GameSkillPro

Lesson DC Circuits · Advanced analysis and integration

Thévenin, Norton, and superposition

Thévenin replaces a network with voltage (E_{Th}) in series with (R_{Th}). Norton uses current (I_N) in parallel with (R_N). Superposition solves multi-source circuits one source at a time, then adds effects algebraically. Shortcuts when the load changes or the diagram is heavy.

1

What these theorems are for

Imagine a power source and internal network feeding a load that changes often. Instead of re-analyzing the whole network each time, replace everything seen from the load terminals with a simple equivalent—then only recalculate the load.

That is the daily value of Thévenin and Norton.

2

Thévenin steps

At the two terminals where the load connects:

  1. Remove the load (open the terminals).
  2. \(E_{Th}\) = open-circuit voltage between terminals.
  3. \(R_{Th}\): deactivate independent sources (voltage source → short; current source → open); find resistance looking into terminals.
  4. Equivalent = \(E_{Th}\) in series with \(R_{Th}\). Reconnect load; use Ohm.

Thévenin's theorem reduces the network to one voltage source and one series resistance.

3

Norton steps

Same terminals:

  1. \(I_N\) = short-circuit current between load terminals (replace load with a short).
  2. \(R_N\) same as \(R_{Th}\) with sources deactivated.
  3. Equivalent = current source \(I_N\) in parallel with \(R_N\).

Useful when you think naturally in available short-circuit current.

4

Thévenin ↔ Norton conversion

\[ E_{Th} = I_N imes R_{Th},\quad R_N = R_{Th} \] Convert when the other form makes the next step easier.

5

Superposition theorem

For linear resistive networks with multiple sources:

  1. Keep one source active; deactivate others (V → short, I → open).
  2. Find the desired voltage or current contribution.
  3. Repeat for each source.
  4. Add algebraically (watch signs/directions).

Does not apply blindly to nonlinear devices. In DC resistive circuits, it works.

6

Choosing a tool

NeedUse
Try many different loadsThévenin or Norton
Think in Thevenin voltageThévenin
Think in short-circuit currentNorton
Several sources, one quantitySuperposition or Kirchhoff
Check equivalenceConvert Th ↔ N
7

Toy Thévenin example

12 V source, 4 Ω series, 8 Ω to return from tap. Load \(R_L\) on tap.

  1. Open load: \(E_{Th} = 12 × 8/(4+8) = 8\,\mathrm{V}\) (divider view).
  2. Short source: \(R_{Th} = 4 \parallel 8 = 8/3\,\Omega\).
  3. With \(R_L = 8\,\Omega\): \(I_L = E_{Th}/(R_{Th}+R_L)\).

Faster than full re-analysis for each \(R_L\).

8

Field warnings

  • \(R_{Th}\) is not always "the resistor you see"—deactivate sources and compute what the terminals see.
  • Do not bolt a short across live power terminals to "get Norton" without a controlled procedure.
  • Equivalent is valid only from the same two terminals.
9

Field case

Situation. A lab supply with internal resistance feeds changing test loads. Each time the team redraws mesh equations.

How to think with this lesson.

  • Measure open-circuit V → \(E_{Th}\).
  • Measure loaded V and I with known \(R_L\); estimate \(R_{Th} = (E_{Th}-V_L)/I_L\), or compute from schematic.
  • For each new load: \(I_L = E_{Th}/(R_{Th}+R_L)\).

Conclusion: Thévenin turns "ugly network + variable load" into basic arithmetic.

In the field

Symptom

Load voltage sags more than expected

Where to look

High (R_{Th}): weak source, long leads, fuse resistance

Likely causes

  1. Ignored internal resistance
  2. current-limited supply

What to measure

  1. Open-circuit V ((E_{Th}))
  2. loaded V and I
  3. estimate (R_{Th})

What not to do

  • Short power terminals for Norton current without control

Checklist

  • I find (E_{Th}) with load removed (open terminals)
  • I find (R_{Th}) with sources properly deactivated
  • I relate Norton: (I_N), (R_N = R_{Th})
  • I apply superposition one source at a time
  • I know equivalent is for two specific terminals
  • I estimate (R_{Th}) from loaded sag when needed

Common mistakes

Symptom Typical cause Action
Wrong \(R_{Th}\) Left sources active or all opened V source → short; I source → open
Superposition mismatch Added powers instead of V or I Superpose linear quantities only
Equivalent fails at new tap Changed terminal pair Recalculate for new port
\(E_{Th}\) with load connected Did not open terminals Remove load for open-circuit V