Superposition Theorem

Step 1 — the 12 V source alone. Deactivate the current source by opening it. What remains is a simple divider:

Step 2 — the 2 A source alone. Deactivate the voltage source by shorting it. The 4 Ω is now a shunt, in parallel with the 6 Ω across the current source:

Add. Both contributions flow downward through the 6 Ω, so both are positive in the reference direction:

Check by direct nodal analysis. KCL at \(N\) with both sources present:

so \(I = 12/6 = 2\ \text{A}\;\checkmark\). Note also that the node voltages superpose too: \(7.2 + 4.8 = 12\ \text{V}\;\checkmark\)

Set 8, Problem 9 wrote the whole network as one linear system:

Here \(\mathbf{Y}\) depends only on the resistors and dependent-source coefficients; the independent sources appear only on the right-hand side.

Split the source vector into one term per independent source:

Because \(\mathbf{Y}\) is the same matrix in every case, its inverse distributes across the sum:

And \(\mathbf{V}_n^{(k)}\) is exactly the solution with source \(k\) alone. Branch currents follow by \(\mathbf{i} = \mathbf{G}_b\mathbf{A}^{\mathsf T}\mathbf{V}_n\), itself linear, so they superpose too.

What was required. One thing only: that \(\mathbf{Y}\) not depend on the solution. That is precisely linearity — resistances fixed, and dependent-source coefficients constant. Nothing about topology, planarity or source type entered.

Why dependent sources cannot be deactivated. They live inside \(\mathbf{Y}\), not in \(\mathbf{i}_s\). Removing one changes the matrix, and the argument above needs the same \(\mathbf{Y}\) in every sub-circuit. That is the whole reason for the rule.

Why deactivation takes two forms. Setting \(\mathbf{i}_k = \mathbf{0}\) means the source contributes nothing. For a current source that means \(I = 0\) — an open circuit. A voltage source enters the nodal system only after conversion to Norton form (Set 10, Problem 15), where setting its value to zero leaves \(V = 0\) — a short:

Step 1 — the 12 V source alone. Shorting the 6 V source turns its branch into a bare 2 Ω to the reference, in parallel with the middle 4 Ω:

Step 2 — the 6 V source alone. Shorting the 12 V source leaves its 4 Ω as a shunt, parallel with the middle 4 Ω:

Adding:

Check. KCL at \(N\) with both sources active:

An equal-contribution coincidence. The two sources happen to contribute 3 V each, despite differing by a factor of two in voltage — because the 6 V source sees the stiffer path (2 Ω against 4 Ω). Millman's theorem (Set 10, Problem 11) gives the same result in one line:

Step 1 — the 1 V source alone, with the current source opened. A single series loop of \(1 + 2 = 3\ \Omega\):

Step 2 — the 1 A source alone, with the voltage source shorted. Its 1 Ω becomes a shunt, and the 1 A drawn out of \(N\) is supplied by the two resistors in parallel. The 2 Ω branch supplies

Opposite to the reference direction, so \(I'' = -\tfrac13\ \text{A}\).

Adding:

Check. KCL at \(N\):

The node sits at exactly zero, so no current flows in the 2 Ω \(\checkmark\)

What a zero result does not mean. The branch is not idle — it carries \(\tfrac13\) A one way from one source and \(\tfrac13\) A the other from the second, and those currents are physically real in the sense that removing either source reveals them. What is zero is only the sum.

With every source deactivated but one, the loop resistance is the same each time:

Step 1 — the 2 V source alone (4 V shorted, 2 A opened):

Step 2 — the 4 V source alone (2 V shorted, 2 A opened). It drives the loop the opposite way:

Step 3 — the 2 A source alone (both voltage sources shorted). Transform it with its parallel 4 Ω into an 8 V source in series with 4 Ω, which leaves the same 12 Ω loop:

Adding:

Check without superposition. The net driving voltage round the loop is \(2 - 4 - 8 = -10\ \text{V}\), so

From Problem 1, \(I' = 1.2\ \text{A}\), \(I'' = 0.8\ \text{A}\) and \(I = 2.0\ \text{A}\).

But the true power is

Nearly double. Adding the powers understates the answer by 11.52 W — not a rounding error but a structural one.

Where the missing term is. Expand the square rather than assuming it distributes:

And \(12.48 + 11.52 = 24\ \text{W}\;\checkmark\). The cross term is exactly what superposition of powers discards.

