Questions: Phasor Notation and Complex Impedance

Question 1

The complex impedance of a capacitor is Z_C = 1/(jωC). What does this formula imply about capacitor behavior at very low frequencies compared to very high frequencies?

• A. At low frequencies Z_C is small (near zero), so capacitors pass low-frequency signals; at high frequencies Z_C is large, blocking them
• B. At high frequencies Z_C is small (near zero), so capacitors pass high-frequency signals; at low frequencies Z_C is large, blocking them
• C. Z_C is constant across all frequencies, so capacitors behave identically at low and high frequencies
• D. At resonance frequency only, Z_C becomes real (purely resistive); at all other frequencies it blocks all signals equally

Answer: (B)

As ω → 0 (DC), Z_C = 1/(ωC) → ∞: the capacitor is an open circuit, blocking DC. As ω → ∞, Z_C → 0: the capacitor is nearly a short circuit, passing high-frequency signals freely. Physically, a capacitor cannot sustain DC current (charge just accumulates), but at high frequencies charge oscillates so rapidly that effective impedance is very small. This high-pass character is the opposite of an inductor, whose impedance Z_L = jωL increases with frequency.

Question 2

A student states: 'The phasor V = 5∠30° is the voltage in the circuit.' What correction does this statement require?

• A. The phasor should include the imaginary unit j to represent time-varying behavior
• B. V = 5∠30° is a complex number encoding amplitude (5) and phase offset (30°) at frequency ω; the actual time-domain voltage is v(t) = 5cos(ωt + 30°), recovered by multiplying by e^(jωt) and taking the real part
• C. Phasors only represent current, not voltage; different notation is required for voltage
• D. The angle 30° must be converted to radians before the phasor can be used in any calculation

Answer: (B)

A phasor is a compact representation, not the signal itself. V = 5∠30° tells you amplitude (5) and phase offset (30°) relative to a reference sinusoid at frequency ω. To recover the time-domain signal you must know ω and compute v(t) = Re{V · e^(jωt)} = 5cos(ωt + 30°). The phasor analysis 'freezes' the e^(jωt) factor that is common to every voltage and current in steady-state single-frequency circuits.

Question 3

Phasor analysis is only valid when all sources in the circuit operate at the same single frequency, because the transformation relies on the e^(jωt) factor being identical for every signal.

• T. True
• F. False

Answer: True

The phasor method works by dividing out e^(jωt) from every voltage and current in KVL and KCL. This is only valid if all signals share the same ω — otherwise different e^(jωt) factors cannot be canceled. For multi-frequency sources (a DC offset plus an AC component, or a distorted waveform), you must use superposition: decompose into frequency components, solve each phasor problem separately, and add the time-domain results.

Question 4

An inductor has higher impedance at low frequencies than at high frequencies, so inductors block low-frequency signals and pass high-frequency ones.

• T. True
• F. False

Answer: False

An inductor's impedance is Z_L = jωL, which increases with frequency. At low ω, Z_L is small — the inductor is nearly a short circuit and passes low-frequency signals. At high ω, Z_L is large — the inductor opposes rapid current changes and blocks high-frequency signals. Inductors are low-pass elements; capacitors are high-pass elements. These complementary behaviors are why LC circuits create selective resonance at the frequency where the two impedances are equal in magnitude.

Question 5

Explain why differentiation in the time domain corresponds to multiplication by jω in the phasor domain, and how this transforms circuit equations.

Model answer: Any steady-state signal has the form A cos(ωt + φ) = Re{A e^(jφ) e^(jωt)}. Taking the time derivative gives jω · A e^(jφ) e^(jωt) — simply multiplied by jω. Since phasors factor out e^(jωt), differentiation becomes multiplication by jω in the phasor domain. For an inductor (v = L di/dt), this gives V = jωL · I, so Z_L = jωL. For a capacitor (i = C dv/dt), it gives I = jωC · V, so Z_C = 1/(jωC). Differential equations governing reactive elements become algebraic equations.

Without phasors, AC circuit analysis requires solving differential equations and manually tracking trig identities and phase shifts. Phasors reduce this to complex arithmetic, letting every DC technique — Ohm's law, nodal analysis, mesh analysis, Thévenin equivalents, voltage dividers — apply directly in the frequency domain. The only additional step is converting back to the time domain at the end, which usually means reading off the amplitude and phase from the final phasor.