Millman, Reciprocity, Substitution and Tellegen

The statement. Any number of practical voltage sources in parallel are equivalent to a single source

The derivation is Set 12, Problem 15 in three lines. Transform every branch to Norton form:

Current sources in parallel add, and so do conductances:

Converting back gives \(V_{eq} = I_{eq}/G_{eq}\), which is the theorem.

Applying it:

Check against Set 10, Problem 11, which combined the same pair by Norton conversion: \(I_{eq} = 4 + 1.5 = 5.5\ \text{A}\) in parallel with 2 Ω, giving \(V = 11\ \text{V}\;\checkmark\)

Reading the formula. It is a weighted average with the conductances as weights, so:

Here the 3 Ω branch carries twice the weight of the 6 Ω, pulling the result to 11 V rather than the unweighted 10.5 V.

A plain resistor is a source of zero volts. That is the whole trick — it contributes nothing to the numerator but its full conductance to the denominator:

The equivalent resistance:

Check by nodal analysis, which is what Millman is:

Notice what the passive branch did. Without it the two sources alone would give

The 40 Ω drags the result down by 13 V, towards zero — because it is a branch insisting on 0 V with a weight of \(1/40\). Loading a source always pulls its terminal voltage towards zero, and Millman quantifies by how much.

A coincidence worth not over-reading: the answer 80 V equals one of the source voltages exactly. Nothing significant follows — it is arithmetic, and the 80 V branch happens to be carrying zero current as a result.

Apply Set 8's dictionary to Millman's statement — voltage ↔ current, parallel ↔ series, \(G \leftrightarrow R\):

The derivation, again by transformation. Convert each branch to Thévenin form, add in series, convert back:

Applying it:

Check by direct transformation. The two branches become 6 V + 3 Ω and 30 V + 6 Ω; in series that is 36 V + 9 Ω, and converting back gives \(36/9 = 4\ \text{A} \parallel 9\ \Omega\;\checkmark\)

Reading it. The equivalent current is a resistance-weighted average, so here the 6 Ω branch dominates and the answer sits at 4 A, closer to 5 A than to 2 A. As always the result lies between the extremes.

Case 1 — an ideal voltage source in one branch. Then \(R_k = 0\) and \(G_k = \infty\), so both sums diverge. Taking the limit:

The formula does give the right answer in the limit — the ideal source simply fixes the terminal voltage and every other branch becomes irrelevant. But two such branches with different voltages give \(\infty/\infty\), and the circuit genuinely has no solution (Set 3).

Case 2 — the branches are not all in parallel. Millman requires every source to share the same pair of terminals. A source buried inside a ladder does not qualify.

Case 3 — dependent sources. Millman's derivation transformed each branch independently, which assumes each \(V_k\) is a known constant. A branch whose source depends on the terminal voltage cannot be treated this way — its contribution to the numerator is itself unknown.

Case 4 — non-linear branches. A diode has no Thévenin equivalent, so it cannot enter the sum at all.

Summary:

The honest assessment. Millman applies only to the single-node case, where nodal analysis would give one equation anyway. Its value is speed on a specific, common shape — a set of parallel supplies feeding one bus — not generality.

The statement. In a linear network containing only bilateral elements and a single independent source, interchanging the positions of the source and the response leaves the response unchanged.

Set up the mesh system. Set 4 wrote the network as

Writing \(\mathbf{Z} = \mathbf{R}^{-1}\), the current in mesh \(m\) due to a source in mesh \(n\) is \(i_m = z_{mn}v_n\).

The key fact is that \(\mathbf{R}\) is symmetric. Its off-diagonal entry \(R_{mn}\) is minus the resistance shared between meshes \(m\) and \(n\) — and "shared between \(m\) and \(n\)" is the same statement as "shared between \(n\) and \(m\)":

The inverse of a symmetric matrix is symmetric, so

which is the theorem. The two experiments give identical readings because they are governed by the same matrix entry.

What the proof required. Only the symmetry of \(\mathbf{R}\) — and that in turn requires every element to be bilateral, meaning its behaviour is the same in both directions. Resistors, inductors and capacitors all are.

What breaks the symmetry:

A dependent source contributes to \(R_{mn}\) without a matching contribution to \(R_{nm}\), because it acts in one direction only. Problem 8 works an example.

Inverting the mesh matrix gives the transfer impedances:

Symmetric, as Problem 5 requires.

Source in mesh 1, response in mesh 3:

Source in mesh 3, response in mesh 1:

Identical, exactly as the theorem promises.

What is not claimed. Only the one interchanged pair matches. Compare the full solution vectors from Set 11, Problem 11:

Here the whole vector reverses, but only because this particular network happens to be symmetric end-to-end. Reciprocity alone guarantees the two bold entries — nothing more.

