FOUNDATION · CHAPTER 3 OF 7

Linearity: the one idea behind the theorems

This chapter explains what makes an element and a circuit linear, and shows that proportionality, superposition and Thevenin's theorem all follow from one fact: the circuit equations are linear. You need it to know when a theorem may be used and when it quietly fails.

Updated 12 Oct 2026

What you will be able to do

You will say exactly what makes an element, and then a circuit, linear. You will prove proportionality and superposition from the circuit equations and derive Thevenin's theorem from superposition. You will see why a dependent source keeps a circuit linear and a diode does not, and what each condition of a theorem is for.

Before you start

Chapter 1, Charge, current, voltage and power: what the quantities mean
Chapter 2, Why R, L and C behave as they do
Series and parallel resistors and the voltage divider
Solving two or three simultaneous linear equations
Matrices: product, inverse and determinant
The straight line y = mx + c
Differentiation and integration, including those of a sum
The exponential function e^x

What linear means

A relation is linear when it passes two tests. Homogeneity: scale the cause and the effect scales by the same factor. Additivity: add two causes and the effects add. A circuit is linear when every element passes both, so you check elements one at a time. Independent sources, set from outside, are the causes and are not tested.

LINEAR ELEMENT

f (a x_{1} + b x_{2}) = a f (x_{1}) + b f (x_{2})

Holds for all numbers a, b and all inputs $x_{1}$ , $x_{2}$ . x is the cause (a current or voltage waveform) and f(x) the effect. Putting a = b = 0 gives f(0) = 0: the graph passes through the origin.

A dependent source is an element whose value is a constant times its control, a voltage $v_{x}$ or current $i_{x}$ elsewhere in the circuit. Throughout, the current enters the + terminal of an element's voltage, so $p = v i$ is the power absorbed ( $p < 0$ means it delivers power).

Element	Law	Linear?
Resistor	$v = R i$ , R constant	Yes
Capacitor	$i = C d v / d t$ , C constant	Yes: the derivative of a sum is the sum of the derivatives
Inductor	$v = L d i / d t$ , L constant	Yes, for the same reason
Dependent source	$v = μ v_{x}$ , $i = β i_{x}$ , $i = g v_{x}$ or $v = r i_{x}$ ; constant gains ( $μ$ , $β$ no unit, g in S, r in ohm)	Yes: a constant times its control
Saturated inductor	flux linkage not proportional to i	No (Chapter 2)
Diode	i grows exponentially with v (law in the last section)	No

WATCH OUT FOR

A straight line is not always linear: $v = 5 + 2 i$ (V, A) gives 5 V at $i = 0$ . Test: zero cause must give zero effect. This element is a 5 V source, a cause, in series with a linear 2 ohm resistor.

Superposition from the equations

Choose one node as the reference (ground); the voltage of every other node relative to it is a node voltage. At each other node write Kirchhoff's current law (KCL, Chapter 1): the currents leaving the node sum to zero. Replace each branch current by its element law: a resistor's current is its conductance $1/ R$ (siemens, S) times a difference of node voltages, and a current source's value is a known term. A voltage source has no current law, so its current becomes one more unknown and the source adds one row: its two node voltages differ by the source value. Every row is then linear in the unknowns, with each source value on the right-hand side.

NODE EQUATIONS OF A LINEAR CIRCUIT

G x = s

For resistors and sources: a KCL row per non-reference node and a row per voltage source. The unknowns, node voltages (V) and voltage-source currents (A), are collected together in $x$ . G holds conductances (S), and 1, $- 1$ or 0 where a voltage source enters. $s$ holds source currents (A) in KCL rows and source voltages (V) in source rows.

PROPORTIONALITY AND SUPERPOSITION

x = G^{- 1} s = G^{- 1} s_{1} + G^{- 1} s_{2} + \dots

Needs every element linear and $det G \neq = 0$ . $s_{k}$ is $s$ with every source but source k at zero, so $s = s_{1} + s_{2} + \dots$ . A zero voltage source is a short circuit (0 V at any current), a zero current source an open circuit (0 A at any voltage). Scaling every source by K scales $x$ by K: proportionality.

A capacitor adds $C d v / d t$ to a KCL row and an inductor adds $\frac{1}{L} \int_{0}^{t} v d t$ , with v the element's voltage. First split any initial capacitor voltage or inductor current off as an independent source, so each element starts from zero. The rows then hold derivatives and integrals, not one matrix, but every term is still linear, since the derivative and the integral of a sum are the sums of each. So if $x_{1}$ answers $s_{1}$ and $x_{2}$ answers $s_{2}$ , then $x_{1} + x_{2}$ answers $s_{1} + s_{2}$ , and superposition carries over.

