Theorems in the Frequency Domain

The general argument. Every theorem in Sets 9 to 14 was derived from exactly two facts:

Set 20, Problem 16 proved Kirchhoff's laws hold for phasors, and impedance is by construction a linear relation \(\mathbf{V} = \mathbf{Z}\mathbf{I}\). Both premises survive, so every conclusion must.

The audit:

Why maximum power is the exception. It is not a consequence of Kirchhoff and linearity alone — it is an optimisation, and what is being optimised is average power, a quadratic quantity involving the phase between voltage and current. The load now has two free parameters:

Two free parameters give two optimisation conditions. Problem 14 derives them.

Two caveats that apply to all of them:

The second was already true in DC (Set 11, Problem 16) and remains true here for the same reason. It is worth restating because AC problems more often involve several sources.

Transform at \(\omega = 10\ \text{rad/s}\):

The two branch impedances:

Current division. The current into a branch is proportional to the other branch's impedance:

Transform back, restoring the frequency that the phasor set aside:

Note the phase. The output leads the input by 54°, and its magnitude is less than half the source current — neither of which a resistive divider could produce in combination. A resistive division always preserves phase and gives a fraction between 0 and 1; here the fraction is a complex number of magnitude 0.465 and angle \(+54.5°\).

Element impedances at \(\omega = 200\):

Combine the parallel pair:

Voltage division across the inductor:

The two reactances cancel exactly in the denominator — \(+j20\) from the inductor and \(-j20\) from the parallel combination — leaving a purely real total of 70 Ω.

A note on the reference. The original source of this problem quoted the answer as \(17.14\sin(200t+90°)\), which is the same waveform only if the source is also read as a sine. Following the cosine convention fixed in Set 20, both source and answer are stated as cosines here. Mixing the two references within one problem is the fastest route to a 90° error.

Why the output is at exactly \(90°\). Because the denominator came out purely real, the division ratio is \(j20/70\) — a pure imaginary number, contributing exactly \(+90°\). That cancellation is a coincidence of the chosen frequency; at any other \(\omega\) the angle would not be a round number.

Convert both sources to cosine reference:

The two sources are 90° apart, which is what makes this problem interesting.

KCL at the node:

Collect the \(\mathbf{V}_x\) terms:

The two reactive admittances cancel exactly. The node's total admittance is purely the conductance 0.04 S — the inductor and capacitor, being equal and opposite at this frequency, contribute nothing to it.

The right-hand side does not cancel, because the two sources differ:

The first term is rotated \(-90°\) by the inductor, the second \(+90°\) by the capacitor — and since the sources were already 90° apart, the results land 180° from each other in one component and reinforce in the other.

Solve:

Note the magnitude: 70.7 V from two 20 V sources. The reactances have vanished from the node's admittance while still delivering current into it, which is a partial resonance — and it is the same magnification effect as Set 20, Problem 8, arriving by a different route.

Impedances at \(\omega = 100\ \text{rad/s}\):

The three mesh equations:

Divide each row through to reduce the numbers — row 1 by 10, row 2 by \(j10\), row 3 by 10:

Dividing row 2 by \(j10\) converts \(-j30 \to -3\) and \(-j130 \to -13\) — a legitimate operation that destroys the matrix's symmetry but greatly simplifies the arithmetic. The symmetry check of Set 21 must therefore be applied before any such scaling.

Solving the system:

This difference is the capacitor's current, since the capacitor is the branch shared by meshes 2 and 3.

The capacitor voltage:

A plausibility note. The capacitor's reactance is 200 Ω while the resistances are 20 and 10 Ω, so the circuit is strongly reactive and a large capacitor voltage is expected. The 56 V result sits comfortably between the two source amplitudes, which is reassuring without being a proof.

Start from the known branch. The capacitor's impedance at \(\omega = 2\) is

The resistor shares that voltage, being in parallel:

In phase with the capacitor voltage, and lagging the capacitor current by exactly 90° — as it must, since the two elements share a voltage.

KCL for the source current, summing the two branch currents:

Convert to rectangular first — this is the step that must not be skipped:

The magnitudes did not add. \(0.5 + 2 = 2.5\), but the answer is 2.06. The two currents are 90° apart, so

Pythagoras applies exactly because the phase difference is 90°, which happens whenever a resistor and a capacitor share a voltage.