The reason, stated properly. Superposition is a consequence of linearity, and \(P = I^2R\) is quadratic in the response. A quadratic function does not distribute over sums:

The one exception. The cross term vanishes when \(I'I'' = 0\) — that is, when at most one contribution is non-zero — or, in AC circuits, when the two contributions are orthogonal, as sinusoids at different frequencies are:

This is why average powers at different harmonics do add — the basis of the RMS summation used throughout Sets 26 and 33.

The correct procedure is therefore always the same: superpose the current or voltage, obtain the total, and only then square it.

The dependent source \(2V_x\) is active in both sub-circuits, and \(V_x\) is recomputed in each — its value differs between them, since it is a response.

Step 1 — the 10 V source alone, with the 2 A source opened. The network reduces to a single series loop carrying \(I_1\), so \(V_{x1} = 2I_1\). KVL:

Step 2 — the 2 A source alone, with the 10 V source shorted. KCL at the top node, whose voltage is \(V_{x2}\):

The current in the 5 Ω branch is then

Adding:

Check by direct nodal analysis with both sources active:

Note that \(V_x\) superposes too: \(V_{x1} = 2I_1 = 20/3\) and \(V_{x2} = 20/3\), summing to \(40/3\;\checkmark\)

Step 1 — the 6 V source alone, with the 1 A source opened. The network becomes a single loop, so \(i_0' = i'\), and KVL with the dependent source still present gives

Step 2 — the 1 A source alone, with the 6 V source shorted. The current source sits between the two branches, so a supermesh is needed (Set 5). The constraint and the supermesh KVL:

Substituting the constraint:

Adding both responses:

Check by solving the full circuit directly. The 1 A source gives \(i - i_0 = 1\), and the supermesh KVL including the 6 V source gives \(3i_0 + 5i = 6\). Substituting \(i = i_0 + 1\):

Three independent sources means three sub-circuits, with the dependent source \(4i\) present in all of them.

Step 1 — the 8 A source alone (2 A opened, 10 V shorted). The supermesh KVL and the KCL constraint:

Step 2 — the 2 A source alone (8 A opened, 10 V shorted), with \(i'' = i_2 + 2\):

Step 3 — the 10 V source alone (both current sources opened). A single loop:

Adding:

Check. The three contributions to \(i\) are \(-16, 4, 10\), so \(i = -2\ \text{A}\); those to \(i_1\) are \(-8, 2, 10\), so \(i_1 = 4\ \text{A}\). The overall supermesh KVL \(-i + 2i_1 = 10\) then reads

and \(V_1 = 3(-2) = -6\ \text{V}\;\checkmark\)

Step 1 — the 4 V source alone, with the 2 A source opened. The remaining loop gives \(V_x' = -2I'\), and KVL round it:

Step 2 — the 2 A source alone, with the 4 V source shorted. KCL at the top node, whose voltage is \(V_x''\):

Adding:

Check directly. KCL at the node with both sources active:

And \(V_x\) superposes: \(V_x' = -2(-0.8) = 1.6\) plus \(V_x'' = -2.4\) gives \(-0.8\ \text{V}\;\checkmark\)

The full source vector is \((-120, 0, 120)^{\mathsf T}\). Split it into one term per source:

Left source alone. Solving \(\mathbf{R}\mathbf{i} = (-120,0,0)^{\mathsf T}\):

Right source alone. Solving \(\mathbf{R}\mathbf{i} = (0,0,120)^{\mathsf T}\):

Note that \(\mathbf{i}''\) is \(\mathbf{i}'\) reversed end-to-end and negated — the numerical signature of the network's antisymmetry.

Adding:

What the symmetry argument was asserting. Each source alone drives a substantial current through the middle mesh — \(\mp 0.6667\) A, not a small quantity. The middle mesh is idle only because those two contributions are equal and opposite:

Symmetry guarantees the cancellation without computing either term, which is why Set 4 got the answer in two lines. Superposition shows what was being cancelled.

Reciprocity (Set 4, Problem 19) supplies the reason the two solutions mirror each other: the matrix is symmetric, so the response at mesh 3 to a source at mesh 1 equals the response at mesh 1 to the same source at mesh 3 — visible here as \(-0.1667\) and \(+0.1667\).

The test. Superposition pays when deactivating sources simplifies the topology — when each sub-circuit can be solved by inspection while the original needs simultaneous equations. It does not pay when the sub-circuits are as hard as the original, only more numerous.

Applying it:

Why Problem 5 was a poor case. Its network is a single loop, and in a series loop the sources already add — \(2 - 4 - 8 = -10\ \text{V}\) over 12 Ω. Deactivating a source changes nothing structurally, so three sub-circuits reproduce what one KVL gives.

