Reduce a source and resistor divider to its Thevenin and Norton equivalents, find them from two bench measurements, or convert one to the other. The Thevenin equivalent calculator then shows what any load gets and where maximum power transfer happens.
Optional: internal resistance of the source.
Leave blank for a plain source with series resistance.
Optional: resistor between the divider node and the output.
Thevenin voltage Vth
5 V
Vth = Vs × R2 / (Rs + R1 + R2) and Rth = R3 + (Rs + R1) ∥ R2. The source is treated as ideal behind Rs.
⚠ The load pulls the output 33.3 % below its open-circuit voltage. For a reference or an ADC input, keep RL above 100 × Rth (under 1 % droop) or buffer the output.
The same source into other loads
| Load | Load voltage | Load current | Load power | Share of maximum | Efficiency from Vs |
|---|---|---|---|---|---|
| 0.1 × Rth = 500 Ω | 454.5 mV | 909.1 µA | 413.2 µW | 33.1 % | 4.33 % |
| 0.5 × Rth = 2.5 kΩ | 1.667 V | 666.7 µA | 1.111 mW | 88.9 % | 13.3 % |
| 1 × Rth = 5 kΩ | 2.5 V | 500 µA | 1.25 mW | 100 % | 16.7 % |
| 2 × Rth = 10 kΩ | 3.333 V | 333.3 µA | 1.111 mW | 88.9 % | 16.7 % |
| 10 × Rth = 50 kΩ | 4.545 V | 90.91 µA | 413.2 µW | 33.1 % | 7.58 % |
| 100 × Rth = 500 kΩ | 4.95 V | 9.901 µA | 49.01 µW | 3.92 % | 0.971 % |
Vth is the open-circuit output, Vs × R2 ÷ (R1 + R2), and Rth is the resistance looking back with the source shorted, R1 ∥ R2. A 10 V source with a 10 kΩ / 10 kΩ divider is 5 V behind 5 kΩ, so its Norton current is 1 mA, a 10 kΩ load gets 3.333 V, and a 5 kΩ load would draw the maximum possible power, 1.25 mW.
The divider mode models a source Vs with internal resistance Rs, a series resistor R1 to a node, a shunt resistor R2 from that node to ground, and an optional series resistor R3 from the node to the output. The load RL connects from the output to ground.
Any load then sees a simple series circuit:
The maximum power transfer formula says load power peaks when RL equals Rth:
From two measurements on a real source, the open-circuit voltage Voc plus either a loaded voltage V_L across a known RL or the short-circuit current Isc:
10 V drives 100 Ω in series with an 800 Ω shunt, then 200 Ω to a 1 kΩ load. With the load removed no current flows in the 200 Ω, so Vth = 10 × 800 ÷ 900 = 8.889 V. Shorting the source, Rth = 200 + 100 ∥ 800 = 288.89 Ω. The 1 kΩ load gets 6.897 V and 6.897 mA. The most any load could draw is 68.4 mW (at 288.9 Ω), and at that point only 27.6 % of the source’s power reaches the load, because the 800 Ω shunt also draws current.
A network that reduces to 11.2 V behind 800 mΩ has a Norton current of 14 A. A 2 Ω load gets 8 V and 4 A, or 32 W. A 0.8 Ω load gets the maximum, 39.2 W; 0.5 Ω and 1.1 Ω get 37.11 W and 38.22 W, both less.
A source reads 5 V open circuit and 4.8 V with a 1 kΩ resistor across it. The resistor draws 4.8 mA, so Rth = 1 kΩ × (5 ÷ 4.8 − 1) = 41.67 Ω. Its short-circuit current would be 120 mA, and the most it can deliver to any resistor is 150 mW.
Every Thevenin source follows the same curves; only Vth and Rth scale them. Efficiency here is inside the equivalent (power in RL over power from Vth); a real divider is less efficient because its shunt resistor also draws current.
| RL / Rth | V_L / Vth | I_L / In | P_L / P_max | Efficiency |
|---|---|---|---|---|
| 0.01 | 0.99 % | 99 % | 3.92 % | 0.99 % |
| 0.1 | 9.09 % | 90.9 % | 33.1 % | 9.09 % |
| 0.2 | 16.7 % | 83.3 % | 55.6 % | 16.7 % |
| 0.5 | 33.3 % | 66.7 % | 88.9 % | 33.3 % |
| 1 | 50 % | 50 % | 100 % | 50 % |
| 2 | 66.7 % | 33.3 % | 88.9 % | 66.7 % |
| 5 | 83.3 % | 16.7 % | 55.6 % | 83.3 % |
| 10 | 90.9 % | 9.09 % | 33.1 % | 90.9 % |
| 100 | 99 % | 0.99 % | 3.92 % | 99 % |
For an ideal source the divider’s Thevenin resistance is R1 ∥ R2. The last column is the smallest load that keeps the output within 1 % of its unloaded value (99 × Rth).
