A complex number that represents a sinusoidal AC voltage or current of a given frequency by its magnitude (usually the RMS value) and phase. v(t) = 100√2 sin(ωt + 30°) is written 100∠30° V. Differentiation becomes multiplication by jω and integration becomes division by jω, so AC circuit calculations become algebra like DC calculations.
Japanese name: フェーザ
Phasor appears in 8 practice questions across 1 topic, including Circuit Analysis: AC, Linear Circuits and Transients.
Sine waves of the same angular frequency ω differ only in magnitude and phase. A phasor drops the part e^(jωt) that rotates with time and keeps only the magnitude and phase as a complex number V = |V|∠θ = |V|(cos θ + j sin θ). The original waveform is recovered by multiplying V by √2 and e^(jωt) and taking the imaginary (or real) part. Whether the magnitude is the RMS or the peak value is a matter of convention; the RMS value, convenient for power calculations, is common.
The coil voltage v = L di/dt becomes V = jωL I as phasors, and the capacitor current i = C dv/dt becomes I = jωC V. This is where the impedances jωL and 1/(jωC) come from. Kirchhoff's laws, Thevenin's theorem and the superposition theorem all work directly with complex numbers, so AC circuits can be solved as systems of linear equations.
A phasor diagram draws voltage and current phasors as arrows in the complex plane. It is easiest to read when the current is taken as the reference (the real axis) for a series circuit and the voltage for a parallel circuit. Before calculating anything, it shows how the voltages in an RLC circuit form a right triangle and how, in power-factor correction, the capacitor current cancels the lagging current.
Answer: B. 100∠30° V
Method 1 (from the definition): the magnitude of the phasor is the RMS value, the peak value 100√2 V divided by √2, which is 100 V. The angle is the initial phase π/6 rad = 30°, so the phasor is 100∠30° V. Method 2 (check with complex numbers): 100∠30° = 100(cos 30° + j sin 30°) = 86.6 + j50 V; multiplying by √2 and e^(jωt) and taking the imaginary part gives back the original v(t). In circuit calculations it is usual to use the RMS value, not the peak value, as the magnitude of a phasor.
Answer: A. 250 var
Method 1 (from the power-factor angle): the current lags the voltage by 30°, so the active power is P = VI cos 30° ≈ 433 W and the reactive power is Q = VI sin 30° = 100 × 5 × 0.5 = 250 var. Method 2 (complex power): the complex power is S = V × (conjugate of I) = 100∠0° × 5∠30° = 500∠30° ≈ 433 + j250 VA; the real part is the active power and the imaginary part is the reactive power. The imaginary part is positive, so the load takes 250 var of lagging (inductive) reactive power. The apparent power is |S| = 500 VA.
They can be used when all voltages and currents are sinusoids of the same frequency and the circuit is linear. A distorted waveform is first split into sinusoids of each frequency with a Fourier series, each frequency is calculated with phasors, and the results are superposed. For transients such as the moment a switch is closed, the Laplace transform is used.
Replacing s in the Laplace transform with jω turns the s-domain impedances sL and 1/(sC) into the phasor impedances jωL and 1/(jωC). Phasors can be seen as a special case of the Laplace transform that deals only with the sinusoidal steady state.
Method 1 (from the definition): the magnitude of the phasor is the RMS value, the peak value 100√2 V divided by √2, which is 100 V.
Which of the following expresses the sinusoidal AC voltage …Method 2 (phasor diagram): IL and IC point in opposite directions, so only their difference of 3 A remains, and it combines at a right angle with the 4 A of IR to form a 3-4-5 right triangle.
In the parallel RLC circuit in the figure, the RMS currents…Method 2 (phasor diagram): at f0, IL and IC are equal in size and opposite in direction, so they cancel and only IR remains in the source current.
In the parallel RLC circuit in the figure, the source frequ…Method 2 (phasor diagram): taking the current as the reference, VR = 10I is in phase, VL = 30I leads by 90° and VC = 20I lags by 90°, so only the difference 10I remains as the reactive part.
In the series RLC circuit in the figure, R = 10 Ω, ωL = 30 …A line current is the phasor difference of two phase currents, and its magnitude is √3 times the phase current: 10√3 ≈ 17.3 A.
A Δ-connected load with a 20 Ω resistor in each phase, as i…As an Amazon Associate I earn from qualifying purchases.
RMS value Impedance Reactive power Admittance Resonant circuit Three-phase AC Transfer function Reactance Thevenin's theorem Mesh analysis