In an AC circuit, as the frequency ω increases, what happens to the magnitude of a capacitor's impedance?
AIt increases, because higher frequency means the capacitor charges and discharges faster, opposing more current.
BIt decreases, because |Z_C| = 1/(ωC) and magnitude shrinks as ω grows.
CIt stays the same; impedance is a fixed property of the component regardless of frequency.
DIt becomes purely real at high frequencies because the imaginary part cancels.
Z_C = 1/(iωC), so |Z_C| = 1/(ωC). As ω increases, the denominator grows and impedance magnitude shrinks toward zero — a capacitor becomes a near short-circuit at high frequencies. This is physically intuitive: at high frequencies, the capacitor charges and discharges so quickly that it passes AC current easily. At low frequencies (ω → 0), Z_C → ∞, meaning the capacitor blocks DC entirely. This frequency-dependence is the whole point of using capacitors as frequency-selective filters.
Question 2 Multiple Choice
An inductor has impedance Z = iωL. What does the phase angle of this impedance tell you about the relationship between voltage and current?
AVoltage and current are in phase, so all power delivered to the inductor is dissipated as heat.
BVoltage leads current by 90°, meaning the inductor stores and returns energy each cycle rather than dissipating it.
CCurrent leads voltage by 90°, so the inductor behaves capacitively at all frequencies.
DThe inductor has zero resistance, so it dissipates no power and the phase angle is irrelevant.
Z = iωL has argument (phase) = +90°. In the phasor representation, voltage is proportional to Z times current; the +90° phase means voltage leads current by a quarter cycle. This 90° phase difference is the signature of a purely reactive element: power flows into the inductor for half a cycle and returns to the source during the other half, so average power dissipation is zero. Option C (current leads voltage) describes a capacitor, where Z_C has phase −90°.
Question 3 True / False
In AC circuit analysis, complex impedances combine using the same series and parallel combination rules as DC resistances.
TTrue
FFalse
Answer: True
This is the central payoff of the phasor/impedance formalism. Because Kirchhoff's voltage and current laws hold for phasors (complex amplitudes), the same algebraic derivations that give Z_series = Z₁ + Z₂ and 1/Z_parallel = 1/Z₁ + 1/Z₂ for resistors apply equally to complex impedances. The only difference is that the arithmetic involves complex numbers instead of real ones. This turns AC circuit analysis — which would otherwise require solving systems of differential equations — into routine algebra.
Question 4 True / False
A circuit with a high power factor draws large peak currents because it stores most of the energy supplied to it.
TTrue
FFalse
Answer: False
This reverses the relationship. A *low* power factor (cos φ near 0) is the problematic case: voltage and current are nearly 90° out of phase, so large peak currents flow back and forth between source and circuit without delivering useful real power. P = (1/2)V₀I₀ cos φ, so when cos φ is small, you need large V₀I₀ to deliver modest P. A high power factor (cos φ near 1) means voltage and current are nearly in phase, so most of the current flow does useful work — exactly the efficient case. Low power factor is the engineering problem to fix.
Question 5 Short Answer
Why does representing AC voltages and currents as complex phasors turn differential equations into algebraic equations?
Think about your answer, then reveal below.
Model answer: Differentiating a complex exponential Ae^(iωt) simply multiplies it by iω — a constant. So d/dt → ×iω and ∫dt → ÷iω. This means the inductor's constitutive relation V = L dI/dt becomes V = iωL · I in phasor form (dividing out the common e^(iωt) factor), defining impedance Z_L = iωL. Similarly, the capacitor's I = C dV/dt becomes I = iωC · V, giving Z_C = 1/(iωC). The differential relationships become multiplicative ones, turning the circuit's system of differential equations into a system of linear algebraic equations solvable by standard circuit techniques.
The magic is the derivative rule for exponentials: d/dt[e^(iωt)] = iω·e^(iωt). Since all voltages and currents at a single frequency share the same e^(iωt) time factor, that factor cancels everywhere, leaving only the complex amplitudes (phasors) and multiplicative factors of iω. This reduces the problem from calculus to algebra — the key reason why impedance-based AC analysis is so powerful and so widely used in electrical engineering.