Understanding Phasor Representation of Alternating Voltage

Quantities that can be completely represented by a number of units are termed scalar quantities. Some examples of scalar quantities are temperature, volume, and resistance. Other quantities, such as force and velocity, must have a direction indicated and the number of units stated if they are to be completely specified. Such quantities can be graphically represented by a line that has a length proportional to the number of units, with the line drawn at an angle concerning some reference direction. This line is termed a vector, and the quantity it represents is classified as a vector quantity.

 

Phasors

A phasor is a vector rotating at a constant angular velocity. The sinusoidal output voltage produced by the simple generator can be represented by the phasor diagram shown in Figure 1. Here OA is a rotating vector (or phasor) with a constant angular velocity and a length that represents the peak output voltage E_m. The arrowhead identifies the end of the phasor that moves, and the other end is the axis of rotation. By convention, the direction of phasor rotation is taken counterclockwise.

 

Figure 1. Phasor representation of a sine wave. Phasor OA rotates in a counterclockwise direction, and its length equals the peak voltage E_m. At any angle θ, OA sinθ equals the instantaneous value of the sine wave. Image used courtesy of Amna Ahmad

 

The instantaneous value of the generated voltage represented by the phasor depends on the angle of the phasor with respect to the zero level. At time t = 0 on the waveform in Figure 1, the angle is θ = 0, and the instantaneous voltage level is zero. At t₁, the angle is θ_1, and the instantaneous level is e₁, as illustrated in the figure.

At t₁,                           \(e_{1}=(OA)sin\theta_{1}=E_{m}sin\theta_{1}\)

At t₂,                           \(e_{s}=E_{m}sin\theta_{2}\)

And at t₃,                         \(e_{3}=E_{m}sin\theta_{3}\)

At each of these instants, the voltage level is represented by the stopped phasor (or vector) OA₁ at an angle θ₁, OA₂ at an angle θ₂, and OA₃ at an angle θ₃. Any sinusoidal AC voltage or current can be represented by a phasor, and instead of degrees, the angles may be expressed in radians.

Example 1

Draw the phasor diagrams for the following voltages:

\[(a)V_{1}=15Vsin35\degree,(b)V_{2}=10Vsin\,sin\,60\degree,(c)V_{3}=12Vsin\,135\degree\]

Solution

See Figure 2. 

 

Figure 2. Phasor diagrams for Example 1. Image used courtesy of Amna Ahmad

 

Addition And Subtraction of Phasors

Phasor Addition

In Figure 3, two sinusoidal waveforms (A and B) are shown together, with the phasor representing each. It is seen that waveform B has a larger amplitude than waveform A. So, phasor OB is longer than phasor OA. Also, waveform B starts to grow positively from its zero level φ^o after the beginning of waveform A. Consequently, waveform B lags waveform A by φ^o, so the phasor OB is shown φ^o behind OA. It can also be said that waveform A leads waveform B by φ^o.

 

(a) Waveform 

(b) Phasors

Figure 3. Two waveforms and their phasors. OB is larger than OA because waveform B has a higher peak value than waveform A. Waveform A leads waveform B by the angle φ; consequently, phasor OA leads phasor OB by φ. Image used courtesy of Amna Ahmad

 

The instantaneous level of waveform A can be written as

\[e_{A}=OA\,sin\,\theta\]

and the instantaneous level of waveform B is

\[e_{B}=OBsin(\theta-\varphi)\]

Now consider Figure 4(a), in which the two waveforms are shown to be outputs from each of two series-connected generators. The resultant output of the two in series is the phasor sum of the two waveforms. Figure 4(b) shows how the instantaneous levels of A and B may be added to give the resultant waveform C. At t₁, a₁+b₁ gives point 1 on waveform C. At t₂, a₂+b₂ gives point 2 on waveform C, and so on. It is seen that waveform C lags waveform A by an angle α^o. So, waveform C is represented by phasor OC lagging α^o behind OA [see Figure 4(c)].

 

(a) Two sine wave generators in series

(b) Addition of instantaneous levels of two sine waves

(c) Phasor addition

(d) Phasor addition

Figure 4. To determine the resultant output from two series-connected AC generators, the instantaneous levels of the waveforms can be added together. Alternatively, phasors OA and OB, representing each voltage, can be added graphically to obtain the resultant OC. Image used courtesy of Amna Ahmad

 

Another method of obtaining the resulting waveforms A and B is illustrated in Figure 4(c). A parallelogram is constructed, with AC drawn equal to and parallel with OB and BC equal to and parallel with OA. The resultant diagonal of the parallelogram (OC) represents the amplitude of waveform C, and the angle α^o is the angle by which waveform C lags waveform A. Another approach to obtaining the same result is shown in Figure 4(d). Here, OA, identified as E₁, is first drawn horizontally. Then E₂ (which is OB) is drawn from the end of E₁ and at an angle φ with respect to E₁. The resultant (E₃ or OC) is found by drawing a line from O to the end of E₂. The angle of E₃ with respect to E₁ is again seen to be -α^o.

