Circuit Reduction: Thevenin & Norton (IPhO)

IPhO circuit questions often hide a simple equivalent behind a “busy” diagram. Thevenin/Norton reduction replaces any linear two-terminal network with one source plus one resistance, so you can compute load current, maximum power, or time constants fast.

1. Definitions (Must Know)

• Thevenin equivalent at terminals A–B: an ideal voltage source Vₜₕ in series with Rₜₕ.
• Norton equivalent at terminals A–B: an ideal current source I_N in parallel with R_N.
• Core equalities for linear circuits:
  Vₜₕ = V_oc, I_N = I_sc, R_N = Rₜₕ, Vₜₕ = I_N Rₜₕ.
• “Turn off” independent sources (for finding Rₜₕ):
  • Independent voltage source → short circuit.
  • Independent current source → open circuit.
  • Dependent sources are not turned off (they stay active).
• With a load R_L connected to a Thevenin source:
  I_L = Vₜₕ/(Rₜₕ + R_L), V_L = VₜₕR_L/(Rₜₕ + R_L).

2. Key Ideas (What Earns Marks)

• The first mark is often just choosing the correct two terminals and stating “equivalent seen from A–B.”
• Use the method that minimizes algebra:
  • Vₜₕ: open-circuit voltage (often a divider or a quick KCL/KVL).
  • Rₜₕ: easiest by source-killing and series/parallel; if dependent sources exist, use a test source.
  • Rₜₕ = V_oc/I_sc is fast when I_sc is easy to compute (linear networks).
• Convert Thevenin ↔ Norton when it turns series into parallel (or vice versa).
• Treat Rₜₕ as the “one resistance that matters” for:
  • maximum power transfer,
  • RC/RL time constants (the resistance seen by C or L with sources turned off).
• Sanity checks are not optional:
  • R_L → 0 should give V_L → 0.
  • R_L → ∞ should give I_L → 0.

3. Detailed Explanations

A. Finding Vₜₕ (open-circuit voltage).
Remove the load between A and B and compute V_AB. Common shortcuts:

• Voltage divider: if A is the node between series resistors.
• Superposition: compute contributions from each independent source separately and add.
• Symmetry: equal branches often mean equal node potentials (so V_AB = 0).

B. Finding Rₜₕ (resistance seen from the terminals).
Method 1 (no dependent sources): turn off independent sources, then reduce by series/parallel (and occasional Δ–Y if needed).

Method 2 (dependent sources present): keep dependent sources active, turn off independent sources, apply a test source at A–B:

Method 3 (linear networks): compute I_sc with A–B shorted and use

C. Conversions and quick algebra.
Thevenin → Norton:

Norton → Thevenin:

D. Two high-yield consequences.

• Maximum power transfer to a resistive load:
  P_L = (Vₜₕ² R_L)/((Rₜₕ + R_L)²), max at R_L = Rₜₕ.
• Time constants: to find τ, compute the resistance seen by the reactive element with independent sources turned off:
  τ = RₜₕC (RC), τ = L/Rₜₕ (RL).

4. Common Mistakes

• Finding Rₜₕ without removing the load first.
• Turning off dependent sources (don’t).
• Turning off sources the wrong way (current source shorted, voltage source opened).
• Mixing sign conventions: defining V_AB one way and using the opposite in Vₜₕ.
• Using Rₜₕ = V_oc/I_sc on a circuit that is not linear (e.g. contains a diode without linearization).
• Forgetting internal resistances (real batteries, meters, coils).

5. Exam Tips

• Draw the final equivalent circuit explicitly and label Vₜₕ and Rₜₕ.
• Write “open circuit ⇒ I = 0 in this branch” whenever it kills a term.
• Keep your answer symbolic until the end; many olympiad problems simplify only after cancellations.
• Do at least one limiting-case check using R_L → 0 and R_L → ∞ (it catches algebra slips fast).
• If the question mentions “maximum power,” “optimal load,” or “time constant,” think Thevenin immediately.

6. Worked Examples

Example 1 (classic divider seen by a load).
A source E drives R₁ in series with R₂. The load connects across R₂ (terminals are the top and bottom of R₂). Find the Thevenin equivalent seen by the load and then the load current I_L.

Example 2 (Thevenin between the midpoints of a bridge).
A source E is connected between a top node and a bottom node. Left branch: R₁ (top to node A) then R₂ (node A to bottom). Right branch: R₃ (top to node B) then R₄ (node B to bottom). Find the Thevenin equivalent between A and B.

Example 3 (maximum power to a load).
A linear network has Thevenin equivalent (Vₜₕ,Rₜₕ). Find the R_L that maximizes power delivered to the load and the corresponding Pₘₐₓ.

7. Mind Stretchers

• Dependent sources can produce an effective Rₜₕ that is negative (active circuits). Try a test-source calculation and interpret what “negative resistance” would do to a load.
• If the network contains a nonlinear element (diode, filament lamp), Thevenin still works after linearization about an operating point: replace the element by its small-signal resistance.
• Infinite ladders can often be solved by self-similarity: write an equation for the equivalent resistance and solve it, then wrap the result into a Thevenin equivalent for the load.

8. Practice