To calculate Thevenin resistance (Rth), you must determine the equivalent resistance looking into two specific terminal nodes of a linear circuit. The most direct method is to deactivate all independent sources (short-circuit voltage sources, open-circuit current sources) and calculate the remaining equivalent resistance. If the circuit contains dependent sources, or if you prefer an algebraic approach, you calculate the open-circuit voltage (Voc) divided by the short-circuit current (Isc) at those same terminals. This simplification reduces any complex linear network into a single voltage source and a single series resistor, making load analysis trivial.

The Core Thevenin Resistance Formula & Symbol Definitions

While the topological 'source zeroing' method relies on series/parallel reduction rules, the universal algebraic formula for Thevenin resistance applies to any linear network, including those with dependent sources where zeroing fails. This is derived from the relationship between the Thevenin and Norton equivalent circuits.

Rth = Voc / Isc

For circuits requiring a test source (often used in small-signal transistor models or op-amp feedback networks where independent sources are zeroed but dependent sources remain), the formula is expressed as:

Rth = Vtest / Itest

Table 1: Symbol Definitions for Thevenin Resistance Calculations
Symbol Parameter Unit Definition & Measurement Condition
Rth Thevenin Resistance Ohms (Ω) The equivalent internal resistance of the network as seen from the output terminals.
Voc Open-Circuit Voltage Volts (V) The potential difference across the terminals when no load is connected (I = 0). Also known as Thevenin Voltage (Vth).
Isc Short-Circuit Current Amperes (A) The current flowing through a zero-resistance wire placed directly across the output terminals (V = 0). Also known as Norton Current (In).
Vtest Test Voltage Volts (V) An arbitrary DC voltage (often 1V) applied to the terminals after zeroing all independent sources.
Itest Test Current Amperes (A) The resulting current drawn from the test voltage source.

Rearranged Forms: Solving for Unknown Variables

In bench testing and circuit debugging, you rarely solve for Rth in isolation. More often, you know the Thevenin resistance of a driver stage and need to predict voltage sag or current limits. Here are the rearranged algebraic forms for the primary Voc/Isc relationship:

  • Solving for Open-Circuit Voltage: Voc = Rth × Isc
  • Solving for Short-Circuit Current: Isc = Voc / Rth
  • Solving for Test Current (Test Source Method): Itest = Vtest / Rth
  • Solving for Test Voltage (Test Source Method): Vtest = Rth × Itest

Worked Example 1: Source Zeroing in a Mixed Resistor Network

The source zeroing method is the fastest way to find Rth for circuits containing only independent sources and passive resistors. Let us track the units and topology step-by-step.

Circuit Parameters: A 15V DC voltage source (Vs) is in series with R1 (10Ω). This connects to Node A. From Node A, R2 (15Ω) connects to Ground. Also from Node A, R3 (20Ω) connects to Terminal X. Terminal Y is connected directly to Ground. We need to find Rth looking into terminals X and Y.

  1. Deactivate the independent source: Replace the 15V voltage source (Vs) with a short circuit (a wire with 0Ω resistance). The node that was connected to Vs+ is now effectively connected to Ground.
  2. Identify the topology from X to Y: Imagine injecting a test current into Terminal X. It must flow through R3 to reach Node A. From Node A, the current has two paths to Ground (Terminal Y): it can flow down through R2, or it can flow backward through R1 and through the shorted Vs to Ground.
  3. Calculate the parallel equivalent: Because both R1 and R2 connect Node A to Ground, they are in parallel.
    R1||2 = (R1 × R2) / (R1 + R2)
    R1||2 = (10Ω × 15Ω) / (10Ω + 15Ω)
    R1||2 = 150Ω² / 25Ω = 6Ω
  4. Calculate the total series equivalent: The current flowing through the parallel combination must first pass through R3. Therefore, R3 is in series with R1||2.
    Rth = R3 + R1||2
    Rth = 20Ω + 6Ω = 26Ω

Worked Example 2: The Voc/Isc Method with a Current Source

When dealing with current sources or complex meshes, calculating Voc and Isc independently often prevents topological errors.

Circuit Parameters: A 5A independent current source (Is) points upward from Ground to Node A. R1 (8Ω) is in parallel with the current source (connected between Node A and Ground). R2 (12Ω) connects Node A to Terminal X. Terminal Y is Ground. Find Rth across X and Y.

