Thévenin and Norton Equivalent Calculator

Replace a source and its resistor network with one voltage source and one resistance — or one current source and the same resistance — and see exactly what the equivalent predicts under any load: voltage, current, power, and the load that draws the most power from it.

Last updated 24 Sep 2026

Thévenin / Norton equivalent

Network → Vth, Rth, In and the load's behaviour

What do you have?

Source voltage Vs (V)

Unit for all the resistances

Source’s own resistance Rs0 for an ideal source. A 9 V battery is roughly 1 Ω when new and several ohms when flat; a bench supply is milliohms.

R1 — in series, from the source to the output node

R2 — from the output node to ground

Thévenin voltage Vth (V)The open-circuit voltage at the terminals — what a perfect voltmeter reads with nothing else connected.

Thévenin / Norton resistanceThe resistance looking back into the terminals with every ideal voltage source shorted and every ideal current source opened. Rth and Rn are the same number.

Norton current In (A)The short-circuit current at the terminals.

Load resistance RL (0 = no load)Whatever you are going to connect. Leave it at 0 for the open-circuit figures only.

On the left, the network you actually have — a source, whatever resistance it has of its own, and a divider — with its two output terminals. On the right, the Thévenin equivalent that behaves identically at those terminals, with your load on it. Both are live. The left half is only drawn when the page is working from a network.

9VExample

12 V into R1 = 100 Ω and R2 = 300 Ω, with a 150 Ω load

The equivalent, and what it predicts

from a divider: V_th = V_s × R2 ÷ (R_s + R1 + R2), R_th = (R_s + R1) ∥ R2
source transformation: I_N = V_th ÷ R_th, V_th = I_N × R_N, R_N = R_th
under a load: V_L = V_th R_L ÷ (R_th + R_L), I_L = V_th ÷ (R_th + R_L), P_L = I_L² R_L
maximum power: R_L = R_th, P_max = V_th² ÷ 4R_th, efficiency = R_L ÷ (R_L + R_th)

V th: the open-circuit voltage at the terminals
R th: the resistance into the terminals with every ideal voltage source shorted and every ideal current source opened
I N: the short-circuit current at the terminals, Vth ÷ Rth
R L: whatever you connect. The equivalent is exact for all of them, and only at these two terminals

Worked example

12 V into R1 = 100 Ω and R2 = 300 Ω, with a 150 Ω load
V_th is the unloaded output: 12 × 300 ÷ 400 = 9 V
R_th is R1 ∥ R2 with the source shorted: 100 × 300 ÷ 400 = 75 Ω
Norton: I_N = 9 ÷ 75 Ω = 120 mA, across the same 75 Ω
With the 150 Ω load: V_L = 9 × 150 ÷ (75 + 150) = 6 V, I_L = 40 mA, P_L = 240 mW
Efficiency = 150 ÷ 225 = 66.67% — the other 33.33% is heating R_th
The most power this network can give anything is 270 mW, at a load of 75 Ω — and at that load the efficiency is 50%

What the same equivalent does into different loads

The page’s own equivalent — Vth = 9 V, Rth = 75 Ω — into loads from a tenth of Rth to ten times it. The power peaks at RL = Rth and is flat around it: half Rth and twice Rth both give 88.9% of the maximum. Efficiency, by contrast, climbs all the way, which is why maximum power and good efficiency are different goals.
Load	RL	Load voltage	Load current	Load power	Of the maximum	Efficiency
0.10 × R_th	7.5 Ω	0.818 V	109.1 mA	89.3 mW	33.1%	9.1%
0.25 × R_th	18.8 Ω	1.800 V	96.0 mA	172.8 mW	64.0%	20.0%
0.50 × R_th	37.5 Ω	3.000 V	80.0 mA	240.0 mW	88.9%	33.3%
1.00 × R_th	75.0 Ω	4.500 V	60.0 mA	270.0 mW	100.0%	50.0%
2.00 × R_th	150.0 Ω	6.000 V	40.0 mA	240.0 mW	88.9%	66.7%
4.00 × R_th	300.0 Ω	7.200 V	24.0 mA	172.8 mW	64.0%	80.0%
10.00 × R_th	750.0 Ω	8.182 V	10.9 mA	89.3 mW	33.1%	90.9%

What the equivalent is, and when matching is the wrong idea

Thévenin’s theorem says that any network of linear resistances and ideal sources, looked at through two terminals, is indistinguishable from a single voltage source behind a single resistance. Not approximately — exactly, for every load you could ever connect. Norton’s theorem says the same network is equally indistinguishable from a current source with a resistance across it, and the two are related by nothing more than Ohm’s law: I_N = V_th ÷ R_th, with the same resistance in both.

