Any tangle of sources and resistors behaves like one battery with one resistor. Find those two numbers and you are done.
The idea: the socket in your wall
Behind a power socket sits a wire in the wall, a fuse box, a street transformer, a substation, a power station. An enormous amount of equipment.
When you plug in a lamp, how much of that do you need to know? Two numbers:
🔌 All that matters about a socket
Its voltage with nothing plugged in230 V. This is V_th.
How much it sags under loadPlug in a heater and the voltage dips. That dip reveals the resistance of everything behind the socket. This is R_th.
Those two numbers describe the entire grid, as far as your lamp can tell. That is Thevenin's theorem: any network, however complicated, behaves exactly like one voltage source in series with one resistance.
Why bother
For a single load it is more work than solving the circuit directly. The value appears when the load changes.
Reduce the network ONCE, then for any load:
I = V_th / (R_th + R_L)
Ten different loads, ten divisions. No re-solving.
The method
1. REMOVE the load. Leave a gap, label it a–b.
2. FIND V_th — the voltage across that empty gap.
Nothing is drawing current, so this is often a
simple series calculation.
3. FIND R_th — kill every source (battery → wire,
current source → gap), then find the resistance
looking INTO a–b.
4. REBUILD as V_th in series with R_th, reconnect
the load, and use Ohm's law.
Worked example
A 24 V source with 4 Ω in series feeds a point where a 12 Ω goes to ground. An 8 Ω load hangs across that point. Find the load current.
Step 1 — remove the load
Take the 8 Ω out and leave a gap labelled a–b. It may feel wrong to remove the thing you are asked about, but Thevenin describes the rest of the circuit as seen from that gap. The load is the visitor; we are measuring the house.
Step 2 — find V_th
With the load gone there is only one path: source → 4 Ω → 12 Ω → back. A plain series loop.
I = 24/(4 + 12) = 1.5 A
The gap sits across the 12 Ω, so:
V_th = 1.5 × 12 = 18 V
💡 Check: 18 V is less than 24 V, as it must be — the 4 Ω drops the rest. A V_th larger than the source is always an error.
Step 3 — find R_th
Replace the 24 V source with a wire. Now stand at a–b and ask what a meter would read looking back in.
Trace it: from a back to b there are two routes — one through the 4 Ω, one through the 12 Ω. Both start at the same point and end at the same point, so they are in parallel.
R_th = (4 × 12)/(4 + 12) = 48/16 = 3 Ω
This step feels strange because the circuit does not look like a parallel pair until you redraw it with the battery as a wire. So redraw it, on paper, every time. Fifteen seconds, and it is where nearly every lost mark in this topic comes from.
Step 4 — reconnect the load
The whole network is now 18 V with 3 Ω:
I = 18/(3 + 8) = 18/11 = 1.636 A
What the reduction bought you
Same network, three different loads:
R_L = 2 Ω → 18/(3+2) = 3.600 A
R_L = 8 Ω → 18/(3+8) = 1.636 A
R_L = 20 Ω → 18/(3+20) = 0.783 A
One division each. The network is finished with forever.
💡 Exam questions often ask for two or three loads precisely to test whether you understood this, rather than just following the procedure.
Exam notes
📝 Where marks are won and lost
Redraw for R_thWith sources killed. The single highest-value habit in this topic.
Label a–bAnd keep the labels consistent through every diagram.
Sanity-check V_thIt cannot exceed the source voltage.
Typical weight8–10 marks, often with a second part asking for a different load or for maximum power.
Syllabus points
Vth, Rth; equivalent circuit (numerical)
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