Thevenin, Norton and Superposition With easy examples.

Thevenin's theorem replaces any linear two-terminal network with a voltage source (Vth) in series with a resistance (Rth). Norton's uses a current source (IN) in parallel with a resistance (RN). Superposition says the response with several sources equals the sum of responses with each source acting alone.

These three theorems matter because real circuits are messy — but a sensor, a battery with internal resistance, or a whole amplifier stage can all be reduced to one source and one resistor, making load calculations and maximum power transfer almost trivial.

Key Concepts

• Vth: the open-circuit voltage across the terminals, with the load removed.
• Rth: the equivalent resistance from the terminals with sources turned off — voltage sources shorted, current sources opened.
• IN: the short-circuit current between the two terminals.
• Source transformation: Vth = IN × RN and RN = Rth.
• Superposition rule: take one source at a time, turn the others off (voltage → short, current → open), solve, then add results.
• Linearity condition: all three theorems apply only to linear circuits — resistors, linear dependent sources and independent sources. They fail for circuits with diodes or saturated transistors.

Formulas

• Thevenin voltage: Vth = open-circuit voltage. For a divider (Vs, series R1, R2 across terminals): Vth = Vs × R2 / (R1 + R2).
• Thevenin resistance: Rth = resistance from the terminals with sources zeroed. For the divider: Rth = R1 ∥ R2 = (R1 × R2) / (R1 + R2).
• Norton current: IN = Vth / Rth, the short-circuit current at the terminals.
• Maximum power transfer: RL = Rth gives Pmax = Vth² / (4 × Rth).

Worked Example

A 12 V source connects in series with a 4 Ω resistor, then a 6 Ω resistor to complete the loop. Find the Thevenin equivalent across the 6 Ω resistor, and the maximum power a load could draw.

1. Remove the 6 Ω resistor mentally — the open-circuit voltage across its terminals is the divider output: Vth = 12 × 6 / (4 + 6) = 72 / 10 = 7.2 V.
2. Short the 12 V source and look back from the terminals: the 4 Ω and 6 Ω resistors are now in parallel, so Rth = (4 × 6) / (4 + 6) = 24 / 10 = 2.4 Ω.
3. Norton equivalent: IN = 7.2 / 2.4 = 3 A in parallel with 2.4 Ω. Maximum power needs RL = 2.4 Ω: Pmax = 7.2² / (4 × 2.4) = 5.4 W.
4. Check with the original circuit: reconnecting the 6 Ω load gives current 12 / 10 = 1.2 A and load voltage 7.2 V — exactly the Thevenin prediction.

Common Misconceptions

• "Superposition works for power too." It does not. Power depends on the square of voltage or current — superpose voltages and currents first, then compute power from the total.
• "Turn off a source means remove it." Turning off means replace with its internal impedance: a voltage source becomes a short circuit (a wire), a current source becomes an open circuit (a break). Deleting the component entirely changes the resistance network.
• "Thevenin works for any circuit." Only linear ones. A circuit with a diode, a relay or a transistor in saturation cannot be reduced this way.

Real-World Applications

• Sensors: a sensor plus wiring reduces to Vth and Rth for choosing amplifier input impedance.
• Battery modelling: a real battery is an ideal voltage source in series with its internal resistance — a Thevenin model.
• Maximum power transfer: audio amplifiers and RF antennas are matched to their loads using Rth for loudest, most efficient output.

Thevenin gives Vth in series with Rth, Norton gives IN in parallel with the same resistance, and superposition builds the total response source by source. Together they turn complicated circuits into one-source problems.