Questions: Phasor Conversion and Representation

Question 1

A circuit has two voltage sources: V₁ = 10cos(100t + 30°) V and V₂ = 5cos(200t + 45°) V. How should you find the total voltage using phasors?

• A. Add the phasors directly: V̅_total = 10∠30° + 5∠45° = 15∠37.5° V
• B. You cannot add these phasors directly — they operate at different frequencies. Solve the circuit separately at 100 rad/s and at 200 rad/s, then add the resulting time-domain signals
• C. Convert both to rectangular form and add: (10cos30° + 5cos45°) + j(10sin30° + 5sin45°)
• D. Phasors cannot represent this circuit at all since superposition does not apply in AC circuits

Answer: (B)

Phasors are only valid for single-frequency steady state. The phasor representation freezes the common e^(jωt) factor — but if two sources have different ω values, there is no common factor to freeze. You must perform two separate phasor analyses (one at ω=100, one at ω=200), find the contributions from each source independently, convert both results back to time domain, and then add the time-domain expressions. Superposition is valid; phasor addition across different frequencies is not.

Question 2

An inductor carries a sinusoidal current i(t) = 2cos(1000t + 30°) A. The inductor has L = 0.01 H. What is the phasor voltage across the inductor?

• A. V̅ = 2∠30° V — the voltage phasor equals the current phasor for inductors
• B. V̅ = jωL × Ī = j(1000)(0.01) × 2∠30° = j10 × 2∠30° = 20∠120° V
• C. V̅ = L × dĪ/dt = 0.01 × j × 2∠30° = 0.02∠120° V
• D. V̅ = 2∠(30°+90°) = 2∠120° V — inductors only shift phase by 90°

Answer: (B)

In the phasor domain, the inductor voltage is V̅ = jωL × Ī. Here ω = 1000 rad/s, L = 0.01 H, so jωL = j10. Multiplying: j10 × 2∠30° = 10∠90° × 2∠30° = 20∠120° V. The key transformation is that differentiation in time domain (v = L di/dt) becomes multiplication by jω in phasor domain — this is what converts the differential equation to algebra. Option D gets the phase right but loses the magnitude scaling by ωL.

Question 3

A phasor representation of a sinusoidal voltage captures most of the information needed to reconstruct the original time-domain signal.

• T. True
• F. False

Answer: False

A phasor captures amplitude and phase but NOT frequency. The phasor V̅ = Vm∠φ tells you the peak magnitude Vm and the phase shift φ, but to reconstruct v(t) = Vm·cos(ωt + φ) you also need to know ω. In phasor analysis, ω is assumed known and fixed throughout (single-frequency constraint); it is carried implicitly as context, not stored in the phasor itself. This is why phasors only work for single-frequency circuits — without a specified ω, the phasor is incomplete.

Question 4

In phasor domain, adding two voltages is done by adding complex numbers, which is the same as adding their magnitudes directly.

• T. True
• F. False

Answer: False

Phasor addition is vector addition in the complex plane — you add real parts together and imaginary parts together. Adding magnitudes directly (scalar addition) ignores phase and gives the wrong answer whenever the phases differ. For example, 10∠0° + 10∠90° = (10 + 0j) + (0 + 10j) = 10 + 10j = 10√2 ∠45°, not 20∠45°. Only when two phasors are exactly in phase (same angle) does the magnitude of the sum equal the sum of the magnitudes. The misconception arises from everyday intuition about adding 'voltages' without thinking about phase.

Question 5

Explain why differentiation in the time domain becomes multiplication by jω in the phasor domain, and why this matters for AC circuit analysis.

Model answer: A sinusoid v(t) = Vm·cos(ωt + φ) can be written as Re[Vm·e^(jφ)·e^(jωt)]. Differentiating: dv/dt = Re[jω·Vm·e^(jφ)·e^(jωt)]. The e^(jωt) factor is just the rotating phasor base; multiplying Vm·e^(jφ) by jω gives jω·V̅ in the phasor domain. Since j = e^(jπ/2), multiplication by jω scales the magnitude by ω and rotates the phase by 90°. This matters because inductor voltage (v = L·di/dt) and capacitor current (i = C·dv/dt) involve derivatives — replacing differentiation with jω multiplication turns these differential equations into algebraic equations V̅ = jωL·Ī and Ī = jωC·V̅, enabling Ohm's law and KVL/KCL to work algebraically in the frequency domain.

This is the core payoff of phasor analysis. Without it, solving an AC circuit would require setting up and solving coupled differential equations for each branch. With it, you treat every element as having a complex impedance (Z_R = R, Z_L = jωL, Z_C = 1/jωC), apply Kirchhoff's laws as complex algebra, and solve with standard algebraic techniques. The transformation is a kind of structured simplification that works precisely because all signals share the same frequency.