Instead of graphically adding the two phasors, the results may be obtained mathematically. Referring to Figure 5, each phasor is resolved into horizontal and vertical components (by trigonometry). The arithmetical sum of the horizontal components is then the horizontal component of the resultant. Similarly, the arithmetical sum of the vertical components is the vertical component of the resultant.

 

Figure 5. The resultant of two phasors can be determined by first resolving each into its horizontal and vertical components. The two horizontal quantities are added together, and the two vertical quantities are added to give the horizontal and vertical components of the resultant. Image used courtesy of Amna Ahmad

 

Horizontal component of E₁ = OA

Horizontal component of E₂ = OD = E₂ cos φ

Total of horizontal components = OE = OA + OD = E₁+E₂ cos φ

Vertical component of E₁ = 0

Vertical component of E₂ = OF = -E₂ cos (90°-φ) = -E₂ sin φ

Total of vertical components = 0 + OF= -E₂ sin φ

The negative sign here indicates that the vertical component is measured down from the horizontal rather than up.

Resultant

\[E_{3}=OC-\sqrt{(OE)^{2}+(OF)^{2}}\]

Angle

\[=tan^{-1}(\frac{OF}{OE})\]

Example 2

Two sinusoidal quantities are given as v₁ = 120 sin θ and v₂ = 75 sin (φ - 30°). Determine the results obtained when v₁ and v₂ are added.

 

Figure 6. The sum of two phasors, v₃ = v₁ + v_2. Image used courtesy of Amna Ahmad

 

Solution

Referring to Figure 6:

Horizontal component of v₁ = 120

Horizontal component of v₂ = 75 cos 30^o ≈ 65

Horizontal component of v₃ = 120 + 65 = 185

Vertical component of v₁ = 0

Vertical component of v₂ = -75 sin 30^o = -37.5

Vertical component of v₃ = 0 – 37.5 = -37.5 

\[v_{3}=\sqrt{185^{2}+37.5^{2}}=188.8\]

\[Angle=tan^{-1}{\Bigg[}\frac{v_{3}\,vertical\,component}{v_{3}\,horizontal\,component}{\Bigg]}\] 

\[=tan^{-1}\frac{-37.5}{185}=-11.5\degree\]

\[v_{3}=188.8sin(\theta-11.5\degree)\]

 

Phasor Subtraction

Sometimes instead of phasor addition, phasor subtraction is required. An example of this occurs with the circuit shown in Figure 7(a), where.

E₃ = E₁-E₂

Figure 7(b) shows the process of phasor subtraction. E₂ is shown as θ^o ahead of E₁. For subtraction, E₂ is reversed to become -E₂, as illustrated. The resultant (E₃) is then the phasor sums of E₁ and -E₂. Again, the results may be found mathematically or by either of the two graphical methods described.

 

(a) E₃ is the phasor difference between E₁ and E₂ 

(b) Determination of E₃

Figure 7. The phasor difference of two phasors, E₁ and E₂, can be determined by summing E₁ and -E₂. Image used courtesy of Amna Ahmad

 

Phasor addition and subtraction are not limited to two phasor quantities. The resultant of any number of phasor quantities can be found; some of these may require phasor addition, while others may have to be subtracted.

Example 3

Determine v₁ - v₂ for the two quantities specified in Example 2.

Solution

Referring to Figure 8

Horizontal component of v₁ = 120

Horizontal component of -v₂ =75 cos (180^o - 30^o) ≈ -65

Horizontal component of v₄ = 120 - 65 = 55

Vertical component of v₁ =0

Vertical component of -v₂ = 75 sin (180^o - 30^o) = 37.5

Vertical component of v₄ = 0 + 37.5 = 37.5 

\[v_{4}=\sqrt{55^{2}+37.5^{2}}=66.6\]

\[Angle=tan^{-1}\frac{-37.5}{55}=34.3\degree\]

\[v_{4}=66.6sin(\theta+34.3\degree)\]

 

Figure 8. Difference between the two phasors, v₄ = v₁ – v_2. Image used courtesy of Amna Ahmad

 

Takeaways of Phasor Representation of Alternating Voltage

The phasor representation of alternating voltage is a mathematical technique to simplify AC circuit analysis. It involves representing the time-varying voltage or current as a single complex number or phasor, which has a magnitude and phase angle. By using phasor diagrams, the behavior of AC circuits can be analyzed in a much more efficient and intuitive manner.

 

Featured image used courtesy of Adobe Stock