A circuit has two sinusoidal voltage sources: one at 60 Hz and one at 1000 Hz. Which approach correctly applies phasor analysis?
ARepresent both sources as phasors and solve the combined node equations simultaneously
BAnalyze the circuit at 60 Hz with phasors, then separately at 1000 Hz, then add the time-domain results
CUse the magnitudes of both phasors in KVL without tracking phase angle
DAverage the two frequencies and perform a single phasor analysis at 530 Hz
Phasor analysis assumes a single sinusoidal frequency — impedances of inductors (jωL) and capacitors (1/jωC) are defined at a specific ω and take different values at different frequencies. With two source frequencies, the correct method is superposition: analyze the circuit once using only the 60 Hz source (with 60 Hz impedances), then again using only the 1000 Hz source (with 1000 Hz impedances), and sum the resulting time-domain waveforms. Combining them in a single phasor analysis yields incorrect impedance values.
Question 2 True / False
When writing KCL in the phasor domain for a node connected to a capacitor, it is valid to use the time-domain expression i = C dv/dt alongside phasor currents from other branches in the same equation.
TTrue
FFalse
Answer: False
Phasor analysis requires all quantities in an equation to be in the same domain. In the phasor domain the capacitor relationship becomes I = jωC·V — a complex algebraic equation, not a differential equation. Mixing phasor voltages with the time-domain expression i = C dv/dt produces a mathematically inconsistent equation. You must commit to one domain throughout the analysis.
Question 3 Short Answer
What information does the transfer function H(jω) = V_out(jω) / V_in(jω) encode, and why is it more useful than a single time-domain solution?
Think about your answer, then reveal below.
Model answer: H(jω) is a complex function of frequency encoding both the amplitude ratio |H(jω)| (how the circuit scales the input magnitude) and the phase shift ∠H(jω) (how much the output leads or lags the input) at every frequency ω. A single time-domain solution gives the response to one specific input. H(jω) characterizes the circuit's behavior across all sinusoidal inputs simultaneously, enabling filter design and frequency-response analysis without re-solving the circuit.
Once H(jω) is known, the phasor output for any sinusoidal input at frequency ω is simply V_in · H(jω). This is far more powerful than solving a differential equation anew for each input. The magnitude and phase of H(jω) as functions of frequency are the Bode plot — the standard tool for filter design and stability analysis.