Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Phasor and Phasor Diagram

4.7.4

Phasor and Phasor Diagram

A sinusoidal voltage or current can be represented geometrically by a phasor: a vector of fixed length, rotating anticlockwise about the origin at a constant angular velocity $\omega$ equal to the alternating quantity's own angular frequency. A phasor is drawn so that its length equals the peak value ( $V_m$ or $I_m$ ) of the corresponding alternating quantity, its angular velocity equals the angular frequency $\omega$ , and its projection onto any fixed vertical axis at a given instant gives the INSTANTANEOUS value of that quantity at that same instant.

A phasor diagram shows several such phasors and the phase relations between them. For $v=V_m\sin\omega t$ , the phasor $\vec{OA}$ has length $V_m$ and makes an angle $\omega t$ with the horizontal reference (positive x-) axis, and its vertical projection at any instant, $V_m\sin\omega t$ , is exactly the instantaneous voltage at that instant; as $\vec{OA}$ rotates anticlockwise, one full rotation of the phasor traces out one complete cycle of the voltage waveform. For a current $i=I_m\sin(\omega t+\phi)$ flowing in the same circuit, represented by a second phasor $\vec{OB}$ , the angle $\phi$ between $\vec{OA}$ and $\vec{OB}$ (measured in the direction of rotation) is the phase angle between voltage and current. If $\vec{OB}$ sits AHEAD of $\vec{OA}$ in the rotation direction, the current LEADS the voltage by $\phi$ ; if instead current lags, the relation is written $i=I_m\sin(\omega t-\phi)$ . Phasor dia …

Figure 4.38Phasor diagram for an alternating voltage $v=V_m\sin\omega t$

What this figure shows. A vector $\vec{OA}$ of length $V_m$ is drawn from the origin, making an angle $\omega t$ with the horizontal (reference) axis, alongside a sine-wave graph of v against $\omega t$ . The figure demonstrates the defining property of a phasor: the vertical (y-axis) projection of $\vec{OA}$ at any instant equals the corresponding point $V_m\sin\omega t$ on the sine-wave graph directly beside it, and as $\vec{OA}$ is imagined rotating anticlockwise about O with angular velocity $\omega$ , one complete rotation of the phasor traces out exactly one complete cycle of the voltage waveform -- establishing the length-equals-peak-value, angle-equals-phase, projection-equals-instantaneous-v …

Figure 4.39Phasor diagram and wave diagram showing current leading voltage

What this figure shows. Two phasors, $\vec{OA}$ (length $V_m$ , representing the voltage) and $\vec{OB}$ (length $I_m$ , representing the current), are drawn from the same origin with $\vec{OB}$ positioned an angle $\phi$ AHEAD of $\vec{OA}$ in the anticlockwise (rotation) direction, alongside matching sine-wave graphs of v and i that show the current curve crossing zero and reaching its peaks a fraction $\phi/\omega$ of a cycle EARLIER than the voltage curve. The figure defines what it means for current to 'lead' voltage by phase angle $\phi$ : the current phasor sits ahead of the voltage phasor in the direction of rotation, so the current's corresponding waveform features (zero-crossings, peaks) all occur earlier in time than the voltage's -- with the opposite case, curren …

Misc Example 4.19Peak current, frequency, period and an instantaneous value from a given AC equation

Worked out. The alternating current equation $i=77\sin(314t)$ is given, and the peak current, frequency, time period, and instantaneous value at t = 2 ms are all required. Comparing directly with the general form $i=I_m\sin\omega t$ : the peak current is simply $I_m=77$ A, and the angular frequency $\omega=314$ rad/s gives frequency $f=\omega/2\pi = 314/6.28 = 50$ Hz and time period $T=1/f=1/50=0.02$ s. At t = 2 ms, the phase angle is $314\times2\times10^{-3}=0.628$ rad $\approx36^{\circ}$ , giving instantaneous current $i=77\sin36^{\circ}=77\times0.5878\approx45.26$ A. This four-part problem is a complete drill in extracting every basic AC descriptor (peak, frequency, period, and an instantaneous value) directly from a sing …

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