The method worth noting is working backwards. Given one branch quantity, every other quantity in a series–parallel network follows by element laws and Kirchhoff, with no simultaneous equations at all. This is Set 11's proportionality method, and it transfers unchanged — though the "scaling constant" is now complex.

Write KCL at both nodes, treating the dependent source as a source and substituting afterwards:

Solving the pair gives

A correction worth recording. An earlier version of this problem quoted \(1\angle{-36.87°}\) — the same magnitude, exactly 180° away. Substituting each candidate back into node 1 settles it:

A 180° error is the easiest one to make and the hardest to notice: the magnitude is right, the answer looks reasonable, and only substitution exposes it. It arises from a source's assumed direction, or from moving a term across an equals sign.

Why 180° errors are systematically dangerous in AC work:

Only the last works, which is why Set 21, Problem 12 insisted on it.

The dependent source's effect. Its term \(-2\mathbf{V}_1\) appears in node 2's equation but has no counterpart in node 1's, so the admittance matrix is asymmetric and the circuit is not reciprocal — Set 21, Problem 17.

Because a dependent source is present, \(\mathbf{Z}_{Th}\) cannot be found by deactivating sources and reducing. The reliable route is \(\mathbf{V}_{oc}/\mathbf{I}_{sc}\) — Set 9's method, unchanged.

1Open-circuit voltage. With A–B open, one mesh current flows:

Note the sign: \(-5\mathbf{I} \times (-j5) = +j25\mathbf{I}\). Two negatives, and the dependent source reinforces rather than opposes.

2Short-circuit current. With A–B shorted, two currents flow:

From (2): \(j35\mathbf{I} = j5\mathbf{I}_N\), so \(\mathbf{I}_N = 7\mathbf{I}\). Substituting into (1):

3The Thévenin impedance:

Positive real part, as a check confirms it should be here — Set 21, Challenge C2 guarantees this for passive networks, and although a dependent source is present, it has not driven the resistance negative in this case.

Note how the dependent source distorts the result. The passive elements alone would give \(j10 \parallel (-j5)\) in some combination with 100 Ω — nothing resembling \(3.47+j4.65\). The dependent source has made the network behave as though it contained a resistance that is not there.

Solve for the open-circuit condition. From the second equation:

Substituting into the first:

The Thévenin voltage:

The Thévenin impedance:

Stop and examine that. The real part is negative, and the angle exceeds 90° in magnitude. Set 21, Challenge C2 proved that for a passive network

so this result would be impossible — if the network were passive. The asymmetry of the mesh equations (\(Z_{12} = j4\) but \(Z_{21} = -j2\)) confirms a dependent source is present, and the proof's assumption fails at exactly that point.

Complete the calculation. Reconnecting the 2 Ω load:

Note that the total resistance is \(-0.64 + 2 = 1.36\ \Omega\), still positive — the load's resistance more than cancels the source's negative one, so the combination is stable.

What if the load resistance were below 0.64 Ω? The total resistance would go negative and the circuit would be unstable — the steady-state phasor solution would be a mathematical fiction, since the transient would grow rather than decay. Set 19, Challenge C2 established this criterion, and it applies to any network containing an active element.

Set up the node equations with the load removed, including the controlling relation as a fourth equation:

Treating the controlling current as a fourth unknown, with its defining relation as a fourth equation, avoids substitution errors — worth doing whenever the controlling variable is awkward to express.

In matrix form:

Solving gives the controlling current, and from it the short-circuit current:

Apply the Norton equivalent. The source current divides between \(\mathbf{Z}_{Th}\) and the 1 Ω load, and the load voltage is the current times the parallel combination:

Check by the Thévenin route, which must agree:

1Transform the voltage source to a current source:

The 5 Ω moves from series to parallel. Note that dividing by a real impedance leaves the phase unchanged; had the series element been reactive, the current source's phase would have shifted.

2Combine the now-parallel impedances:

3Transform back to a voltage source:

Here the multiplication is by a complex impedance, so the phase changes — from \(-90°\) to \(-63.4°\).

4Now a single series loop remains. Voltage division across the 10 Ω:

Why transformation was the right method here. The alternative — mesh or nodal analysis — would have needed a 2×2 complex system. Transformation reduced the circuit to a single loop by two mechanical steps, and a single loop needs only one division.