  1. Calculate Open-Circuit Voltage (Voc): Leave terminals X and Y open. Because the circuit is open at X, zero current flows through R2. By Ohm's Law, the voltage drop across R2 is 0A × 12Ω = 0V. Therefore, the voltage at Terminal X is exactly the voltage at Node A. All 5A from the source must flow through R1.
    VA = Is × R1
    VA = 5A × 8Ω = 40V
    Voc = 40V
  2. Calculate Short-Circuit Current (Isc): Place a short circuit (0Ω wire) between Terminal X and Terminal Y (Ground). This places R2 in direct parallel with R1. We can use the current divider rule to find how much of the 5A flows through the shorted path (which is through R2).
    Isc = Is × [ R1 / (R1 + R2) ]
    Isc = 5A × [ 8Ω / (8Ω + 12Ω) ]
    Isc = 5A × (8Ω / 20Ω)
    Isc = 5A × 0.4 = 2A
  3. Apply the Thevenin Formula:
    Rth = Voc / Isc
    Rth = 40V / 2A = 20Ω
  4. Verification via Zeroing: If we open the 5A current source (deactivate it), R1 and R2 are simply in series looking into X-Y. Rth = 12Ω + 8Ω = 20Ω. The algebraic method matches the topological method perfectly.

Practical Assumptions, Unit Traps, and Realistic Magnitudes

When the Formula Applies (and When It Fails)

Thevenin's theorem relies on the assumption of linearity and bilateral symmetry. It applies perfectly to networks of resistors, capacitors, inductors, and linear dependent sources. It fails if your load or network contains non-linear components like diodes, incandescent bulbs (whose resistance changes with temperature), or transistors operating in large-signal switching modes. For AC circuits, the exact same logic applies, but resistance (R) is replaced by complex impedance (Z), requiring phasor math.

Unit Mistakes That Break the Math

Warning: The mA and kΩ Trap

The most common bench mistake is mixing base units. If your Voc is measured in Volts, but your DMM reads Isc in milliamps (mA), dividing them directly yields kilo-ohms (kΩ), not ohms. Always convert to base SI units (Volts, Amperes, Ohms) before executing the division, or explicitly track the prefix: 40V / 2mA = 20kΩ.

Realistic Answer Magnitudes

If your calculation yields 0.0001Ω for a signal circuit, or 50,000Ω for a car battery, you have made a topological error. Use this reference table to sanity-check your results:

Table 2: Expected Thevenin Resistance Magnitudes by Application
Application Domain Typical Rth Range Physical Reason
RF & High-Speed Signal Lines 50Ω or 75Ω Strict impedance matching to prevent signal reflections.
Audio Line Outputs 100Ω to 600Ω Low enough to drive cables without high-frequency roll-off.
DC Power Supplies (Bench) 0.01Ω to 0.5Ω Extremely low to maintain voltage regulation under heavy load.
Sensor Bridges (Strain Gauges) 1kΩ to 10kΩ Higher resistance minimizes self-heating (I²R losses) in the gauge.
MOSFET Gate Drivers 1Ω to 50Ω Low enough to charge gate capacitance quickly, high enough to prevent ringing.

Frequently Asked Questions

How to calculate Thevenin resistance with dependent sources?

You cannot simply zero out dependent sources (like a current-controlled current source in a transistor model). Instead, zero all independent sources. Then, apply a 1V test voltage source (Vtest) across the output terminals. Use nodal or mesh analysis to calculate the current (Itest) that the 1V source supplies to the network. Finally, apply Rth = Vtest / Itest. Because Vtest is 1V, the numerical value of Rth is simply 1 / Itest.

What is the difference between Thevenin resistance and Norton resistance?

Mathematically and physically, there is zero difference. Rth and Norton resistance (Rn) are the exact same value. The distinction is purely in how the equivalent circuit is drawn: Thevenin uses a voltage source in series with Rth, while Norton uses a current source in parallel with Rn. They are source-transformed versions of the identical network.

Can Thevenin resistance be negative?

Yes, but only in circuits containing active dependent sources (like op-amps or transistor amplifiers with specific feedback topologies). A negative Thevenin resistance implies that as you draw more current from the terminals, the voltage across them actually increases. This is a mathematical model used in oscillator design and active load circuits; you will never measure a negative resistance with a standard multimeter on a passive board.

How do I measure Thevenin resistance physically on a PCB?

Never measure resistance on a live board. First, completely de-energize the circuit and discharge all capacitors. To measure Rth directly, set your multimeter to the Ohms (Ω) range and place the probes across the two output terminals. Note that this only works accurately if the circuit contains no active dependent sources or semiconductors that alter their resistance based on bias voltages. For active circuits, you must measure Voc (live), then measure Isc (live, using an ammeter in series with a short), and calculate Rth algebraically.