The middle mesh illustrates the point. Its current is 0.667 A in both experiments, but that is \(z_{21} = z_{23}\), which reciprocity does not assert — it is a consequence of this network's physical symmetry, not of the theorem.

The nodal statement. A current source \(I\) injected at node \(p\) produces a voltage at node \(q\) equal to the voltage it would produce at \(p\) if injected at \(q\).

The proof is identical in structure to Problem 5, using Set 8's nodal system:

Set 8, Problem 9 built \(\mathbf{Y} = \mathbf{A}\mathbf{G}_b\mathbf{A}^{\mathsf T}\), which is symmetric by construction whenever \(\mathbf{G}_b\) is diagonal — that is, whenever every branch is a plain bilateral element.

Hence \(\left(\mathbf{Y}^{-1}\right)_{qp} = \left(\mathbf{Y}^{-1}\right)_{pq}\), and the two voltages agree.

The two forms compared:

The physical requirement is the same in both. The source must be ideal and the meter must be ideal — a real ammeter with resistance, or a real voltmeter that draws current, becomes part of the network and must be identical in both experiments for the comparison to hold.

What is not reciprocal is the mixed case: a voltage source in one place and a voltage response elsewhere. That ratio is a transfer ratio, not a transfer impedance, and it is not symmetric. Reciprocity always exchanges a source of one kind for a response of the complementary kind.

The modified matrix is no longer symmetric — the dependent source acts in one direction only:

The two transfer impedances now differ:

A factor of three apart — reciprocity fails, and not marginally.

The responsible class of elements is the non-bilateral ones. An element is bilateral when its behaviour is unchanged by reversing it:

Note that mutual inductance is reciprocal, which is not obvious. Set 26 will show that \(M_{12} = M_{21}\) — the flux coupling from coil 1 to coil 2 equals that from 2 to 1 — so a transformer preserves the symmetry of the matrix and reciprocity survives.

The practical significance of non-reciprocity. A circulator or isolator is deliberately non-reciprocal: it passes a signal one way and absorbs it the other, which is exactly what protects a transmitter from its own reflected power. Building one requires a magnetised ferrite — no arrangement of ordinary passive components can do it, precisely because they are all bilateral.

Use 1 — halving the work. Characterising an \(n\)-port requires \(n^2\) transfer impedances, but symmetry means only \(n(n+1)/2\) are independent:

For a passive network the saving approaches half as \(n\) grows. Set 35 uses this constantly — a reciprocal two-port has \(z_{12} = z_{21}\), reducing four parameters to three.

Use 2 — measuring the unmeasurable. If one point of a circuit is inaccessible — buried in a package, or at the far end of a cable — reciprocity lets the measurement be made the other way round. The answer is guaranteed identical.

Use 3 — antenna theory. An antenna's receiving pattern equals its transmitting pattern, which is a direct consequence. This is why antennas are specified by a single radiation pattern rather than two, and why a good transmitting antenna is automatically a good receiving one.

Use 4 — a check on analysis or measurement. If a network is known to be passive and bilateral, then any computed or measured matrix must be symmetric. An asymmetry means an error — in the arithmetic, the wiring, or the assumption that the network is passive.

Use 5 — the adjoint method, which Problem 16 develops. Sensitivity of one output to every element in a network can be found from two analyses rather than one per element, and reciprocity is what makes the second analysis usable.

The statement. If a branch of a network carries current \(I\) at voltage \(V\), that branch may be replaced by either

without altering any voltage or current elsewhere in the network.

The proof is almost trivial, and that is the point. The rest of the network interacts with the branch only through the pair \((V, I)\) at its terminals. Kirchhoff's laws involve branch voltages and currents, never element types:

So any element presenting the same \((V, I)\) satisfies exactly the same equations, and the original solution remains a solution.

The condition: the new circuit must still have a unique solution. The original solution is a solution of the substituted circuit — but if that circuit is degenerate, it may have others, and the substitution is then meaningless.

Two failure cases:

These are the same degeneracies Set 3 met with ideal sources, arriving by a different route.

Note what is not preserved. The substituted branch dissipates differently: a resistor carrying \(I\) at \(V\) absorbs \(VI\), while an ideal source presenting the same pair supplies it. Terminal behaviour is identical; internal behaviour is not — the same limitation as Thévenin's (Set 9, Problem 20) and source transformation's (Set 12, Problem 10).

Replace the 6 Ω load by a 4.8 V ideal source. Every node voltage in the ladder is unchanged — 12, 7.2 and 4.8 V — because the rest of the network sees the same terminal voltage as before, and its own equations are untouched.

