circuit to 0 and find the combined impedance seen from the terminals; this is Zth. If the current Isc with the terminals shorted is known, Zth = Vth/Isc also works. This method can be used even for circuits with dependent sources.
Circuits with dependent sources
In circuits containing dependent sources proportional to a current or voltage in the circuit, such as the small-signal equivalent circuit of a transistor, the dependent sources must not be set to 0. Find Zth from the ratio of the open-circuit voltage to the current with the terminals shorted, or set only the independent sources to 0, apply a test voltage to the terminals and take the ratio to the current that flows in. With dependent sources, Zth can even appear negative.
Norton's theorem and maximum power
The same circuit can also be represented as a current source Isc in parallel with Zth (the Norton equivalent), and the two convert into each other through Vth = Zth × Isc. The Thevenin equivalent also shows at once when the power consumed by the load is maximum: at DC when the load resistance equals Rth, and at AC when the load impedance is the complex conjugate of Zth; the maximum power is Vth²/(4Rth).
Vth is the open-circuit output voltage: 10 V × 4/(6 + 4) = 4 V. Rth is the resistance seen from the output with the voltage source shorted; R1 and R2 then appear in parallel, giving 6 × 4/(6 + 4) = 2.4 kΩ. With this equivalent, the current into a load RL is simply 4 V/(2.4 kΩ + RL). In circuits containing current sources, those are opened when finding Rth.
Method 1 (conjugate matching): choosing ZL as the complex conjugate of Zth, 3 − j4 Ω, cancels the reactances, so the circuit impedance is just a 6 Ω resistance. The current is 10/6 ≈ 1.67 A and the load power is (10/6)² × 3 ≈ 8.3 W. In general, Pmax = Vth²/(4Rth) = 100/12 ≈ 8.3 W. Method 2 (maximize the formula): with ZL = RL + jXL, P = Vth² RL/((3 + RL)² + (4 + XL)²); first set XL = −4 Ω to remove the reactive part of the denominator, then choose RL = 3 Ω, which maximizes RL/(3 + RL)², giving the same result.
FAQ
Can Thevenin's theorem be used for any circuit?
It applies to linear circuits. Circuits with nonlinear elements such as diodes can be treated as linear only for small changes around an operating point. Also, only the voltage-current relation at the terminals is equivalent; the power dissipated inside the circuit is not.
What happens to the terminal voltage when a load is connected?
Vth is divided between Zth and the load. For example, connecting a 2.4 kΩ load to a circuit with Vth = 4 V and Rth = 2.4 kΩ halves the terminal voltage to 4 × 2.4/(2.4 + 2.4) = 2 V. If the load is much larger than Rth, the terminal voltage stays close to Vth.
Thevenin's theorem in the explanations
With impedances and phasors, the same Thevenin's theorem as for DC applies directly to AC.
Thevenin's theorem says a linear circuit, seen from two terminals, is equivalent to a voltage source equal to the open-circuit voltage Vth in series with the internal resistance Rth (measured with sources removed).