Any circuit with a source and a load can be replaced by a single voltage source in series with a single resistance. Work through the real example below, one clear step at a time — no clutter, no rushing.
📐The rule: to find Rth, short every voltage source. To find Vth, remove the load and find the open-circuit voltage at the terminals. Never remove the load when finding Rth — remove the sources instead.
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Watch it happen — one step at a time
Circuit: a 10 V source (10 Ω) and a 20 V source (20 Ω) both feed a 40 Ω motor. Click through each step below.
Step 0 · The circuit
Two sources sharing one load
A 10 V source (through 10 Ω) and a 20 V source (through 20 Ω) both connect to a 40 Ω motor between terminals A and B.
What we want to find:The current through the motor
Step 1 · Remove the load
Take the motor out
Thevenin's theorem replaces everything except the load with one simple equivalent. So the very first move is to disconnect the load and look at the open terminals A–B.
Only one loop is left now:E1 → R1 → R2 → E2 → back
Step 2 · Find R_th
Short both sources, then look in from A–B
Replace every voltage source with a plain wire. Watch what that does below: both resistors end up starting at the same node A and ending at the same node B — that's the actual definition of "parallel" (not just "drawn side by side").
R_th = (R1 × R2) ÷ (R1 + R2)= (10×20) ÷ (10+20)
R_th = 200 ÷ 30= 6.67 Ω
Step 3 · Find V_th
Put the sources back — find the loop current
With the load still removed, there's exactly one loop. The two sources oppose each other around it, so the net driving voltage is the difference between them.
🤔 Wait — weren't R1 and R2 in parallel a moment ago?
Yes — but only in Step 2, where we switched the sources off (shorted) to test the resistance. That's a completely different scenario to this one.
Now the sources are back on. The load is still disconnected, so there is only one route the current can take: out of one source, through R1, along the wire, through R2, into the other source, and back. One route means one path in sequence — that's what "series" means. Nothing about the resistors themselves changed; what changed is which question we're asking.
Step 2 — finding R_th
Sources OFF. Testing resistance from A–B. Current has two possible routes (via R1 or via R2) → parallel.
Step 3 — finding V_th
Sources ON, load OFF. Current has only one possible route (through both) → series.
I = (E2 − E1) ÷ (R1 + R2) = (20−10) ÷ 30I = 0.33 A
V_th = E2 − (I × R2) = 20 − (0.33×20)V_th = 13.4 V
Step 4 · Build the equivalent
Everything becomes ONE source, ONE resistor
That's the whole point of Thevenin's theorem: no matter how complicated the original network was, it collapses down to just two numbers.
Thevenin equivalent:V_th = 13.4 V, R_th = 6.67 Ω
Step 5 · Reconnect the load
Now bring the motor back
The equivalent circuit drives the real load exactly as the original (much more complicated) circuit would — but now it's just one loop.
Total R = R_th + R_L = 6.67 + 40= 46.67 Ω
I_load = V_th ÷ Total R = 13.4 ÷ 46.67= 0.286 A
Step 1 of 6
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Worked Example 2 — different values, same method
New circuit: a 5 V source (4 Ω) and a 15 V source (6 Ω) both feed a 7.6 Ω load between terminals A and B. Click through — same steps as Example 1.
Step 0 · The circuit
Two sources sharing one load
A 5 V source (through 4 Ω) and a 15 V source (through 6 Ω) both connect to a 7.6 Ω load between terminals A and B.
What we want to find:The current through the load
Step 1 · Remove the load
Take the load out
Disconnect the load and look at the open terminals A–B — this is what we're about to build an equivalent circuit for.
Only one loop is left now:E1 → R1 → R2 → E2 → back
Step 2 · Find R_th
Short both sources, then look in from A–B
With both sources replaced by plain wire, R1 and R2 both run from A to B — that means they're in parallel.
R_th = (R1 × R2) ÷ (R1 + R2)= (4×6) ÷ (4+6)
R_th = 24 ÷ 10= 2.4 Ω
Step 3 · Find V_th
Put the sources back — find the loop current
🤔 Same question as Example 1 — why series now, not parallel?
Step 2 asked "how does current get from A to B with the sources off?" — two routes, parallel. This step asks "with the sources back on and the load still off, what's the one and only current path?" — one route, in sequence, so series. The resistors haven't changed; the question has.
I = (E2 − E1) ÷ (R1+R2) = (15−5) ÷ 10I = 1 A
V_th = E2 − (I×R2) = 15 − (1×6)V_th = 9 V
Check via E1: 5 + (1×4)= 9 V ✓
Step 4 · Build the equivalent
Everything becomes ONE source, ONE resistor
Same collapse as Example 1 — a complicated two-source network reduces to just two numbers.
Thevenin equivalent:V_th = 9 V, R_th = 2.4 Ω
Step 5 · Reconnect the load
Now bring the load back
The equivalent circuit drives the real load exactly as the original circuit would.