Or replace it by a 0.8 A ideal source. Again nothing changes: the network upstream delivers 0.8 A at whatever voltage results, and that voltage is 4.8 V.

Why the current-source substitution is safe here. Node 3 still has the 3 Ω connecting it to the rest of the ladder, so it is not fed only by current sources — the degeneracy of Problem 10 does not arise.

Now the assumed-current method. Set 11, Problem 15 solved this ladder by assuming \(I_{\text{load}} = 1\ \text{A}\), propagating backwards to find that this requires a 25 V source, then scaling by \(20/25 = 0.8\).

Substitution is what licenses that first step. Assuming a load current means replacing the load with a 1 A source — a legitimate substitution — and then asking what source voltage is consistent with it:

Two theorems, one method: substitution makes the assumption legal, proportionality makes the scaling legal.

Why this works only for a ladder. The propagation backwards requires each stage to determine the next unambiguously. A bridge has no such ordering, so the assumed value cannot be propagated — the same obstruction that defeated source transformation in Set 12, Problem 13.

Recall the proof from Set 11, Problem 13. A network \(N\) drives a load through terminals carrying \((V, I)\). The first move is:

That step is the substitution theorem. Nothing else in the proof needs it, and nothing else could replace it.

Why it is needed. Superposition applies to independent sources. Before the substitution, the load is a resistor — not a source — so there is nothing to superpose it with. Substitution converts the load into a source, after which the network has two source groups and superposition applies:

Why the proof would otherwise be circular. Without substitution one might argue: "the load sees \(V_{TH}\) behind \(R_{TH}\), so the current is \(V_{TH}/(R_{TH}+R_L)\)" — which assumes the very equivalence being proved. Substitution breaks the circle by removing the load from the argument entirely, replacing it with a known source whose value is whatever the current happens to be. The proof never needs to know that value.

The chain of dependencies across Part 1:

Only the first two are primitive. Everything else follows.

Note that substitution needs less than superposition does. Superposition requires linearity; substitution requires only Kirchhoff's laws, so it holds in non-linear networks too — a diode branch may be replaced by a source matching its operating point. That is exactly the load-line method, and it is why substitution survives where every other theorem here fails.

The statement. For any network graph with \(b\) branches, let \(\mathbf{v}\) be any set of branch voltages satisfying KVL and \(\mathbf{i}\) any set of branch currents satisfying KCL, both on that graph. Then

The proof. Set 8, Problem 5 established that KVL is equivalent to the branch voltages lying in the range of \(\mathbf{A}^{\mathsf T}\), and KCL to the currents lying in the null space of \(\mathbf{A}\):

Therefore

Three lines, and the element laws were never mentioned.

What it assumes about the elements: nothing whatever.

The last row is the remarkable one, and Problem 15 exploits it.

The obvious reading — conservation of power. If \(\mathbf{v}\) and \(\mathbf{i}\) do come from the same network, then \(v_ki_k\) is the power absorbed by branch \(k\), and the theorem says the total absorbed power is zero — sources supply exactly what the rest consumes.

The node voltages from Set 6 are \(V_a = 8\), \(V_b = 0\), \(V_c = 4.5\) and \(V_d = 3\ \text{V}\). The branch quantities follow:

The source entry is negative because, with the passive sign convention applied consistently, it absorbs \(-8\) W — that is, supplies 8 W.

Check Kirchhoff first, since Tellegen depends on nothing else:

Now the sum:

The five resistors absorb 8 W between them, exactly what the source delivers.

The 18 Ω arm is worth noting. It absorbs only 125 mW of the 8 W — the bridge is close enough to balance that its diagonal arm carries little. Set 13, Problem 10 showed it is nevertheless collecting 83.5% of the maximum power available to it.

What this verification does and does not establish. It confirms the arithmetic of the earlier sets — a genuine check, since an error in any node voltage would break the sum. What it does not do is confirm anything physical: the sum would vanish for any numbers obeying KCL and KVL, even nonsensical ones.

Solve the second network. With \(\hat{V}_a = 4\), \(\hat{V}_b = 0\), nodal analysis gives

Note that \(\hat{V}_c < \hat{V}_d\) here, the reverse of the first network — the diagonal current flows the other way.

The cross sums. Pairing the voltages of network 1 with the currents of network 2, branch by branch:

and pairing the other way round:

Both vanish exactly, to the precision of the arithmetic.

Why this is surprising, and why it is not. The two networks have different sources, different resistors, different node voltages and different currents. Nothing physical connects them. But Problem 13's proof used only

and both hold, because both networks share the same incidence matrix \(\mathbf{A}\) — the same graph. The elements never entered the proof, so they cannot affect the conclusion.