| R1 (top) | R2 (bottom) | Vth / Vs | Rth | Load for < 1 % droop |
|---|---|---|---|---|
| 10 kΩ | 10 kΩ | 0.5000 | 5 kΩ | 495 kΩ |
| 100 kΩ | 100 kΩ | 0.5000 | 50 kΩ | 4.95 MΩ |
| 10 kΩ | 4.7 kΩ | 0.3197 | 3.197 kΩ | 317 kΩ |
| 100 kΩ | 27 kΩ | 0.2126 | 21.26 kΩ | 2.1 MΩ |
| 1 MΩ | 330 kΩ | 0.2481 | 248.1 kΩ | 24.6 MΩ |
| 4.7 kΩ | 10 kΩ | 0.6803 | 3.197 kΩ | 317 kΩ |
Rth is what the next stage sees. A 100 kΩ over 27 kΩ battery-sense divider is 2.551 V behind 21.26 kΩ, so an input with 1 MΩ of resistance reads 2.08 % low. A sampling ADC is worse than its DC input resistance suggests: each conversion pulls a burst of charge from the source. Keep the load at least 100 times Rth, put a capacitor at the ADC pin, or buffer the divider.
Maximum power transfer is a design goal only where power is scarce and efficiency is not, such as a weak signal source, an antenna feed or a solar panel at its maximum power point. For supplies and amplifiers you want the opposite: a load much larger than the source resistance, so little power is lost inside. Remember that the power dissipated in Rth of an equivalent is not the power dissipated in the real circuit; only the load figures carry over.
The theorem needs a linear network. Batteries, regulators and transistor stages can be described by a Thevenin equivalent only around one operating point, and their internal resistance changes with current, temperature and state of charge. Measure them with a test load close to the current you care about.
Any linear network of sources and resistors, seen from two terminals, behaves exactly like its Thevenin equivalent circuit: one voltage source Vth in series with one resistance Rth. Vth is the open-circuit voltage at the terminals and Rth is the resistance between them with every source turned off.
Remove the load and take the open-circuit voltage at the terminals. For a divider the Thevenin voltage formula is Vth = Vs × R2 / (R1 + R2), so 10 V into 10 kΩ / 10 kΩ gives 5 V.
Remove the load, replace voltage sources with shorts and current sources with opens, and work out the resistance looking back into the terminals. For a divider fed from an ideal source the Thevenin resistance formula is R1 ∥ R2; any series output resistor adds to it. For a real circuit, use the From measurements tab: give it the open-circuit voltage and a loaded reading or the short-circuit current.
Keep the resistance and set the Norton current to the short-circuit current, In = Vth / Rth. A 11.2 V source behind 800 mΩ becomes 14 A in parallel with 800 mΩ. Going back, Vth = In × Rn. The calculator above runs the Thevenin to Norton conversion in either direction.
A load equal to the Thevenin resistance. The load then gets half of Vth and a power of Vth² / (4 Rth): 39.2 W for 11.2 V behind 0.8 Ω. Loads of 0.5 Ω and 1.1 Ω on the same source get 37.11 W and 38.22 W.
No. At RL = Rth only half of the power leaves the equivalent source, so efficiency is at most 50 %, and less when part of Rth is a shunt resistor. Power supplies, audio amplifiers and batteries are run with a load much larger than their source resistance to be efficient.
Measure the open-circuit voltage, then the voltage across a known load, and use Rth = RL × (Voc / VL − 1). If the output drops from 5 V to 4.8 V across 1 kΩ, Rth = 41.67 Ω. Adjusting the load until the voltage halves gives RL = Rth directly.
R1 ∥ R2, because the ideal supply is a short to ground for this purpose. A 100 kΩ over 27 kΩ divider from 12 V is 2.551 V behind 21.26 kΩ; an input with 1 MΩ resistance reads it 2.08 % low.
Yes, at one frequency at a time, with impedances in place of resistances. Maximum power into a complex load then needs the load to be the complex conjugate of the Thevenin impedance, not just equal in magnitude. This Thevenin calculator covers resistive networks only.
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