03 · Foundational

Linearity & superposition

Superposition is not a source-switching trick: it follows from linear circuit equations. Define additivity and homogeneity, decompose a two-source network into source contributions, keep dependent sources active, and learn why voltage/current superpose but power does not.

130–165 minFoundational

In this lesson

Define homogeneity and additivity and use them to recognize a linear circuit relation.
Derive the superposition principle from a linear system of circuit equations.
Deactivate independent voltage/current sources by setting their imposed values to zero while keeping dependent sources active.
Solve a two-independent-source resistor network by superposition and verify the sum against the direct solution.
Apply superposition in a circuit that contains a linear dependent source.
Explain rigorously why branch voltage and current contributions superpose but resistor power contributions generally do not.
Use source polarity and sign consistently when a source contribution is negative.

Starting question

Why is it legitimate to solve one source at a time and add the answers?

The answer is linearity. Superposition is not a special trick attached to circuit diagrams. It is a theorem about linear equations: if the equations relating independent excitations to a chosen voltage or current response are linear, the response to a weighted sum of excitations is the same weighted sum of the individual responses.

03.1Linearity means homogeneity plus additivity

Let a circuit mapping L take an excitation x to a response y=L(x). The mapping is linear if, for any excitations x1 and x2 and any scalars α and β,

L(αx₁+βx₂) = αL(x₁) + βL(x₂)

This single statement contains two useful tests.

Homogeneity: scaling the excitation scales the response by the same factor.

L(αx)=αL(x)

Additivity: the response to a sum is the sum of the responses.

L(x₁+x₂)=L(x₁)+L(x₂)

An ideal resistor with constant R is linear because V=RI. A controlled source with a constant proportional coefficient, such as io=gmvx, is also linear. A square-law element with i=kv² is not.

03.2The circuit equations make superposition almost unavoidable

After choosing node voltages, mesh currents or another valid set of unknowns, a well-posed linear DC circuit can be written schematically as

A x = B s

Here x is the vector of unknown circuit responses, s is the vector of independent source values, and A and B contain fixed coefficients from resistors and linear dependent-source gains.

If the network has a unique solution,

x = A⁻¹B s

so x is a linear function of s. If the source vector is split into s=s1+s2, then

x(s₁+s₂) = x(s₁) + x(s₂)

This is the superposition principle. “One source at a time” is merely a convenient way of constructing the source-vector pieces.

03.3What “deactivate the other independent sources” really means

To isolate the contribution of one independent source, set every other independent source value to zero while preserving the rest of the model.

Independent ideal voltage source: set its imposed voltage to 0 V → replace by an ideal short.
Independent ideal current source: set its imposed current to 0 A → replace by an ideal open.
Linear dependent source: keep its controlling equation active.
Resistors and other linear elements: keep them unchanged.

This is analytical source deactivation, not physically turning equipment off. The zero-source replacements follow directly from the source terminal laws established in chapter 01.

When the source of interest is negative relative to its defined reference, keep that sign. Superposition adds signed contributions.

03.4Direct solution of a two-source network

Figure 3.1 — Two independent sources act on the same node

The left source fixes 12 V and reaches VOUT through 3 kΩ. The right source fixes 6 V and reaches the same node through 6 kΩ. A 2 kΩ resistor connects VOUT to ground.

verified topology

The one question is: what VOUT results when both excitations act simultaneously in this linear network?

Let V denote VOUT. Kirchhoff's current law with currents leaving V gives

(V−12 V)/(3 kΩ) + (V−6 V)/(6 kΩ) + V/(2 kΩ) = 0

Multiply by 6 kΩ:

2(V−12 V)+(V−6 V)+3V=0

6V=30 V

V=5 V

This direct solution is our reference answer. Superposition must reproduce it exactly—not approximately.

03.5Contribution from the 12 V source alone

Keep the 12 V source at its specified value. Set the independent 6 V source to 0 V, which replaces that source by a short. The 6 kΩ branch therefore becomes a resistor from V to ground.

(V⁽¹⁾−12 V)/(3 kΩ) + V⁽¹⁾/(6 kΩ) + V⁽¹⁾/(2 kΩ) = 0

Multiplying by 6 kΩ gives

2(V⁽¹⁾−12 V)+V⁽¹⁾+3V⁽¹⁾=0

V⁽¹⁾=4 V

The superscript labels the source contribution. It does not mean “voltage squared” and it is not a new physical node.

03.6Contribution from the 6 V source alone—and signed addition

Now set the independent 12 V source to 0 V and keep the 6 V source active:

V⁽²⁾/(3 kΩ) + (V⁽²⁾−6 V)/(6 kΩ) + V⁽²⁾/(2 kΩ) = 0

2V⁽²⁾+(V⁽²⁾−6 V)+3V⁽²⁾=0

V⁽²⁾=1 V

Add the two signed contributions:

V = V⁽¹⁾+V⁽²⁾ = 4 V+1 V = 5 V

Figure 3.2 — Decompose the response, not the physical circuit

Each box is an analysis problem. In the first, only the 12 V source keeps its value; in the second, only the 6 V source keeps its value. The signed VOUT contributions are then added.

The result matches the direct solution exactly.

