Thevenin's Theorem
Thevenin's theorem is the most powerful simplification tool in DC circuit analysis. It states that any combination of voltage sources, current sources, and resistors — no matter how complex — can be reduced to a single voltage source (Vth) in series with a single resistance (Rth), from the perspective of any two output terminals. Once you have the Thevenin equivalent, analysis of any load connected to those terminals becomes trivial: it is just a two-element series circuit. The theorem does not change anything inside the original network — it changes only what a load sees from the outside. This distinction is worth keeping clearly in mind as you work through the procedure.
Thevenin's theorem: any linear network reduces to a single voltage source Vth in series with a single resistance Rth, as seen from two output terminals.
View LargerThe Theorem and Its Physical Meaning
Vth = the open-circuit voltage at the terminals (no load connected)
Rth = the resistance seen at the terminals with all independent sources deactivated
The critical phrase is "from the perspective of two external terminals." Inside the original network, nothing changes — the theorem only describes what any load connected to those two terminals will experience. Whatever load you attach, it receives exactly the same current and voltage as it would from the full original network. The Thevenin equivalent is a different circuit that is externally identical.
The theorem applies to linear networks — those built from resistors, linear dependent sources, and independent sources. Non-linear components like diodes and transistors operating in their non-linear region cannot be included directly, but small-signal AC models of transistors are linear, so Thevenin applies to small-signal amplifier analysis.
The three-step Thevenin reduction: find Vth (open-circuit voltage), find Rth (resistance with sources deactivated), then assemble the equivalent and connect any desired load.
View LargerWhy Source Deactivation is Valid
A common question is: why is it valid to deactivate (short or open) sources when finding Rth? The answer comes from the superposition principle.
In any linear network, the response at the terminals can be written as:
Vterminal = (contribution from all internal sources) + (contribution from the terminal voltage itself if a test source is applied)
The Thevenin resistance is defined as the ratio of a small test voltage applied at the terminals to the current it drives, with all internal sources set to zero. Setting sources to zero means:
- An ideal voltage source has zero volts — which means it is a short circuit (zero resistance, zero voltage drop).
- An ideal current source has zero current — which means it is an open circuit (infinite resistance, zero current).
With internal sources zeroed, only the resistive network remains, and the resistance seen looking into the terminals is Rth. This is physically the "output impedance" of the network — the resistance the terminal voltage must overcome to drive current into the network when the internal sources are not contributing.
The Three-Step Procedure
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Remove the load — leave the terminals open. Find Vth.
With no load connected, no current flows out of the terminals. Calculate the voltage appearing across the open terminals using any method: series-parallel reduction, voltage divider formula, KVL, KCL, or node analysis. This voltage is Vth. -
Deactivate all independent sources. Find Rth.
Replace every ideal voltage source with a wire (short circuit). Replace every ideal current source with an open circuit (remove it). Then calculate the resistance seen looking into the open terminals using series-parallel reduction. If the network has only independent sources (no dependent sources), this is always a straightforward series-parallel calculation. -
Assemble the Thevenin equivalent and analyse any load.
Draw Vth in series with Rth between the two output terminals. Connect any load RL and analyse:
IL = Vth / (Rth + RL)
VL = IL × RL = Vth × RL / (Rth + RL)
Worked Examples
Example 1 — Voltage Divider Network (standard case)
Step 1 — Find Vth (open-circuit voltage at A−ground):
With no load, R1 and R2 form a voltage divider:
Vth = 12 × R2/(R1 + R2) = 12 × 4000/(8000 + 4000) = 12 × 1/3 = 4.00 V
Step 2 — Find Rth (short the 12 V source, look into terminals):
With the source shorted, R1's left end connects to ground. Both R1 and R2 now connect between node A and ground → they are in parallel:
Rth = R1 ∥ R2 = (8000 × 4000)/(8000 + 4000) = 32,000,000/12,000 = 2667 Ω
Step 3 — Load analysis with RL = 2 kΩ:
IL = Vth/(Rth + RL) = 4.00/(2667 + 2000) = 4.00/4667 = 0.857 mA
VL = 0.857 mA × 2000 Ω = 1.714 V
Power of the Thevenin approach: Once Vth and Rth are known, any other load can be analysed instantly:
RL = 10 kΩ: VL = 4.00 × 10000/(2667 + 10000) = 3.16 V
RL = 100 Ω: VL = 4.00 × 100/(2667 + 100) = 0.145 V
RL = ∞ (open): VL = Vth = 4.00 V
Example 2 — Finding the current through a specific resistor using Thevenin
Strategy: Treat R3 as the "load." Find the Thevenin equivalent of everything else seen from R3's terminals (node A and ground, with R3 removed).
