What is the fundamental reason that phasor analysis converts differential equations (from capacitors and inductors) into algebraic equations?
APhasors use complex numbers, which obey special algebraic rules that bypass calculus entirely
BAll signals in a single-frequency circuit share the same e^(jωt) factor, so differentiation in time corresponds to multiplying the phasor by jω — a purely algebraic operation
CPhasors work in the frequency domain where time does not exist, making differential equations irrelevant
DCapacitors and inductors become equivalent to resistors in phasor analysis, removing all reactive behavior from the equations
The key is that when all signals have the same frequency ω, differentiating a sinusoid v(t) = Re{V·e^(jωt)} gives d/dt[Re{V·e^(jωt)}] = Re{jω·V·e^(jωt)}. The e^(jωt) factor is common to every signal in the circuit and cancels out of all equations — what remains is multiplication by jω, a purely algebraic operation. The differential equation i = C·dv/dt becomes I = jωC·V in the phasor domain. This is not a trick or approximation; it is an exact consequence of assuming sinusoidal steady-state with a single frequency.
Question 2 Multiple Choice
An engineer analyzes a circuit driven by two AC sources: one at 60 Hz and one at 180 Hz. What problem arises if she tries to apply phasor analysis to both sources simultaneously?
APhasor analysis cannot handle sources with different phases, so sources at the same frequency with phase offsets must also be treated separately
BThe 180 Hz source is the third harmonic of the 60 Hz source, so they automatically combine and phasor analysis works normally
CPhasor analysis assumes all signals share a single frequency; at two different frequencies, signals cannot be combined in a single phasor diagram — superposition across two separate single-frequency analyses is required
DThe two frequencies create a time-varying impedance for capacitors and inductors, making the circuit nonlinear
Phasors encode amplitude and phase only — they strip out the e^(jωt) factor. This is only valid when every signal in the circuit has the same ω. If two sources have different frequencies, their phasors cannot be combined directly because they correspond to different e^(jωt) factors. The correct approach is superposition: analyze the circuit twice (once for each frequency using that frequency's impedances), then add the time-domain responses. At 60 Hz, a capacitor has impedance 1/(j·2π·60·C); at 180 Hz, it has impedance 1/(j·2π·180·C) — these are different, so the circuits are effectively different problems.
Question 3 True / False
Phasor analysis is an approximation that gives slightly different answers than solving the time-domain differential equations directly.
TTrue
FFalse
Answer: False
Phasor analysis gives exactly the same answers as direct time-domain differential equation solution for linear circuits in sinusoidal steady state. The derivation is rigorous: a sinusoid is the real part of a complex exponential (Euler's formula), and for linear circuits, the real-part operation and the circuit equations commute. Solving in the phasor domain and then recovering the time-domain answer (attach e^(jωt), take real part) is mathematically identical to solving the differential equations directly. Phasor analysis is a transformation technique, not an approximation — it trades calculus for algebra while preserving exactness.
Question 4 True / False
A sinusoidal voltage v(t) = 20·cos(500t − 45°) volts is fully described by the phasor V = 20∠−45° when the circuit's operating frequency is known.
TTrue
FFalse
Answer: True
The phasor encodes the two pieces of information that distinguish one sinusoid from another in a single-frequency circuit: amplitude (20 V) and phase (−45°). The angular frequency ω = 500 rad/s is a shared circuit parameter, not carried in the phasor itself. To recover the time-domain signal: v(t) = Re{20∠−45° · e^(j500t)} = 20·cos(500t − 45°). As long as ω is known, the phasor contains complete information about the signal.
Question 5 Short Answer
A capacitor has impedance Z_C = 1/(jωC) in phasor analysis. Explain where the 'jω' comes from — derive it from the capacitor's time-domain current-voltage relationship.
Think about your answer, then reveal below.
Model answer: The time-domain relationship is i(t) = C·dv/dt. In phasor analysis, if v(t) = Re{V·e^(jωt)}, then dv/dt = Re{jω·V·e^(jωt)}, so i(t) = Re{C·jω·V·e^(jωt)} = Re{I·e^(jωt)} where I = jωC·V. Solving for the impedance V/I = 1/(jωC). The jω factor arises directly from differentiating the complex exponential e^(jωt): d/dt[e^(jωt)] = jω·e^(jωt). The factor j represents the 90° phase lead of current over voltage in a capacitor; the ω represents that the current magnitude grows with frequency (a capacitor passes high-frequency signals more easily than low-frequency ones).
This derivation shows that phasor impedances are not definitions or conventions — they are exact consequences of applying calculus to complex exponentials and then stripping the common e^(jωt) factor. The same derivation for an inductor gives V = L·di/dt → Z_L = jωL, capturing that inductors block high frequencies. The frequency dependence baked into j/ω and jω is what makes reactive circuit design possible.