These sums are not powers. The quantity \(v_k\hat{\imath}_k\) multiplies a voltage from one circuit by a current from another; it has the units of watts and no physical meaning at all. The name quasi-power records exactly that.

What must match, and what need not:

The second network could contain diodes and the result would still hold.

The problem to be solved. A network has \(b\) elements, and we want \(\partial V_{\text{out}}/\partial R_k\) for every one of them — to know which components need tight tolerances. The direct approach perturbs each in turn:

For a circuit with a thousand components that is a thousand solutions of the network.

The adjoint method. Build a second network \(\hat{N}\) on the same graph, with a unit source applied at the output port and every original source deactivated. Solve it once.

Apply quasi-power both ways. Let the original network be perturbed so that branch quantities change by \(\Delta v_k\) and \(\Delta i_k\). The perturbed voltages still satisfy KVL and the perturbed currents still satisfy KCL, so Problem 15's relation applies to them:

Subtracting gives the central identity:

Every term involves one quantity from the original network and one from the adjoint. Substituting the element laws — \(\Delta v_k = R_k\Delta i_k + i_k\Delta R_k\) for a resistor — collapses the sum to

Read what that says. The sensitivity to element \(k\) is simply the product of the current in that branch in the original network and the current in the same branch in the adjoint network. Both are already known:

The connection to reciprocity. For a reciprocal network the adjoint is the original network with the source moved to the output port — which is precisely Problem 7's interchange. Non-reciprocal networks need a genuinely different adjoint, with every dependent source transposed, and that is where the general theory earns its keep.

The statement. If a resistance \(R\) carrying current \(I\) is changed to \(R + \Delta R\), the resulting change in every branch current equals the response of the network — with all independent sources deactivated — to a single compensating voltage source

inserted in that branch, opposing the original current.

The derivation, in three steps. Let the new current in the branch be \(I' = I + \Delta I\). By substitution (Problem 10), replace the altered resistor by a voltage source equal to its actual drop:

The first term is what an unchanged \(R\) carrying \(I'\) would drop. So the branch behaves as the original resistance in series with an extra source of \(I'\Delta R\):

Now superpose (Set 11). The network with its original sources gives the old solution; the compensating source alone, with the originals deactivated, gives the change:

where \(R_{TH}\) is the Thévenin resistance seen by the branch. Using \(I\) rather than \(I'\) in the numerator, with \(\Delta R\) retained in the denominator, makes the expression exact — as Problem 18 confirms numerically.

Three theorems in one derivation:

The compensating source:

It drives the deactivated network, which presents \(R_{TH}\) in series with the new branch resistance:

Verify directly from the Thévenin equivalent, \(V_{TH} = 2700/239\ \text{V}\):

Exact agreement — compensation is not an approximation.

Testing several changes:

The compensation formula is exact at every step size. The first-order estimate — which drops \(\Delta R\) from the denominator — degrades badly for large changes, overstating the effect by 28% at \(\Delta R = 4\ \Omega\).

Why keeping \(\Delta R\) in the denominator matters. Many textbook statements of the theorem omit it and are therefore only valid for small perturbations. The full form costs nothing extra and is exact.

Take the limit of the compensation result as the change becomes small:

Evaluating for the ladder:

The normalised sensitivity is often more useful, being dimensionless:

A 1% increase in the load produces a 0.425% decrease in its current.

The 5% tolerance:

So the load current lies within about \(0.8 \pm 0.017\ \text{A}\). Checking exactly with \(\Delta R = +0.3\ \Omega\) gives \(\Delta I = -0.01664\ \text{A}\), or \(-2.08\%\) — the linear estimate is good for tolerances of this size.

Reading the sensitivity formula. Since \(|S| = R/(R_{TH}+R) < 1\) always, the current is less sensitive than the resistance in fractional terms — and the stiffer the source, the less sensitive still:

The design use. Sensitivities identify which components need close tolerance. A branch with \(|S|\) near zero can use a cheap component; one near unity determines the accuracy of the whole circuit. Problem 16's adjoint method computes this for every element at once.

What each requires, ordered from weakest assumption to strongest:

Reciprocity is the most restrictive: it fails for any circuit containing an amplifier, which is most of them.

What they share. None is independent — each follows from Kirchhoff's laws plus, in three cases, linearity:

How they differ in kind from Sets 9 to 13. Those theorems reduce a network to something smaller; these do not:

Which to reach for:

The honest ranking. Tellegen is the deepest and the most computationally valuable; substitution is the most general and the most used without acknowledgement; compensation is the most practical; Millman is a convenience; reciprocity is elegant but applies to a shrinking class of circuits.