Electronics · Circuit theory · Beginner

Thevenin’s theorem: simplify a circuit at its load

Replace a linear network with one equivalent voltage source and one series resistance, then use that simpler model to analyze a connected load.

By Amit Kumar Das About the author Published

Who this is for: Learners familiar with Ohm’s law, series and parallel resistors, and basic voltage-divider analysis.

What you will learn: how to calculate Thevenin voltage and resistance for a simple loaded divider, then use them to find load voltage and current.

Tested example: The worked values were checked against the calculator’s default output in a local browser preview. The model assumes an ideal DC source and positive resistances.

From the perspective of two output terminals, a linear circuit can be represented by an ideal voltage source Vth in series with a resistance Rth. The original network is not physically changed; the equivalent is a simpler model for analyzing a load attached at those terminals.

The diagram uses the same example values as the calculation below. The left-hand circuit is the original loaded divider; the right-hand circuit is the equivalent seen by RL.

Original circuit
Loaded voltage divider circuit A voltage source feeds resistor R1. At the output node, resistor R2 and load resistor RL connect to ground. + Vs 12 V R1 · 1 kΩ R2 2 kΩ RL 1 kΩ Vout
Equivalent seen by the load
Thevenin equivalent circuit The original source and divider are replaced by a Thevenin voltage source in series with resistance Rth, connected to load RL. + Rth Vth 8 V Rth 667 Ω RL Vload 4.8 V

Find Vth and Rth step by step

  1. Remove RL. Leave the output terminals open. The voltage across those terminals is the open-circuit voltage, Vth.
  2. Calculate Vth. With the load removed, R1 and R2 form a voltage divider: Vth = Vs × R2 / (R1 + R2).
  3. Find Rth looking into the terminals. Deactivate the ideal independent voltage source by replacing it with a short circuit. R1 and R2 are then in parallel: Rth = R1 × R2 / (R1 + R2).
  4. Reconnect RL. The simplified circuit is Vth in series with Rth and RL. Its load voltage is Vload = Vth × RL / (Rth + RL).

Shorting the ideal voltage source is a circuit-analysis step for finding resistance; it does not mean shorting a real power supply in a physical circuit.

Worked example

Use Vs = 12 V, R1 = 1 kΩ, R2 = 2 kΩ, and RL = 1 kΩ. The source and resistors are treated as ideal:

Vth = 12 × 2 / (1 + 2) = 8 V
Rth = 1 × 2 / (1 + 2) = 0.667 kΩ
Vload = 8 × 1 / (0.667 + 1) ≈ 4.8 V
Iload = Vload / RL = 4.8 / 1 = 4.8 mA

The unloaded divider output is 8 V. Connecting a 1 kΩ load draws current and lowers the output to about 4.8 V. The Thevenin model makes that loading effect straightforward to calculate.

Assumptions and limits

  • This example uses a single ideal DC voltage source and positive resistances.
  • The basic theorem applies to linear networks. Circuits with nonlinear elements need additional care and cannot generally be reduced to this fixed model for every operating point.
  • The calculator is specifically for this loaded divider topology. It is not a general-purpose schematic builder, circuit simulator, or component safety checker.
  • Real resistor tolerances, source resistance, power ratings, and frequency-dependent behavior are not modeled.

Try changing the load. Compare the unloaded Thevenin voltage with the voltage delivered to different load resistances.

Open the Thevenin calculator

Continue learning on the electronics resources page or explore the interactive calculator.