Step 1 — Find Vth with R3 removed:
With R3 open, node A is fed by a voltage divider of R1 and R2:
Vth = 15 × 2000/(1000 + 2000) = 15 × 2/3 = 10.00 V
Step 2 — Find Rth (short the 15 V source):
With the source shorted, R1 connects from node A to ground (left end of R1 is now grounded). R2 also connects from node A to ground. They are in parallel:
Rth = R1 ∥ R2 = (1000 × 2000)/(1000 + 2000) = 2,000,000/3000 = 667 Ω
Step 3 — Current through R3 from the Thevenin equivalent:
IR3 = Vth/(Rth + R3) = 10.00/(667 + 3000) = 10.00/3667 = 2.727 mA
Verify using full circuit analysis:
R2 ∥ R3 = (2000 × 3000)/(2000 + 3000) = 1200 Ω
Itotal = 15/(1000 + 1200) = 15/2200 = 6.818 mA
VA = 6.818 mA × 1200 Ω = 8.182 V
IR3 = 8.182/3000 = 2.727 mA ✓
Example 3 — T-network: a more complex reduction
Step 1 — Find Vth (open-circuit at B with RL removed):
With RL open, no current flows through Rc (since B is open, nothing draws current through Rc). Therefore the voltage at node A is:
VA = Vs × Rb/(Ra + Rb) = 24 × 2000/(4000 + 2000) = 24 × 1/3 = 8.00 V
Because no current flows through Rc, there is no voltage drop across Rc, so VB = VA:
Vth = 8.00 V
Step 2 — Find Rth (short the 24 V source, look into terminals B and ground):
With Vs shorted: Ra's left end is now connected to ground. Looking from terminal B toward the network:
• Rc connects from B to node A.
• From node A, Ra (grounded left end) and Rb (grounded bottom end) are both connected to ground → Ra ∥ Rb = (4000 × 2000)/(4000 + 2000) = 8,000,000/6000 = 1333 Ω
• From terminal B, we see Rc in series with (Ra ∥ Rb):
Rth = Rc + (Ra ∥ Rb) = 6000 + 1333 = 7333 Ω
Step 3 — With RL = 3 kΩ:
IL = Vth/(Rth + RL) = 8.00/(7333 + 3000) = 8.00/10333 = 0.774 mA
VL = 0.774 mA × 3000 = 2.322 V
Loading error: (8.00 − 2.322)/8.00 = 71% — a very stiff source (Rth = 7.3 kΩ) loaded by a comparable RL produces significant droop. This circuit needs a lower-impedance output stage.
What Does Rth Represent Physically?
The Thevenin resistance Rth is the source impedance — it is the internal resistance that any load must "fight" to extract current from the network. It has several physically meaningful interpretations:
- It determines loading: If RL >> Rth, the load voltage is close to Vth (small loading error). If RL ≈ Rth, the load sees roughly half of Vth.
- It is the resistance of a potentiometer wiper: For any voltage divider, Rth = R1 ∥ R2. As you move between the extremes, Rth varies from 0 at either end to its maximum (R1 ∥ R2) at the mid-point.
- It is the output impedance of the network: If you inject a test current into the open terminals (with internal sources zeroed), the voltage that appears equals Itest × Rth. This is how output impedance is measured in the lab.
- It sets the maximum power transfer condition: Maximum power is transferred to a load when RL = Rth (covered in the Maximum Power Transfer lesson).
- A low Rth means a "stiff" source: A car battery has Rth ≈ 0.01 Ω — it can supply huge currents without its terminal voltage dropping much. A 9 V PP3 battery has Rth ≈ 10 Ω — it droops noticeably under load.
Thevenin and the Loading Effect
The Thevenin resistance provides a precise explanation of the loading effect you observed in the Voltage Dividers lesson:
- If RL ≫ Rth: VL ≈ Vth — negligible loading error (the 10:1 rule from the voltage divider lesson: use RL ≥ 10 × Rth).
- If RL = Rth: VL = Vth/2 — 50% voltage drop. Also the maximum power transfer condition.
- If RL << Rth: VL ≈ 0 — the source is being heavily loaded; almost all voltage drops across Rth itself.
Minimising Rth relative to RL minimises loading. This is why op-amp output stages are designed for near-zero output impedance: Rth ≈ 0 means the load voltage equals Vth regardless of what load is connected. A low Rth is the hallmark of a "stiff" voltage source.
Thevenin Equivalent Calculator
Thevenin Equivalent — Voltage Divider Network
For the most common case: a voltage source Vs with R1 (top) and R2 (bottom) in series. Output taken across R2. Enter Vs, R1, R2, and optionally a load RL.
⚖ Experiment: Measure Vth and Rth Directly
Measure the Thevenin parameters of a real voltage divider by the open-circuit / loaded-voltage method, then verify using a second known load.
- 9 V battery and clip connector
- R1: 10 kΩ; R2: 4.7 kΩ (the divider under test)
- Load resistors: 2.2 kΩ and 10 kΩ
- Breadboard and jumper wires
- Digital multimeter
- Build the divider: battery → 10 kΩ → node A → 4.7 kΩ → ground. Measure V at node A with nothing else connected. This is Vth(measured).
- Calculate the predicted Vth = 9 × 4700/(10000 + 4700) using the calculator. Compare with your measurement.
- Connect the 2.2 kΩ load from node A to ground. Measure VL.
