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Electrical
Load voltage from a Thevenin source
Voltage across a resistive load fed by a Thevenin source: a voltage divider between the source resistance and the load.
- Ref
- ELE-31
- Rev
- v1
- Field
- Electrical
- Result
- VL [V]
The voltage across a load RL driven by a Thevenin source is a potential divider: VL = VthRL/(Rth + RL). It approaches Vth when the load is much larger than Rth, and falls towards zero for a small load. The difference Vth − VL is the voltage lost inside the source.
Voltage across a resistive load fed by a Thevenin source: a voltage divider between the source resistance and the load. It is always below the open-circuit voltage.
Calculator
- Load current1.5 A
- Share of the source voltage, VL/Vth0.5
- Voltage lost inside the source6 V
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 12 V source with 4 Ω internal resistance puts 6 V across a 4 Ω load, 50 % of its open-circuit voltage; the other 6 V is dropped inside the source.
- To hold the voltage within 1 % of Vth the load must be at least 396 Ω, which is 99 times the source resistance: a general rule for connecting a sensor or a supply to a load.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vth Thevenin (open-circuit) voltage | 1 to 100 V | 0.5 | 50 | 49.5 | +10% | +10 % |
| RL Load resistance | 0.1 to 1,000 Ω | 0.2927 | 11.95 | 11.66 | +10% | +4.76 % |
| Rth Thevenin resistance | 0.1 to 100 Ω | 0.4615 | 11.71 | 11.25 | +10% | −4.76 % |
Sweep of Load resistance
| Load resistance (Ω) | Load voltage (V) |
|---|---|
| 0.1 | 0.2927 |
| 0.3162 | 0.8792 |
| 1 | 2.4 |
| 3.162 | 5.298 |
| 10 | 8.571 |
| 31.62 | 10.65 |
| 100 | 11.54 |
| 316.2 | 11.85 |
| 1,000 | 11.95 |
Sweep one variable
Variables and how each one contributes
Load voltage
What the equation gives you. Everything else feeds it.
Thevenin (open-circuit) voltage
The open-circuit (Thevenin) voltage of the source, in volts. Raises the load voltage in direct proportion. Raising it by 1 V (to 13 V) moves the result by +8.33 %, from 6 to 6.5 V.
Thevenin resistance
The Thevenin resistance seen from the load terminals, in ohms: the source’s internal resistance. Lowers the load voltage: more of the source voltage is dropped inside the source. Raising it by 0.5 Ω (to 4.5 Ω) moves the result by −5.88 %, from 6 to 5.647 V.
Load resistance
The load resistance, in ohms. Raises the load voltage: a larger load takes a bigger share of the source voltage. Raising it by 0.5 Ω (to 4.5 Ω) moves the result by +5.88 %, from 6 to 6.353 V.
Worked example with the values above
- Share of the voltage: RL/(Rth + RL) = 4 / 8 = 0.5.
- Multiply the source voltage: 12 × 0.5.
- VL = 6 V.
Assumptions and limits
- A linear circuit that can be replaced by a Thevenin equivalent: a voltage source Vth in series with a resistance Rth.
- Steady DC conditions, or AC with purely resistive impedances.
Common mistakes
- Using the load’s resistance as Rth: Rth belongs to the source side.
- Forgetting that the Thevenin voltage is the open-circuit voltage, measured with the load removed.
- Forgetting the meter itself loads the circuit.
Statistics and uncertainty
The divider ratio depends on the ratio RL/Rth, so tolerance errors in both resistors matter. A reading of VL with a voltmeter of finite resistance acts as an extra load; use a meter with a much larger resistance than Rth.
Where it is used
- The voltage a supply or battery delivers under load.
- Choosing a load or an amplifier input resistance so as not to load a sensor.
Maths you need
- Thevenin equivalent: a source and one resistance that replace the rest of the circuit
- Ohm’s law: V = IR