The theorem
Seen from two terminals a–b, any linear network of sources and resistors behaves exactly like one voltage source Vth in series with one resistor Rth (Thévenin), or one current source IN in parallel with the same Rth (Norton). Analyse the network once, and any load is a one-line calculation.
IL = Vth / (Rth + RL) IN = Vth / Rth
Finding Vth, Rth and IN
For the simulator’s network (V, R1, R2 divider, then R3 to terminal a):
- Vth: remove the load and find the open-circuit voltage. No current flows in R3, so Vth is the R1–R2 divider: Vth = V·R2/(R1 + R2).
- Rth: switch off independent sources (a voltage source becomes a wire, a current source an open circuit) and find the resistance seen into a–b: Rth = R1∥R2 + R3.
- IN: short a–b and find the current, or simply IN = Vth/Rth.
Maximum power transfer
The power in the load, PL = Vth²·RL/(Rth + RL)², peaks when the load matches the source: RL = Rth. The maximum is Vth²/(4·Rth), and at that point the efficiency is only 50%: as much power is lost in Rth as reaches the load. Matched loads suit signals (antennas, audio); power systems run far from matched, for efficiency.
PL,max = Vth² / (4·Rth) at RL = Rth
Worked example (the simulator’s default values)
- V = 9 V, R1 = 220 Ω, R2 = 470 Ω, R3 = 330 Ω, RL = 1 kΩ.
- Vth = 9 × 470/690 = 6.13 V.
- Rth = (220 × 470/690) + 330 = 149.9 + 330 = 479.9 Ω.
- IN = 6.13/479.9 = 12.78 mA.
- IL = 6.13/(479.9 + 1000) = 4.14 mA in the original, Thévenin and Norton circuits alike; matched (RL = 480 Ω) it would draw Pmax = 19.6 mW.
Common mistakes
- Leaving the source in when finding Rth. Replace voltage sources with wires and current sources with open circuits.
- Forgetting R3 (or any series resistor) when combining: Rth here is R1∥R2 + R3, not just R1∥R2.
- Expecting maximum power to mean maximum efficiency. At RL = Rth the efficiency is 50%.
Common questions
How do you find the Thévenin equivalent of a circuit?
Vth is the open-circuit voltage at the terminals. Rth is the resistance seen into the terminals with independent sources switched off (voltage sources shorted, current sources opened). Then the circuit is Vth in series with Rth.
What is the relation between Thévenin and Norton equivalents?
They share Rth, and IN = Vth/Rth. A voltage source in series with a resistor and a current source in parallel with the same resistor look identical from the terminals.
When is maximum power transferred to a load?
When the load resistance equals the Thévenin resistance. The load then receives Vth²/(4·Rth), with 50% efficiency.