Thévenin's Theorem

Equivalent forms

R_{th} = \frac{V_{oc}}{I_{sc}}

V_{th} = V_{oc}

Collapses an arbitrarily complicated linear network into two numbers — the deepest labor-saving trick in circuit analysis.

Symbol	Name	SI unit	Dimension	Range
$I_L$	Load currentoutput Current delivered to the load	A	I	0 – 100
$V_th$	Thévenin voltage Open-circuit terminal voltage	V	M·L²·T⁻³·I⁻¹	0 – 100
$R_th$	Thévenin resistance Equivalent internal resistance	Ω	M·L²·T⁻³·I⁻²	0.01 – 100
$R_L$	Load resistance Resistance connected to terminals	Ω	M·L²·T⁻³·I⁻²	0.01 – 100

Unit systems

Symbol	SI
$I_L$	A
$V_th$	V
$R_th$	Ω
$R_L$	Ω

Where it holds

Holds for any linear, time-invariant network (resistors, linear sources, and — in phasor form — impedances). Fails for nonlinear elements (diodes, transistors in their nonlinear regions) unless linearized about an operating point.

Dimensional analysis

$[V_th]/([R_th]+[R_L]) = V/\Omega = A \checkmark$

Discovery

Léon Charles Thévenin · 1883

A French telegraph engineer, Thévenin published the equivalent-source theorem to simplify telegraph-network calculations — unaware Helmholtz had stated it 30 years earlier. The name stuck.

Try this

However tangled the box of batteries and resistors, anything you plug in sees just one battery behind one resistor. How do you find them?

A network presents 12 V open-circuit and delivers 3 A when shorted. Find its Thévenin equivalent and the current into a 2 Ω load.

Research status: stable

Real-world applications

Modeling battery internal resistance
Sensor and amplifier interfacing
Fault and load analysis
Reducing big SPICE nodes to equivalents

Common misconceptions

You must know the box's internals — two terminal measurements fully characterize it
R_th is found by leaving sources on — you null independent sources (shorts for voltage, opens for current) to find it

Experimental verification

Measure the open-circuit voltage (V_th) and short-circuit current (I_sc);

R_th = V_oc/I_sc

. Predicted currents into arbitrary loads then match measurement for any linear box.

Derivation

By superposition, the terminal voltage of a linear network is an affine function of the terminal current:

V = V_oc - R_th I

.

Identifying

V_th = V_oc

(current zero) and R_th as the slope (the resistance seen with all independent sources nulled) yields the equivalent.

With a load R_L,

I_L = V_th/(R_th + R_L)

.

Limiting cases

R_L \to 0

(short)⟶

I_L \to V_th/R_th = I_sc

The short-circuit current is set entirely by the Thévenin resistance.

R_L \to \infty

(open)⟶

I_L \to 0

, V_terminal

\to V_th

With no load the terminals sit at the open-circuit voltage.

What if…

What if you match the load to R_th?

You deliver maximum power to the load — the maximum-power-transfer condition $R_L = R_th$ .

What if the network has AC sources?

The same theorem holds with impedances: a Thévenin voltage phasor in series with a Thévenin impedance.

1

12 V open, 3 A short, 2 Ω load

Given ·

V oc:: 12
I sc:: 3
R L:: 2

Find · R_th and I_L

Steps

$V_th = V_oc = 12\,\mathrm{V}$
$R_th = V_oc/I_sc = 12/3 = 4 \Omega$
$I_L = V_th/(R_th+R_L) = 12/6 = 2$ A

Result · Equivalent is 12 V behind

4 \Omega

; the

2 \Omega

load draws 2 A.

norton theoremmaximum power transferKirchhoff's Voltage Law→resistors series parallel