Kirchhoff laws in sinusoidal steady state
Derive phasor KCL and KVL from their time-domain conservation laws rather than treating them as new AC rules. Preserve reference signs, prove the same-frequency mapping, distinguish complex sums from magnitude sums, state the Maxwell–Faraday boundary of lumped KVL and separate multi-frequency or nonlinear problems into the representations they actually require.
- Derive time-domain KCL from charge conservation under the lumped-node approximation.
- Prove that same-frequency time-domain KCL becomes the algebraic sum of current phasors.
- Derive phasor KVL from the corresponding same-frequency lumped voltage equation.
- State the Maxwell–Faraday limitation behind naive zero-sum KVL.
- Explain why Kirchhoff equations add complex phasors rather than their magnitudes.
- Keep branch and loop reference signs unchanged when moving between time and phasor domains.
- Solve multi-frequency LTI problems frequency by frequency instead of inventing one mixed-frequency phasor.
- Explain why nonlinear elements preserve conservation while invalidating a single-frequency impedance description.
Do Kirchhoff’s laws become new “AC laws” in phasor analysis? No. Their algebraic form survives because the underlying time-domain conservation equations are linear, and every same-frequency sinusoid carries the same time factor.
The important work is to prove that statement, preserve reference signs, and understand exactly when one phasor equation is insufficient.
04.1KCL begins with charge conservation, not with phasors
For a lumped node with chosen branch-current reference directions, charge conservation gives the exact bookkeeping relation
In the ideal lumped-node approximation the node stores no independent net charge, so dq_node/dt=0. Equivalently, using algebraic signs fixed once by the branch references,
A negative solved current does not invalidate the reference direction; it means the physical current is opposite to the chosen positive reference.
04.2Derive phasor KCL for one frequency
Let every branch current in the equation be sinusoidal at the same angular frequency ω:
Insert these expressions into time-domain KCL and use linearity of the real-part operator:
If Re{Aejωt}=0 for every t, then A=0: writing A=a+jb gives a cosωt−b sinωt=0 for all t, which requires a=b=0. Therefore
That is phasor KCL. Nothing new was postulated.
Figure 4.1 — Why do KCL and KVL keep the same form?
The one question is: what happens to a zero sum when every sinusoid shares the same factor e^{jωt}?
The common time factor can be separated. Conservation is then carried by the sum of complex amplitudes — provided the analysis is performed frequency by frequency.
04.3KVL keeps its form for the same reason — within the lumped model
For a chosen loop orientation and voltage references, the lumped-circuit voltage relation is
If every voltage term is at the same ω, the same common-factor argument yields
But KVL is an approximation of the electromagnetic field problem. Maxwell–Faraday gives
Ordinary lumped analysis assumes the interconnect loop is small enough, and field coupling is represented well enough by the chosen lumped elements, that an unmodelled time-varying magnetic flux through the loop is negligible. At sufficiently high frequency or large physical size, distributed/transmission-line or full electromagnetic descriptions become necessary.
04.4Never add magnitudes when KCL/KVL require phasors
The correct resultant magnitude is 5 mA, not 7 mA. Kirchhoff equations act on signed instantaneous quantities; after the phasor mapping they therefore act on signed complex amplitudes, not on their non-negative magnitudes.
04.5Impedance and admittance enter only after Kirchhoff
At a node, phasor KCL can be combined with element admittances. If branch k connects node voltage v to a known neighbouring voltage vk through admittance Yk, and its positive current reference is from the node toward that neighbour,
Likewise, a loop equation can contain terms Z_k i̲_k. The order of reasoning matters: conservation law first, constitutive relation second.
04.6Different frequencies require separate phasor equations
Suppose a branch current contains two frequencies:
There is no single phasor representing that whole waveform. In a linear time-invariant circuit, superposition allows one KCL/KVL phasor problem at ω₁ and another at ω₂, followed by addition in the time domain.
Do not add 3∠20° at 1 kHz to 2∠40° at 2 kHz as if they belonged to one complex plane problem.
04.7KCL/KVL remain true when the circuit is nonlinear — but one-frequency phasors may not
Charge conservation and the field constraints do not disappear because an element is nonlinear. What fails is the simple one-frequency constitutive relation v̲=Zi̲: nonlinear devices can generate DC, harmonics and mixing products.
A harmonic-balance calculation, for example, enforces KCL/KVL at many harmonic frequencies simultaneously. This is an extension of conservation, not a replacement for it.
04.8Reference signs survive the domain change unchanged
The sign attached to a branch in KCL or a voltage in KVL comes from the chosen physical reference direction/orientation. Transforming a same-frequency sinusoid into its phasor does not reverse that reference.
A common error is to redraw references after seeing the phase or sign of an answer. Do not. Keep the model fixed and interpret negative/phase-shifted results algebraically.
04.9A compact proof workflow
This workflow prevents most sign and “AC Kirchhoff law” misconceptions before numerical calculation starts.
04.10Mastery exercises
- Prove that Re{Aejωt}=0 for all t implies A=0.
- Convert i₁(t)−i₂(t)+i₃(t)=0 into a same-frequency phasor equation.
- At a node, two currents leaving are 2∠0° mA and 2∠120° mA. Find the entering current required by KCL.
- Explain why adding the magnitudes in exercise 3 is wrong.
- State the electromagnetic limitation behind lumped KVL.
- A circuit is driven simultaneously at 1 kHz and 3 kHz. Describe the correct LTI phasor workflow.
- Does a nonlinear diode violate KCL when it produces harmonics? Explain.
- Write the generic nodal equation for branches Yₖ connected from node v̲ to neighbouring nodes v̲ₖ.
Solutions
1. With A=a+jb, the real part is a cosωt−b sinωt. Independence of sine and cosine forces a=b=0.
2. i1−i2+i3=0.
3. Leaving sum =2+j0 +2(cos120°+j sin120°)=1+j1.732=2∠60° mA, so the entering current is 2∠60° mA if defined positive entering, or −2∠60° for the same leaving-reference convention.
4. Magnitudes discard relative phase and algebraic sign.
5. A time-varying magnetic flux through a loop gives ∮E·dl=−dΦ_B/dt; naive zero-sum KVL requires the lumped approximation or explicit modelling of that coupling.
6. Solve one phasor network at 1 kHz and another at 3 kHz, then superpose the resulting real waveforms.
7. No. Charge conservation still holds; the diode’s nonlinear constitutive law couples frequencies, so multiple harmonics may be required.
8. ΣYₖ(v−vk)=0 for currents referenced from the node outward.
Chapter 05 will now turn these rules into a systematic end-to-end procedure for solving sinusoidal RLC networks.