The rule for when it applies. Transformation works whenever a source has an impedance in series (voltage source) or parallel (current source). It fails for an ideal source with nothing in series — which is precisely the situation that forces a supernode, as Set 21, Problem 19 noted.

aSource 1 alone, with source 2 replaced by a short (its \(-j5\) remains):

bSource 2 alone, with source 1 shorted:

Add the phasors — legitimate because both sources share a frequency:

The magnitudes 80 and 60 do not sum to 100 arithmetically — but they do as perpendicular vectors, since the two contributions happen to be exactly 90° apart.

Verify by direct nodal analysis. With both sources active, KCL at the load node:

Magnitude 100 V, confirming the result. The sign difference in the real part traces to the reference direction assumed for source 2 in the nodal equation — a reminder that superposition requires the same reference conventions in every sub-problem.

What superposition does not permit. The power delivered to the load is not the sum of the two sources' individual contributions:

These agree only because the two contributions are exactly 90° apart, making the cross term vanish. At any other phase difference they would not — power superposes only for orthogonal contributions, which is why different frequencies always work and the same frequency generally does not.

1Open-circuit voltage, a simple divider with the load removed:

Note the divider ratio has magnitude 0.707 and angle \(-45°\) — a resistive divider would give a real fraction and no phase shift.

2Thévenin impedance. No dependent sources here, so deactivate the source — replace it by a short — and look back:

Positive real part and angle within \(\pm90°\), as passivity requires.

3Reconnect the 2 Ω load:

Check against the original circuit. With the 2 Ω connected, the source sees \(4 + \left[(-j4) \parallel 2\right]\):

Current division into the 2 Ω:

Note what the equivalent is good for. If the load changes — to 5 Ω, or to a capacitor — the Thévenin equivalent gives the new answer in one line, whereas the direct calculation must be redone entirely. That was Set 9's argument for the theorem and it applies unchanged.

Set up. A source \(\mathbf{V}_{Th}\) behind \(\mathbf{Z}_{Th} = R_{Th} + jX_{Th}\) drives a load \(\mathbf{Z}_L = R_L + jX_L\):

Only the load's resistance absorbs average power; reactance stores and returns it. So with amplitude phasors,

Optimise in two stages, because the two variables decouple.

aThe reactance. \(X_L\) appears only in the denominator, and only as a square. The power is therefore maximised by making that term vanish:

This is the step with no DC analogue, and it is where the conjugate comes from: the load's reactance must be equal and opposite.

bThe resistance. With the reactive term gone, the problem is Set 13's exactly:

Combining:

The load is the complex conjugate: same resistance, opposite reactance.

The maximum power. At the match, the total impedance is purely resistive and equal to \(2R_{Th}\):

The factor 8 belongs to the amplitude convention. With RMS phasors it becomes 4 — Problem 19.

Two readings of the condition:

The first is the AC-specific half. It says the matched circuit is at resonance, which is why matching networks and tuned circuits are the same subject seen from different ends.

aThe first network:

The network is capacitive, so the load must be inductive to cancel it.

bThe second network, working from the inside out:

Add the series branch:

Then the 10 Ω in parallel:

This network is inductive, so the load must be capacitive.

Note the useful shortcut in polar form:

Conjugation preserves the magnitude and negates the angle, so no rectangular conversion is needed once the polar form is in hand.

The practical difficulty. A real loudspeaker is what it is — typically \(8\ \Omega\) with some inductance — and cannot be redesigned to match a source. The realistic approach is either to insert a matching network between them, or to accept a mismatch. Problem 16 quantifies the cost of the latter.

The reactance can no longer be cancelled, so \(X_L = 0\) is forced and only \(R_L\) is free:

Differentiate and set to zero. The numerator of \(dP/dR_L\) gives

A clean and slightly surprising result: the best resistive load is the magnitude of the Thévenin impedance, not its real part.

Evaluating:

Comparing the three candidates:

Note the third row: setting \(R_L = R_{Th}\) — the DC answer — is worse than \(R_L = |\mathbf{Z}_{Th}|\), even though it gives a larger current. The larger resistance more than compensates.

Why the larger resistance wins. The uncancelled \(X_{Th}\) already limits the current, so adding more resistance costs relatively less current than it gains in \(R_L\). In the limit \(X_{Th} \to 0\) the formula reduces to \(R_L = R_{Th}\), recovering Set 13's result as it must.

The practical significance. This is the realistic case. Loads are usually fixed — a loudspeaker, an antenna, a heater — and cannot be chosen freely. The 78% figure here is typical: a restricted match loses something, but not catastrophically, and adding a single reactive element to cancel \(X_{Th}\) recovers the rest.