Why Problem 1 was a good case. Opening the current source left a series divider; shorting the voltage source left a parallel pair. Neither sub-circuit needed an equation, while the original needed a node equation.

Where superposition is genuinely indispensable. Not as a calculation shortcut at all, but where the decomposition is the point:

Let network \(N\) have terminals carrying current \(I\) at voltage \(V\). By the substitution principle, replace the load by an ideal current source of exactly \(I\) — nothing inside \(N\) changes.

The network now contains its own internal sources plus this one external source, so superposition applies with two groups:

First term. The external source deactivated means opened, so \(I = 0\) and the terminal voltage is by definition the open-circuit voltage:

Second term. All internal independent sources deactivated leaves a network of resistors and dependent sources, presenting some resistance \(R_{TH}\). With \(I\) flowing out of the terminal:

Adding gives the Thévenin relation, and since \(I\) was arbitrary the equivalence holds for every load:

Why dependent sources survive both terms. Problem 2 located them inside \(\mathbf{Y}\), not in the source vector. Superposition splits only the source vector, so a dependent source is present — and active — in every term of any superposition. Deactivating one would mean using a different \(\mathbf{Y}\) in one term than another, and the theorem's proof collapses.

This also explains the three routes to \(R_{TH}\) of Set 9, Problem 6. Deactivation computes \(R_{TH}\) as the resistance of the second term's network directly; the test source measures the same network; and \(V_{oc}/I_{sc}\) extracts it from two points on the resulting line.

Problem 2 showed the source vector splits arbitrarily. Nothing in that argument required one source per term — grouping several into one \(\mathbf{i}_k\) is equally valid:

Each group is analysed with all sources outside it deactivated and all sources inside it active.

Applying it to Problem 5. Group the two voltage sources together and take the current source alone.

A2 V and 4 V active, 2 A opened. They oppose in the 12 Ω loop:

B2 A alone, both voltage sources shorted — the same as Step 3 before:

Adding:

Same answer, two sub-circuits instead of three. And the grouped term \(-1\ \text{V}\) is exactly the sum \(1 + (-2)\) of the two separate ones.

When grouping helps most. Group sources that are of the same type and share a loop or node — they combine arithmetically before any circuit analysis. Keep separate any source whose deactivation changes the topology, since that is where the simplification lies.

The principle. With a single source, \(\mathbf{V}_n = \mathbf{Y}^{-1}\mathbf{i}_s\), so multiplying \(\mathbf{i}_s\) by \(k\) multiplies every response by \(k\). So guess a response, propagate it through the network to find what source it implies, then rescale.

Assume \(I_{6\Omega} = 1\ \text{A}\) and work back towards the source. Every step is a multiplication — no equations at all:

The 18 Ω shunt then carries \(9/18 = 0.5\ \text{A}\), so the second series resistor carries \(1 + 0.5 = 1.5\ \text{A}\):

The 15 Ω shunt carries \(15/15 = 1\ \text{A}\), so the total source current is \(1.5 + 1 = 2.5\ \text{A}\):

Rescale. The assumed 1 A requires a 25 V source, but the real one is 20 V:

Every figure matches Sets 2, 4, 6 and 10 exactly — obtained by six multiplications and one division.

Its limitation. The method needs the response to propagate unambiguously backwards, which requires a single source and a ladder structure. A bridge has no such ordering, and two sources would need two independent scalings — at which point superposition proper is the tool.

By superposition. With one 5 V source alone, the other shorted, the diode conducts and the node sits at 0.7 V. The active branch delivers

but the shorted branch's 1 kΩ now runs from the node to the reference and draws \(0.7/1000 = 0.7\ \text{mA}\), so the diode receives \(4.3 - 0.7 = 3.6\ \text{mA}\).

By symmetry the second source alone gives the same, so superposition would predict

Directly. With both sources active the diode still drops 0.7 V, and each 1 kΩ branch delivers

The discrepancy is 1.4 mA, and its origin is exact: superposition counted each branch as a 0.7 mA sink once, in the sub-circuit where its source was shorted. In reality neither branch ever sinks anything — both always source 4.3 mA.

Why the method fails here. The diode's 0.7 V is not proportional to anything — it is a constant, present whatever the current. In matrix terms it contributes to \(\mathbf{Y}\) a term that depends on the solution (whether the diode conducts at all), so \(\mathbf{Y}\) is not the same in every sub-circuit and Problem 2's argument fails at its only step.