Finding the two numbers. V_th is the open-circuit voltage: take the load off and measure. R_th is what an ohmmeter would read into the terminals after the sources have been KILLED — every ideal voltage source replaced by a wire and every ideal current source by a gap. That killing step is the one people get wrong; a voltage source becomes a short, not an open. For the divider on this page that gives R_th = (R_s + R1) ∥ R2, which is 75 Ω for 100 Ω and 300 Ω — smaller than either, as any parallel combination must be. The other route, when the network has dependent sources and you cannot kill them, is R_th = V_open ÷ I_short.

What it buys you. A divider feeding an amplifier input, a sensor feeding a converter, a battery feeding a motor: in every case the question is what happens when you connect something, and the equivalent answers it in one line rather than by re-solving the network for each load. It also tells you immediately what kind of source you have. If the load is much larger than R_th the terminal voltage hardly moves and the thing behaves as a voltage source; if it is much smaller, the current hardly moves and it behaves as a current source. The same number, R_th, is what an analog-to-digital converter’s sampling capacitor has to charge through, and what sets how much a load pulls a voltage divider down.

Maximum power transfer, and why it is usually the wrong goal. Differentiate P_L = V_th²R_L ÷ (R_th + R_L)² and it peaks at R_L = R_th, where the load gets V_th² ÷ 4R_th. But at that load the source resistance carries the same current through the same resistance, so it burns exactly as much power as the load: the efficiency is 50%, always, whatever the numbers. That is a fine trade when the source power is free and the signal is not — an antenna, a microphone, an RF stage — and it is a terrible one for anything that has to supply power. No power supply, battery charger, motor drive or mains transformer is matched to its load; they are all designed with R_th as small as they can make it, so the load sees nearly all of V_th and the efficiency approaches 100%. If somebody tells you to match a power supply to its load, they have confused the two problems. The peak is also broad — half or twice the matched load still delivers 88.9% of the maximum — so there is rarely much to be gained by chasing it exactly.

What the equivalent does not tell you. It is a model of the terminals, not of the circuit. The power dissipated in R_th is not the power dissipated in the real resistors that R_th stands for, and for an unloaded divider it is not even close: the real network here draws current through R1 and R2 whether anything is connected or not, while the equivalent draws nothing. It also assumes linearity, so a diode, a transistor or a saturating inductor has an equivalent only over a small enough range to be treated as straight. For a network with more than one loop that will not reduce to a divider, solve it first — the two-loop mesh solver does the two-loop case, and the delta-wye transformation unpicks a bridge.

Frequently asked questions

How do I find the Thévenin resistance?

Kill every independent source — short every ideal voltage source, open every ideal current source — and work out the resistance looking into the two terminals. For a source feeding a divider that is (Rs + R1) in parallel with R2. If the network has dependent sources, use Rth = open-circuit voltage ÷ short-circuit current instead.

What is the difference between Thévenin and Norton?

Only how the same equivalent is written. Thévenin is a voltage source Vth in series with Rth; Norton is a current source In = Vth ÷ Rth in parallel with the same resistance. Any external measurement gives identical results, so pick whichever makes the rest of the analysis shorter.

What load draws the most power?

A load equal to Rth, drawing Vth² ÷ 4Rth. The peak is broad: half or twice that load still gets 88.9% of the maximum, so matching exactly is rarely worth much effort.

Why is the efficiency only 50% at maximum power transfer?

Because the matched load and the source resistance are equal and carry the same current, so they dissipate equally. That is acceptable when the source power is free and the signal is precious, as in RF and audio source impedances. It is unacceptable for a power supply, which is why supplies are built with the lowest output resistance they can manage, not a matched one.

Does the equivalent tell me the power used inside the original circuit?

No. It is exact at the terminals and nowhere else. An unloaded divider draws real current from its supply while its Thévenin equivalent draws none, and the power in Rth is not the power in the resistors it replaced.

Can I use Thévenin’s theorem on a circuit with a transistor or a diode?

Only over a range small enough for the part to count as linear — which is exactly what a small-signal model is. The theorem needs linearity; a large swing across a diode has no single Thévenin equivalent.

Related calculators

References

Nilsson JW, Riedel SA. Electric Circuits, 11th ed. Pearson, 2019. §4.9 “Source Transformations”, §4.10 “Thévenin and Norton Equivalents”, §4.11 “More on Deriving the Thévenin Equivalent” and §4.12 “Maximum Power Transfer”. Section titles confirmed against Pearson’s own published front matter for the eleventh edition.
Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. §1.2.5–1.2.6: Thévenin’s theorem, the equivalent source resistance, and the rule of thumb that a load should be at least ten times the source resistance for signal work.
Thévenin L. Sur un nouveau théorème d’électricité dynamique. Comptes Rendus des Séances de l’Académie des Sciences, vol. 97, pp. 159–161, 1883. The original statement; Norton’s dual was circulated internally at Bell Laboratories in 1926 and independently published by Hans Ferdinand Mayer the same year, which is why the current-source form is also called the Mayer–Norton theorem.