WORKED EXAMPLE

A 12 V source, negative terminal at ground, in series with R = 4 ohm feeds node a. A 12 ohm resistor joins a to ground. A 3 A current source pushes current into a from ground. Find $v_{a}$ , then check it by superposition.

Solution

KCL at a, currents leaving positive. The source branch carries $(v_{a} - 12) /4$ away, the 12 ohm carries $v_{a} /12$ , and 3 A comes in: $(v_{a} - 12) /4 + v_{a} /12 = 3$ .
Multiply by 12 to clear the fractions: $3 v_{a} - 36 + v_{a} = 36$ , so $4 v_{a} = 72$ and $v_{a} = 18$ V.
Superposition, current source off (open circuit): no current enters a from it, so 12 V divides over 4 and 12 ohm in series: $v_{a} = 12 \times 12/16 = 9$ V.
Voltage source off (short circuit): the 4 ohm now joins a to ground, so the 3 A flows into 4 ohm in parallel with 12 ohm, which is 3 ohm: $v_{a} = 3 \times 3 = 9$ V.
Add the two responses: 9 + 9 = 18 V. It matches because the equation is linear in the source values.

Answer: $v_{a} = 18$ V, made of 9 V from each source.

WATCH OUT FOR

Power does not superpose: $p = v^{2} / R$ is quadratic. Test: ask whether the quantity is linear in the sources. Above, 9 V alone across the 12 ohm gives 6.75 W, but the full 18 V gives 27 W, not 2 × 6.75 = 13.5 W.

Thevenin, dependent sources and the diode

Take a linear network N with independent sources and two terminals, labelled + and −, with v the voltage of + relative to −. Drive the port with a test current source $i$ into the + terminal. By superposition, v is the response to $i$ alone plus the response to N's own sources alone. The first is proportional to $i$ and the second does not depend on $i$ , and that is all Thevenin's theorem needs.

PORT RELATION OF A LINEAR NETWORK

v = a i + b

N is linear and its equations, with the test source in place, have one solution. a = v/i with N's independent sources at zero (ohm); b = v with i = 0 (V). N's equations see a load only through its current i, so a current source of that value can stand in for any load: N's solution is unchanged and the relation holds with the load.

THEVENIN EQUIVALENT

v = v_{oc} - R_{t h} i_{L}

Put i = 0: $b = v_{oc}$ , the open-circuit voltage. With N's sources at zero, $v = a i$ , so $a = R_{t h}$ , the resistance seen at the port. A load draws $i_{L}$ out of the + terminal, so $i = - i_{L}$ . This is the law of a source $v_{oc}$ in series with $R_{t h}$ , which can replace N for a load outside N, even a diode (conditions below).

RESISTANCE WHEN DEPENDENT SOURCES ARE PRESENT

R_{t h} = \frac{v _{t}}{i _{t}}

Set only the independent sources to zero, drive the port with a test source, and divide its voltage $v_{t}$ (V) by its current $i_{t}$ (A) into the + terminal. Dependent sources stay in: they are relations between circuit variables, not inputs.

A dependent source such as $i_{d} = β i_{x}$ puts a constant times a circuit variable into a row, and that variable is itself linear in the unknowns. So the term moves to the left side and changes entries of G, and $G x = s$ keeps its form: every result above survives, provided $det G \neq = 0$ . A dependent source is not switched off in superposition: it is part of the circuit, not a cause. A diode is different.

DIODE LAW

i = I_{s} (e^{v / V_{T}} - 1)

Holds below reverse breakdown. v is the anode-to-cathode voltage (V), i the current into the anode (A), $I_{s}$ the saturation current (A), $V_{T} \approx 25$ mV near room temperature. Doubling v from 0.3 V to 0.6 V multiplies i by about $e^{12} \approx 1.6 \times 1 0^{5}$ . Homogeneity fails, so superposition can fail, and an N with a diode inside in general has no Thevenin equivalent valid for every load.

WORKED EXAMPLE

A 12 V source, negative terminal at ground, in series with 4 ohm feeds terminal a, and a 12 ohm resistor joins a to ground. The port is a (+) and ground (−). Find the Thevenin equivalent, then the current and voltage for a 6 ohm load.

