thevenin circuit calculator

Thevenin Circuit Calculator (Voltage Divider Model)

Calculate the Thevenin equivalent seen at the output node of a two-resistor divider powered by an ideal voltage source.

Circuit model: source Vs feeding R1 in series with R2 to ground. Output terminals are between the R1-R2 junction and ground. Optional RL lets you compute loaded voltage/current.

Source Voltage, Vs (V)

Resistor R1 (Ω)

Resistor R2 (Ω)

Load Resistor, RL (Ω) — optional

What Is a Thevenin Equivalent?

Thevenin’s theorem says that any linear two-terminal circuit can be replaced by a much simpler equivalent: one ideal voltage source in series with one resistor. That pair is called the Thevenin equivalent, and it helps you analyze how different loads behave without re-solving the full network every time.

What This Thevenin Circuit Calculator Computes

This calculator is designed for a practical and common case: a source and resistor divider. It provides:

Vth: open-circuit voltage at the output node.
Rth: equivalent resistance seen looking back into the source network with the ideal voltage source turned off (shorted).
In: Norton current, where In = Vth / Rth.
Loaded values when RL is provided: output voltage, load current, and load power.

Formulas Used

1) Open-Circuit Thevenin Voltage

Vth = Vs × (R2 / (R1 + R2))

This is the standard voltage-divider equation at the junction between R1 and R2 when no external load is attached.

2) Thevenin Resistance

Rth = R1 || R2 = (R1 × R2) / (R1 + R2)

To find equivalent resistance, set the ideal source to zero volts (replace it with a short). From the output node, R1 and R2 then appear in parallel.

3) Norton Current

In = Vth / Rth

The Norton equivalent is the dual form of Thevenin: a current source in parallel with Rth. Both models are interchangeable.

4) Loaded Output (Optional RL)

IL = Vth / (Rth + RL), VL = IL × RL, PL = VL × IL

Once the Thevenin model is known, load analysis becomes a one-step series calculation.

Worked Example

Suppose you enter: Vs = 12 V, R1 = 1 kΩ, R2 = 2 kΩ, RL = 4.7 kΩ.

Vth = 12 × (2000 / 3000) = 8 V
Rth = (1000 × 2000) / 3000 = 666.67 Ω
IL = 8 / (666.67 + 4700) ≈ 1.49 mA
VL ≈ 7.00 V

This quickly tells you the load sees about 7 V instead of the no-load 8 V because of source resistance.

Why Engineers Use Thevenin Models

Fast what-if analysis for changing sensor or load resistance.
Simplified hand calculations and faster design iteration.
Clear understanding of source stiffness and voltage droop.
Direct path to maximum power transfer studies.

Common Mistakes to Avoid

Using zero or negative resistance values (physically invalid for this basic model).
Mixing units accidentally (kΩ entered as Ω or vice versa).
Forgetting that loaded voltage is usually less than open-circuit voltage.
Applying this model directly to nonlinear components without linearization.

Quick FAQ

Can I use this for AC circuits?

Yes, in concept. For AC phasor analysis, resistances become impedances. This calculator is strictly real-valued (DC resistive) for simplicity.

What if my source has internal resistance?

Include that resistance in the network before deriving equivalent values, or fold it into R1/R2 as appropriate.

When does Thevenin fail?

The theorem applies to linear networks. Strongly nonlinear circuits require operating-point analysis or piecewise linear methods.

If you want to test your understanding, try changing only RL while keeping Vs, R1, and R2 fixed. You will see how the Thevenin model makes repeated load analysis almost effortless.