Total R = R_th + R_L = 2.4 + 7.6= 10 Ω
I_load = V_th ÷ Total R = 9 ÷ 10= 0.9 A
Step 1 of 6
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Try it yourself
🎛️ Change the values — everything updates live
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Practice
Question 1 · Find R_th and V_th
A 12 V source (6 Ω) and a 24 V source (12 Ω) both feed a load between terminals A and B. Find R_th.
Using the same circuit (E1=12V/R1=6Ω, E2=24V/R2=12Ω), find the loop current with the load removed, then find V_th.
I = (24−12) ÷ (6+12) = 12 ÷ 18 = 0.667 A V_th = 24 − (0.667×12) = 24 − 8 = 16 V
Question 3 · Reconnect the load
Using R_th = 4 Ω and V_th = 16 V from Questions 1–2, a 16 Ω load is reconnected. Find the load current.
Total R = R_th + R_L = 4 + 16 = 20 Ω I_load = V_th ÷ Total R = 16 ÷ 20 = 0.8 A
📜 The story behind Thevenin's Theorem
Where the idea came from, why it works, and where engineers actually reach for it today.
🕰️In short: the theorem was worked out twice, thirty years apart — first by a German physicist studying muscle tissue, then independently by a French telegraph engineer designing telephone lines. History remembers the second name.
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Where it came from
1853
First discovered — but almost unnoticedGerman scientist Hermann von Helmholtz derived the same result while studying how electrical currents distribute through animal tissue. His paper was physics-focused and aimed at a different audience, so it didn't reach the engineers who could have used it.
1883
Rediscovered by Léon Charles ThéveninThirty years later, and unaware of Helmholtz's work, French telegraph engineer Léon Charles Thévenin published "Extension de la loi d'Ohm aux circuits électromoteurs complexes" — extending Ohm's Law to complex electromotive circuits — while working on measurement problems for France's telegraph network.
1883–1926
Slow to catch onDespite being published, the theorem wasn't widely used for decades. American Telephone and Telegraph engineers picked it up around 1904, but it took until the 1930s–40s for it to become a standard part of circuit-theory teaching and textbooks.
🎓 Who was Léon Charles Thévenin?
Thévenin (1857–1926) was born in Meaux, France. He trained at the École Polytechnique before joining the French Corps of Telegraph Engineers, and later taught at the École Supérieure de Télégraphie. He developed his theorem while working through problems of measurement in electrical circuits — practical telegraph engineering, not abstract theory. He was also a keen violinist. He never married, and the theorem is the only thing that carries his name.
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The logic — why it works
🧩 One "black box", two numbers
Thevenin's theorem says that no matter how many sources and resistors sit behind two terminals, as long as the network is linear, the outside world can never tell the difference between that whole tangled network and a single voltage source in series with a single resistor. From the load's point of view, both look electrically identical.
This works because of superposition: in a linear network, the total effect of several sources is just the sum of each source's effect acting alone. That's what lets a messy multi-source circuit collapse down to one equivalent voltage (V_th) and one equivalent resistance (R_th) — the theorem is really superposition applied in the most useful possible way.
🔓 Why leave the terminals open to find V_th?
With the load disconnected, no current can flow out of the terminals, so whatever voltage appears across that open gap is exactly the voltage the source network would push out under a "do-nothing" load — which is, by definition, the Thevenin voltage. It's the mirror image of Norton's approach, which shorts the terminals instead to find a current.
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Where it's actually used
Once a source network is reduced to V_th and R_th, an engineer can try any load value against it instantly — without re-solving the whole circuit each time. That's the real payoff, and it shows up across several fields:
🔋Battery & power-supply modellingReal batteries and supplies are modelled as an ideal source plus an internal (Thevenin) resistance, which is why the output voltage sags under heavier load.
📶Maximum power transferA load draws the most possible power from a source when its resistance equals R_th — used when matching amplifiers to speakers, or antennas to transmitters.
🔌Electronic circuit designFinding the equivalent output impedance of an amplifier stage so it can be connected confidently to the next stage or a test instrument.
⚡Power systems analysisReducing large sections of a grid to a simple source-and-impedance model for fault-current and stability studies.
🛠️Troubleshooting & testingMeasuring the open-circuit voltage and short-circuit current at two terminals of a sealed "black box" circuit is enough to find its full Thevenin equivalent — no need to open it up.
🎓Teaching circuit analysisIt's usually the first tool students meet for turning an intimidating multi-source circuit into something they can actually reason about.
🟢 Key words on this page
Linear network
A circuit where doubling the input (voltage or current) always doubles the output — no diodes or transistors operating outside their linear range.
Superposition
The principle that in a linear circuit with several sources, the total voltage or current at any point is the sum of the effects of each source acting alone.
Two-terminal network
Any circuit, however complicated inside, that connects to the rest of the world through just two wires — the pair Thevenin's theorem is applied across.
💡Worth remembering: Thevenin's theorem and Norton's theorem describe the exact same circuit fact from two different angles — voltage source versus current source. Whichever one is quicker for the problem in front of you is the one to reach for.