Solution

Remove the load. No current leaves the port, so 12 V divides over 4 and 12 ohm: $v_{oc} = 12 \times 12/16 = 9$ V.
Set the source to zero (a short circuit). The terminals see 4 ohm in parallel with 12 ohm: $R_{t h} = 4 \times 12/16 = 3$ ohm.
Connect the load. Set $v = v_{oc} - R_{t h} i_{L}$ equal to the load's law $v = 6 i_{L}$ (Ohm's law for 6 ohm): $9 - 3 i_{L} = 6 i_{L}$ , so $i_{L} = 9/ (3 + 6) = 1$ A.
The load voltage is $v = 6 \times 1 = 6$ V.
Check on the original network: 12 ohm in parallel with 6 ohm is 4 ohm, which forms a divider with the 4 ohm, so $v = 12 \times 4/ (4 + 4) = 6$ V.

Answer: $v_{oc} = 9$ V, $R_{t h} = 3$ ohm, $i_{L} = 1$ A and $v = 6$ V.

A theorem's conditions are the facts its proof used. Where one fails, that step of the proof fails, and the result may fail with it.

Every element in N is linear. Proportionality uses homogeneity, and adding responses uses additivity.
$det G \neq = 0$ . Each voltage-source row reads $v_{p} - v_{q} = v_{s}$ ; around a loop of ideal voltage sources, adding these rows, each signed along the loop, cancels every node voltage, so the rows are dependent. A node joined only to independent current sources has a KCL row of zeros. Either way $det G = 0$ : no solution or many.
Initial capacitor voltages and inductor currents are split off as independent sources. Otherwise the response to zero sources is not zero, and $f (0) = 0$ fails.
N and the load interact only through the two terminals: no dependent source in N is controlled by a load variable. Then the load affects N only through $i$ .

Check yourself

CHECK YOURSELF 1One or more options are correct

A linear network has two independent sources. With source 1 acting alone (source 2 set to zero) the voltage $v_{x}$ is 3 V. With source 2 acting alone it is $- 1$ V. Select all true statements.

(A)
With both sources acting, $v_{x} = 2$ V.
(B)
Doubling both sources gives $v_{x} = 4$ V.
(C)
With both sources acting, the power absorbed by a 1 ohm resistor with $v_{x}$ across it is 9 + 1 = 10 W.
(D)
Tripling source 1 and setting source 2 to zero gives $v_{x} = 9$ V.

Answer and solution

Answer: (A), (B), (D)

Superposition gives 3 + (−1) = 2 V, so A is true.
Proportionality doubles the whole response to 4 V, so B is true.
The power is $2^{2} /1 = 4$ W. Power is quadratic and does not superpose, so C is false.
Tripling source 1 gives 9 V, and source 2 at zero contributes nothing, so D is true.

CHECK YOURSELF 2

A 20 V source, negative terminal at ground and positive terminal toward a, in series with 5 ohm feeds terminal a. A 20 ohm resistor joins a to ground. A 4 ohm load is connected between a and ground. Find the load current in amperes using the Thevenin equivalent seen at a and ground.

Answer and solution

Answer: $2$ (accepted range $\pm 0.01$ )

Open circuit: 20 V divides over 5 and 20 ohm, so $v_{oc} = 20 \times 20/25 = 16$ V.
Source off (short): $R_{t h} = 5 \times 20/25 = 4$ ohm.
Load current: $i_{L} = 16/ (4 + 4) = 2$ A.

CHECK YOURSELF 3

A network has no independent sources. Between terminal a and ground it has a 1 k ohm resistor in parallel with a dependent current source that draws $g v$ from a to ground, where v is the voltage at a and $g = 1$ mS. Find $R_{t h}$ at the terminals in ohms, using a test current $i_{t}$ entering a.

Answer and solution

Answer: $500$ (accepted range $\pm 0.5$ )

KCL at a: $i_{t} = v /1000 + g v$ .
With g = 0.001 S, $i_{t} = (0.001 + 0.001) v = 0.002 v$ .
So $R_{t h} = v / i_{t} = 1/0.002 = 500$ ohm.

CHECK YOURSELF 4

Which element law is linear? Here v is in volts, i in amperes and t in seconds.

(A)
$i = 3 v^{2}$
(B)
$v = 2 i + 4$
(C)
$i = 4 \times 1 0^{- 6} d v / d t$
(D)
$i = 1 0^{- 9} (e^{v /0.025} - 1)$

Answer and solution

Answer: (C)

A linear law gives zero effect for zero cause and obeys additivity.
A is quadratic and D is exponential, so both fail.
B gives 4 V at i = 0, so it fails the zero test.
C is a capacitor law with constant C, and a derivative is linear, so C is linear.

Linearity: the one idea behind the theorems

What you will be able to do

Before you start

What linear means

Superposition from the equations

Thevenin, dependent sources and the diode

Check yourself

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