Thevenin's Theorem: A Fully Solved Example You Can Actually Follow

Thevenin's theorem gets taught as three steps to memorize. The steps are correct, but they don't mean anything until you've watched them collapse an actual circuit — so here's one, worked all the way through.

What the theorem actually claims

Any linear circuit, viewed from two terminals, behaves exactly like a single voltage source (V_th) in series with a single resistor (R_th) — no matter how many resistors and sources the real circuit has behind those two terminals. If you only care what happens at those two terminals (usually because you're about to connect a load there), you can replace the whole mess with this one equivalent.

The circuit

A 12V source, in series with a 4Ω resistor, connects to a node. From that node, a 6Ω resistor goes to ground, and a 3Ω resistor goes to the output terminal A. The other output terminal, B, is ground. We want the Thevenin equivalent seen from A–B, so we can quickly find the current through any load resistor we later connect there.

Step 1 — Find V_th: remove the load, find the open-circuit voltage at A–B

With nothing connected across A–B, no current flows through the 3Ω resistor (an open terminal means no complete path through it) — so there's no voltage drop across it, and the voltage at A equals the voltage at the node between the 6Ω and 4Ω resistors.

That node's voltage is a simple voltage divider between the 12V source (through the 4Ω) and ground (through the 6Ω):

V_node = 12V × (6Ω / (4Ω + 6Ω)) = 12V × 0.6 = 7.2V

Since no current flows through the 3Ω resistor, V_th = V_node = 7.2V.

Step 2 — Find R_th: kill the sources, find resistance looking back into A–B

"Kill the source" means replace the 12V voltage source with a short circuit (0Ω) — voltage sources become shorts, current sources become open circuits, when finding R_th. With the source shorted, the 4Ω resistor is now just connected from the node to ground, same as the 6Ω already there — so from the node's point of view, 4Ω and 6Ω are in parallel:

R_parallel = (4Ω × 6Ω) / (4Ω + 6Ω) = 24/10 = 2.4Ω

That parallel combination is in series with the 3Ω resistor on the path out to terminal A:

R_th = 2.4Ω + 3Ω = 5.4Ω

Step 3 — Draw the equivalent

The entire original circuit, as seen from A–B, is now just: a 7.2V source in series with a 5.4Ω resistor.

Why this is useful, not just a simplification exercise: say you now connect a 10Ω load resistor across A–B. Instead of re-analyzing the whole original circuit with the load included, you use the equivalent directly:

I_load = V_th / (R_th + R_load) = 7.2V / (5.4Ω + 10Ω) = 7.2 / 15.4 ≈ 0.468 A

If you needed to try five different load values, you'd redo this one-line calculation five times — instead of re-solving the full original circuit five times. That's the entire point of the theorem: do the hard analysis once, then swap loads for free.

The three steps, for your viva

1. V_th = open-circuit voltage across the two terminals of interest (load removed).
2. R_th = resistance looking back into those terminals with all independent sources killed (voltage sources shorted, current sources opened).
3. Redraw as V_th in series with R_th — this equivalent behaves identically to the original circuit for any load you connect.

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