- From these two measurements, calculate the experimental Rth:
Rth = (Vth − VL) × RL / VL
(Derivation: VL = Vth × RL/(Rth+RL), rearranging gives Rth.) - Compare your experimental Rth to the predicted value R1 ∥ R2 = 10k ∥ 4.7k from the calculator.
- Predict VL for the 10 kΩ load using your measured Vth and Rth. Then connect the 10 kΩ load and measure the actual voltage. Compare.
Predicted Vth = 9 × 4700/14700 ≈ 2.88 V. Predicted Rth = 10k ∥ 4.7k ≈ 3.19 kΩ. With 2.2 kΩ load: predicted VL = 2.88 × 2200/(3190 + 2200) ≈ 1.17 V (loading error ≈ 59% — significant because 2.2 kΩ is less than Rth = 3.19 kΩ). With 10 kΩ load: predicted VL = 2.88 × 10000/(3190 + 10000) ≈ 2.18 V (loading error ≈ 24%). The Thevenin model predicts both loaded voltages from just Vth and Rth.
Thevenin's Theorem in Ham Radio
Bias network analysis
The voltage-divider bias network of a transistor amplifier is most efficiently analysed using its Thevenin equivalent. The two bias resistors R1 and R2 produce Vth = VCC × R2/(R1+R2) and Rth = R1 ∥ R2. This Thevenin source then drives the transistor base through Rth in series with VBE and RE. A single KVL equation around the loop gives the emitter current. This is the standard textbook analysis for every BJT common-emitter amplifier stage. Without the Thevenin simplification, solving the full mesh equations for the bias network would be far more complex.
Source impedance in receiver front ends
The antenna feedline, balun, and any attenuator pads ahead of a receiver all contribute to the Thevenin source impedance the receiver input stage sees. If this Rth differs significantly from the receiver's 50 Ω input impedance, signal power is lost (loading effect) and noise figure degrades. The Thevenin model makes it clear that the signal amplitude at the receiver input is Vth × Zin/(Rth + Zin) — a simple voltage divider between Rth and the receiver's input impedance. Matching Rth to Zin maximises signal transfer.
Power supply regulation
A real power supply has an internal resistance — its Thevenin resistance — which causes the output voltage to drop under load. A 13.8 V regulated supply with 0.05 Ω internal resistance drops to 12.8 V at 20 A (the load current of a 100 W transceiver on transmit). Knowing Rth for your supply tells you the voltage droop at full transmit load and whether a heavier-gauge cable (lower series resistance) is needed to maintain adequate supply voltage to the transceiver.
Impedance matching analysis
Maximum power transfer (the lesson after Norton's theorem) requires RL = Rth. This principle underpins antenna matching, interstage coupling, and transmission line termination in RF systems. Without the Thevenin concept, the maximum power condition would lack practical meaning — the theorem provides both the framework and the calculation procedure for any matching problem.
Frequently Asked Questions
Can Thevenin's theorem be applied to circuits with dependent sources?
Yes, but Rth cannot be found by simple deactivation and series-parallel reduction when dependent sources are present. A dependent source cannot be set to zero because it depends on a circuit variable that changes when the circuit changes. Instead, find Rth using: Rth = Voc / Isc (open-circuit voltage divided by short-circuit current). Alternatively, apply a test voltage Vtest at the terminals with all independent sources deactivated and find Rth = Vtest / Itest. This method works for any linear network, with or without dependent sources.
What is the difference between Thevenin's theorem and Norton's theorem?
They produce equivalent models of the same source expressed differently. Thevenin uses a voltage source Vth in series with Rth. Norton uses a current source In = Vth/Rth in parallel with Rn = Rth. Any source can be expressed in either form. Thevenin is more natural when a load connects across output terminals (voltage measurement perspective) or when series-path analysis is simpler. Norton is more natural when loads are in parallel with the source or when the short-circuit current is easier to find directly. Both give identical results for any load.
Does Thevenin's theorem apply to AC circuits?
Yes — for linear AC circuits, Thevenin's theorem applies with all quantities expressed as phasors (complex numbers with magnitude and phase angle). Vth becomes a complex phasor Vth∠θ and Rth becomes a complex impedance Zth = Rth + jXth. The procedure is identical: find the open-circuit voltage phasor and the impedance with sources deactivated. This is the foundation of AC circuit analysis, filter design, and RF network analysis.
How do I measure Rth in a real circuit without removing components?
Measure the open-circuit voltage Voc, then connect a known load RL and measure VL. Then: Rth = (Voc − VL) × RL / VL. This avoids the risk of short-circuiting a live circuit (which the Voc/Isc method would require). The experiment in this lesson uses exactly this approach. Choose RL such that VL is noticeably different from Voc — too large an RL gives very little droop and poor precision in the Rth estimate.
Why is Vth called the "open-circuit voltage"?
Because it is measured (or calculated) with the output terminals open — no load connected. With an open circuit, no current flows out of the terminals, so there is no voltage drop across any resistance in series with the terminals. The full internal source voltage appears at the output. This is the maximum voltage the network can deliver; any load will reduce it by the current-times-Rth voltage drop. The open-circuit condition is the simplest state to analyse and gives Vth directly.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.