03.7Branch currents also superpose

With both sources active and V=5 V, define branch currents from each source node toward V and from V toward ground:

I₁=(12−5)V/(3 kΩ)=7/3 mA≈2.333 mA

I₂=(6−5)V/(6 kΩ)=1/6 mA≈0.167 mA

I_G=5 V/(2 kΩ)=2.5 mA

KCL gives 2.333 mA+0.167 mA=2.5 mA.

We could obtain each branch current by adding its two source contributions just as we did for V. The rule applies to any voltage or current that is a linear response of the source vector, provided its reference direction stays fixed in every subproblem.

03.8Power generally does not superpose

For the 2 kΩ resistor, the actual voltage with both sources active is 5 V, so its power is

P = V²/R = (5 V)²/(2 kΩ) = 12.5 mW

If one incorrectly computes “power due to source 1” from V(1)=4 V and “power due to source 2” from V(2)=1 V, then adds them, one gets

P⁽¹⁾+P⁽²⁾ = (4 V)²/(2 kΩ)+(1 V)²/(2 kΩ)=8.5 mW

which is not 12.5 mW.

The reason is algebraic:

(V⁽¹⁾+V⁽²⁾)²/R = [V⁽¹⁾]²/R + [V⁽²⁾]²/R + 2V⁽¹⁾V⁽²⁾/R

The cross term is missing from the naive power sum. Superpose voltage/current first, then compute power from the total solved voltage and current.

03.9Dependent sources stay active: a compact proof by example

Consider one node V connected through 1 kΩ to independent source S1=10 V, through another 1 kΩ to independent source S2=4 V, through 2 kΩ to ground, and through a downward linear controlled current source id=gV with g=0.5 mS.

With both independent sources active:

(V−10)/1k + (V−4)/1k + V/2k + (0.5 mS)V = 0

3 mS·V=14 mA

V=14/3 V≈4.667 V

For the contribution of S1, set only S2 to zero and keep the controlled-source term:

3 mS·V⁽¹⁾=10 mA

V⁽¹⁾=10/3 V≈3.333 V

For the contribution of S2, set only S1 to zero and again keep the controlled source:

3 mS·V⁽²⁾=4 mA

V⁽²⁾=4/3 V≈1.333 V

V⁽¹⁾+V⁽²⁾=14/3 V

If the controlled source had been manually removed, the coefficient 3 mS would change and the source contributions would no longer describe the original circuit.

03.10Verification workflow and mastery practice

Choose the response voltage or current and fix its reference.
Check that the active element laws used in the model are linear over the intended operating range.
Keep one independent source at a time; set the other independent source values to zero.
Keep every dependent-source relation active.
Solve each subproblem with the same response reference.
Add the signed voltage/current contributions.
Verify the sum against a direct all-sources solution when practical.
Compute power only after reconstructing total voltages and currents.

ClaimDiagnosis

“Turn off all other sources.”Too vague: set independent source values to zero; keep dependent sources active.

“Power contributions add.”Generally false because power is nonlinear in voltage/current.

“A negative contribution is impossible.”It means that source drives the response opposite to the fixed reference.

“Superposition works for any circuit.”It requires the source-to-response relation to be linear within the model.

In the worked two-source circuit, reverse the 6 V source polarity so its node is at −6 V. Find the direct VOUT and both superposition contributions.
Double both independent source voltages in the worked circuit. Predict VOUT using homogeneity before calculating.
Keep the original source values but change RG from 2 kΩ to 3 kΩ. Find VOUT directly and by superposition.
For the original circuit, calculate the total resistor power and verify it equals total source delivery.
Use V(1)=4 V and V(2)=1 V to calculate the missing power cross term in the 2 kΩ resistor.
A diode is modeled by i=Is(exp(v/nVT)−1). May ordinary linear superposition be applied to its DC circuit? Explain.
In the dependent-source example, what incorrect result is obtained if the gV source is opened during each source subproblem?

Solutions

1. Direct KCL gives (V−12)/3k+(V+6)/6k+V/2k=0, hence V=3 V. The 12 V contribution remains +4 V; the reversed 6 V contribution is −1 V; their signed sum is 3 V.

2. The entire source vector doubles in a fixed linear network, so VOUT doubles from 5 V to 10 V.

3. Direct KCL gives (V−12)/3k+(V−6)/6k+V/3k=0. Multiplying by 6k gives 2V−24+V−6+2V=0, hence V=6 V. Contributions are 4.8 V and 1.2 V.

4. R1 absorbs 49/3 mW≈16.333 mW, R2 absorbs 1/6 mW≈0.167 mW and RG absorbs 12.5 mW, total 29 mW. The 12 V source delivers 28 mW and the 6 V source delivers 1 mW.

5. The cross term is 2(4 V)(1 V)/(2 kΩ)=4 mW. Adding it to 8.5 mW gives the correct 12.5 mW.

6. No. The exponential constitutive law is nonlinear, so the source-to-response mapping is not generally linear.

7. Opening the controlled source leaves only 1 mS+1 mS+0.5 mS=2.5 mS of V-dependent conductance. The false contributions become 4 V and 1.6 V, summing to 5.6 V instead of the correct 14/3 V≈4.667 V.

Chapter 04 will use the same terminal-behavior discipline to derive source transformations: when a voltage-source-plus-series-resistance model and a current-source-plus-parallel-resistance model are exactly equivalent at a port.