1Thévenin voltage, by division across the inductor:

2Thévenin impedance, source deactivated:

Note \(|\mathbf{Z}_{Th}| = 7.07\ \Omega\) and the angle is \(45°\) — the parallel combination of equal resistance and reactance always gives this.

3The matched load:

A resistance of 5 Ω in series with a capacitive reactance of 5 Ω. At \(\omega = 1000\ \text{rad/s}\) that would be \(C = 1/(1000\times5) = 200\ \mu\text{F}\).

4From first principles. The total impedance is purely resistive:

5By formula, as a check:

The two agree, which is the point of doing both: the formula is quick but easy to misremember, and the first-principles route is slower but self-verifying.

Where the rest of the power goes. The same 45 W is dissipated in \(R_{Th}\), since the two resistances are equal and carry the same current. Total 90 W, so the efficiency is 50% — the same as Set 13's DC result, and unaffected by the reactive matching.

The statement. In a linear network of bilateral elements, interchanging an ideal voltage source and an ideal ammeter leaves the reading unchanged:

Set 14's theorem with phasors substituted, and it follows from \(\mathbf{Z} = \mathbf{Z}^{\mathsf T}\) exactly as before — Set 21, Problem 17.

What is stronger. The equality is now between complex numbers, so it asserts two things:

Not only is the magnitude reciprocal, so is the phase. A signal takes the same time to traverse the network in either direction — which is a physically meaningful statement about delay, not merely about amplitude.

Which elements preserve it:

A practical use — measurement. If a network is buried in equipment and only one port is accessible from each side, reciprocity halves the measurements needed: the transfer function measured in one direction is guaranteed to hold in the other, so a two-port needs only three independent parameters rather than four.

This is exactly the result Set 35 will use to reduce the parameter count for any reciprocal two-port.

A second use — as a check. Computing a transfer function twice, once in each direction, must give identical results for a passive network. Disagreement means an error in the analysis, and it is a check that requires no additional information about the circuit.

A caution. Reciprocity concerns the ratio of response to excitation, not the response itself. Interchanging source and meter generally changes every other quantity in the network — only the particular ratio is preserved.

The source of the difference is what a phasor's magnitude represents:

The RMS convention was invented precisely so the factor of two disappears and AC power formulas look like DC ones — Set 20, Problem 10.

Propagating to the matched case. At the conjugate match the load voltage is half the Thévenin voltage, so

Both are correct. Neither is complete without stating which convention it assumes — and that omission is the actual problem.

Worked example. Take \(\mathbf{V}_{Th} = 20\ \text{V}\), \(R_{Th} = 5\ \Omega\):

A factor of two — not a rounding difference, and enough to specify the wrong component rating.

How to tell which is meant. Three reliable indicators:

The second is the common one in power engineering: "a 240 V supply" always means RMS.

The safe practice is to avoid the shortcut formula entirely and compute

The explicit \(\tfrac12\) forces the convention to be conscious rather than assumed, and it works at any load, not only the matched one. Problem 17 did the calculation both ways for exactly this reason.

The combined tally. Sets 21 and 22 between them transferred:

Fourteen results transferred untouched. Two sets of work produced one modification and one new phenomenon.

Why the ratio is so lopsided. Because the phasor transform was constructed to preserve exactly the two things all the theorems depend on:

Anything derived from those two survives; anything that also involves optimisation, power, or non-linearity may not.

What remains genuinely unexamined is power itself. Every result so far concerned voltages, currents and impedances. Power has appeared only as an optimisation target in Problems 14 to 17, and even there the analysis stopped at \(P = \tfrac12|\mathbf{I}|^2R_L\) without asking what the reactive part of the circuit is doing.

The question Set 23 must answer. Consider a purely reactive load. Voltage and current are 90° apart, so their product

is positive for half the cycle and negative for the other half, averaging to zero. Yet the current is real, flows in real conductors, and causes real heating in them. Something is being transported that carries no net energy.

Three quantities will be needed where DC needed one:

Their relationship \(S^2 = P^2 + Q^2\) is another right triangle, similar to the impedance triangle of Set 20, Problem 14.

Why this matters commercially. A generator and a cable must be sized for \(S\), while only \(P\) is sold. A load drawing large \(Q\) therefore occupies capacity it does not pay for — which is why industrial tariffs penalise poor power factor, and why power-factor correction capacitors are among the most common components in an electrical installation.