What can be salvaged. Once the diode's state is fixed, it behaves like a 0.7 V source, and a network of resistors and constant sources is linear. So superposition may be applied to the piecewise-linear model provided the 0.7 V drop is treated as a source in its own right — three sources, not two. With one 5 V source alone, the other and the 0.7 V source shorted, the node sits at zero:

With the 0.7 V source alone and both supplies shorted, it drives 0.7 mA out through each 1 kΩ, so its contribution to the diode current is negative:

Correct now, because every source is genuinely independent and \(\mathbf{Y}\) is the same in all three sub-circuits. The naive two-term version failed precisely by leaving the 0.7 V inside the network, where it was counted once per sub-circuit instead of once in total.

Why \(I_{sc}\) superposes. With the terminals shorted, the short is simply another branch of the network, and \(I_{sc}\) is the current in it — an ordinary branch current. Problem 2 established that every branch current superposes:

Nothing about the branch being a short matters. The same argument covers \(V_{oc}\), which is a branch voltage of the open terminal pair.

Checking against Set 10: \(1.2 - 1.5 = -0.3\ \text{mA}\), and the equivalent then gave \(V_0 = -0.286\ \text{V}\) — confirmed there by direct analysis.

Why \(R_N\) does not superpose. It is not a response to anything. By definition it is computed with all independent sources dead, so there is no source to attribute it to:

Superposition splits the source vector, and \(R_N\) lives on the other side of the equation entirely.

The general test. A quantity superposes if and only if it is a linear function of the source vector:

Since \(V_{oc}\) and \(I_{sc}\) both superpose while \(R_{TH}\) does not, the sensible procedure for a multi-source network is to superpose whichever of the two is easier and obtain the resistance separately — exactly what Set 10, Problem 6 did.

The governing equation is linear in \(v\), with two independent inputs — the source and the initial condition:

Zero-input response — the initial condition alone, with the source deactivated \((V_s = 0)\):

Zero-state response — the source alone, starting from an uncharged capacitor \((V_0 = 0)\):

Adding:

which is the standard first-order result of Set 18, here derived as a superposition rather than quoted.

Why the initial condition counts as a source. A capacitor charged to \(V_0\) is equivalent to an uncharged capacitor in series with a \(V_0\) voltage source — so the "two inputs" really are two sources, and ordinary superposition applies. Deactivating the initial condition means \(V_0 = 0\), exactly as shorting a voltage source.

Note that this decomposition is different from the transient-plus-steady-state split, which groups the same solution by time behaviour rather than by cause:

The two analyses. Superposition licenses treating the supply and the signal as separate sources:

In the AC analysis the DC supply is deactivated — shorted — which is exactly why the supply rail is treated as an AC ground.

Why the separation is only approximate. A transistor is not linear. Its collector current follows something like

and an exponential does not superpose: doubling \(v_{BE}\) squares the current rather than doubling it. Problem 16's objection applies in full.

What rescues it. Expand about the bias point \(V_{BE}\) with a small perturbation \(v_{be}\):

With \(V_T \approx 25\ \text{mV}\) and a 10 mV input the quadratic term is around 20% — uncomfortably large, which is why practical small-signal inputs are held to a few millivolts.

The cost of the second-order term is distortion. It generates a component at twice the input frequency, and since it grows as the square of the amplitude, halving the input quarters the distortion while only halving the output — the trade every amplifier designer makes.

So the DC and AC analyses are separate not because superposition holds exactly, but because the circuit has been linearised about its operating point, and superposition then holds for the linearised model.

The single test. Write the network as \(\mathbf{Y}\mathbf{V}_n = \mathbf{i}_s\). Then:

Everything below is a consequence.

What superposes. Branch currents, branch voltages, node voltages, mesh currents, \(V_{oc}\), \(I_{sc}\), and the controlling variables of dependent sources — all linear in \(\mathbf{i}_s\).

What does not.

What the theorem needs. Only that \(\mathbf{Y}\) be constant — fixed resistances and fixed dependent-source coefficients. It does not need planarity, a particular source type, independence of the sources from one another, or any restriction on the load.

What it is worth. As arithmetic, usually less than nodal or mesh analysis (Problem 12). As a principle, it is indispensable: it proves Thévenin and Norton (Problem 13), justifies superposing \(I_{sc}\) (Problem 17), separates bias from signal (Problem 19), splits transient responses by cause (Problem 18), and is the only way to handle sources at different frequencies.

And its deepest consequence. The reason a linear circuit can be described by a transfer function at all is superposition: an arbitrary input is decomposed into components, each is passed through the network independently, and the results are added. Fourier analysis in Sets 33 and 34, and convolution in Set 32, are that idea carried to the limit of infinitely many components.