Questions: AC Kirchhoff's Laws in the Phasor Domain

Question 1

An AC circuit has three impedances in a loop: Z₁ = 10Ω (resistor), Z₂ = j15Ω (inductor), Z₃ = −j5Ω (capacitor), and a source phasor V̅_s = 100∠0° V. Using KVL in the phasor domain, what equation describes the loop?

• A. V̅_s = I̅(Z₁ + Z₂ + Z₃) = I̅(10 + j10) — the phasor current times the total complex impedance
• B. V̅_s = I̅·Z₁ + I̅·Z₂ + I̅·Z₃ only after converting each phasor back to a sinusoidal time-domain expression
• C. KVL doesn't apply directly to AC circuits because voltage and current are out of phase
• D. V̅_s = |Z₁| + |Z₂| + |Z₃| multiplied by the peak current amplitude

Answer: (A)

KVL in the phasor domain works exactly as in DC: the sum of phasor voltage drops around a loop equals the source phasor. Since the same phasor current I̅ flows through each series impedance, the voltage across each element is I̅·Zₙ, and their sum equals V̅_s. The total impedance is Z₁ + Z₂ + Z₃ = 10 + j15 − j5 = 10 + j10 Ω. No time-domain conversion is needed — the phasor domain turns what would be a differential equation problem into straightforward complex algebra.

Question 2

What is the fundamental reason that DC circuit analysis techniques — nodal analysis, mesh analysis, Thévenin equivalents — transfer directly to AC circuits in the phasor domain?

• A. AC circuits are mathematically identical to DC circuits when operating at steady state
• B. The phasor transform converts time-domain differential equations (governing inductors and capacitors) into algebraic equations over complex numbers, restoring the same mathematical structure as DC analysis
• C. KVL and KCL only hold for DC circuits, but phasors allow engineers to approximate them for AC
• D. Phasors eliminate the imaginary parts of impedance, reducing AC circuits to equivalent resistive networks

Answer: (B)

The reason these techniques transfer is structural, not accidental. In the time domain, inductors and capacitors introduce derivatives (v = L di/dt, i = C dv/dt), making circuit equations differential equations. The phasor transform, applied to sinusoidal steady state, converts differentiation to multiplication by jω — turning differential equations into algebraic equations. Once the equations are algebraic, all standard linear circuit analysis techniques (which are fundamentally algebraic methods for solving linear systems) apply directly, with complex numbers instead of real ones.

Question 3

Kirchhoff's voltage and current laws hold for phasors: the sum of phasor voltages around a closed loop is zero, and the sum of phasor currents into a node is zero.

• T. True
• F. False

Answer: True

KVL and KCL are conservation laws — conservation of energy (voltage) and conservation of charge (current) — and they apply to instantaneous values of voltage and current at every moment in time. Since phasors represent sinusoidal steady-state signals and the conservation laws hold at every instant, they must also hold for phasors (the complex amplitudes encoding magnitude and phase). The arithmetic is complex rather than real, but the structure of the equations is identical.

Question 4

Thévenin's theorem can seldom be applied in the phasor domain because the equivalent circuit is expected to capture phase relationships between voltages that a simple phasor source and series impedance cannot represent.

• T. True
• F. False

Answer: False

Thévenin's theorem applies fully in the phasor domain. The Thévenin equivalent consists of an open-circuit phasor voltage V̅_th (a complex number encoding both amplitude and phase) in series with a Thévenin impedance Z_th (also complex). The phase relationships between the original circuit's sources and elements are fully captured in V̅_th and Z_th. The procedure is identical to DC: find the open-circuit voltage phasor and the input impedance with independent sources deactivated. There is no information loss.

Question 5

A student solves an AC steady-state circuit problem in the time domain by writing and solving differential equations. How would solving the same problem in the phasor domain differ, and why is the phasor approach preferred for AC steady-state analysis?

Model answer: In the time domain, inductors and capacitors introduce derivatives (v_L = L di/dt, i_C = C dv/dt), making KVL and KCL yield coupled differential equations. For sinusoidal steady state, these must be solved, then matched to the forcing frequency. In the phasor domain, these derivatives become algebraic multiplications (jωL for an inductor, 1/jωC for a capacitor), turning the differential equations into a linear algebraic system over complex numbers. The same nodal or mesh equations are then solved using standard linear algebra, yielding phasors directly. The phasor approach is preferred because it reduces the mathematical complexity by an entire level — differential equations become algebraic — while capturing all steady-state amplitude and phase information in the complex numbers.

The key insight is that the phasor transform exploits the specific structure of sinusoidal steady state: when inputs are sinusoidal and the circuit is linear and time-invariant, all voltages and currents are sinusoidal at the same frequency. This allows the frequency-dependent part (the derivative operator d/dt) to be replaced by the constant jω. The transform is exact for steady-state analysis — no approximation is made.