# Phasors

As noted earlier in this chapter, the addition of sinusoidal voltages and currents will frequently be required in the analysis of ac circuits. One lengthy but valid method of performing this operation is to place both sinusoidal waveforms on the same set of axes and add algebraically the magnitudes of each at every point along the abscissa, as shown for $c = a + b$ in Fig. 1.
Fig. 1: Adding two sinusoidal waveforms on a point-by-point basis.
This, however, can be a long and tedious process with limited accuracy. A shorter method uses the rotating radius vector shown in Fig. 2,
Fig. 2: Generating a sinusoidal waveform through the vertical projection of a rotating vector
This radius vector, having a constant magnitude (length) with one end fixed at the origin, is called a phasor when applied to electric circuits.
During its rotational development of the sine wave, the phasor will, at the instant t = 0, have the positions shown in Fig. 3(a) for each waveform in Fig. 3(b).
Fig. 3: (a) The phasor representation of the sinusoidal waveforms of Fig. 3(b);
(b) finding the sum of two sinusoidal waveforms of v1 and v2.
Note in Fig. 3(b) that $v_2$ passes through the horizontal axis at $t = 0 s$, requiring that the radius vector in Fig. 3(a) be on the horizontal axis to ensure a vertical projection of zero volts at $t = 0 s$. Its length in Fig. 3(a) is equal to the peak value of the sinusoid as required by the radius vector in Fig. 2. The other sinusoid has passed through $90^\circ$ of its rotation by the time $t = 0s$ is reached and therefore has its maximum vertical projection as shown in Fig. 3(a).
Since the vertical projection is a maximum, the peak value of the sinusoid that it will generate is also attained at $t = 0 s$, as shown in Fig. 3(b). Note also that $v_T =v_1$ at $t =0 s$ since $v_2 = 0 V$ at this instant. It can be shown [see Fig. 3(a)] using the vector algebra $$1 V \angle{0^\circ} + 2 V \angle{90^\circ} = 2.236 V \angle{63.43^\circ}$$ In other words, if we convert $v_1$ and $v_2$ to the phasor form using $$v = V_m \sin(wt + \theta) = V_m \angle{\pm \theta}$$ and add them using complex number algebra, we can find the phasor form for $v_T$ with very little difficulty. It can then be converted to the time domain and plotted on the same set of axes, as shown in Fig. 3(b). Figure 3(a), showing the magnitudes and relative positions of the various phasors, is called a phasor diagram. It is actually a "snapshot" of the rotating radius vectors at $t = 0 s$.
In the future, therefore, if the addition of two sinusoids is required, they should first be converted to the phasor domain and the sum found using complex algebra. The result can then be converted to the time domain.
Fig. 4: Adding two sinusoidal currents with phase angles other than 90 degrees.
The case of two sinusoidal functions having phase angles different from 0 and 90 appears in Fig. 4. Note again that the vertical height of the functions in Fig. 4(b) at $t = 0 s$ is determined by the rotational positions of the radius vectors in Fig. 4(a). Since the rms, rather than the peak, values are used almost exclusively in the analysis of ac circuits, the phasor will now be redefined for the purposes of practicality and uniformity as having a magnitude equal to the rms value of the sine wave it represents. The angle associated with the phasor will remain as previously described as the phase angle.
In general, for all of the analyses to follow, the phasor form of a sinusoidal voltage or current will be $$\text{V} =V \angle {\theta} \,\text{and}\, \text{I} =I \angle {\theta}$$ where V and I are rms values and v is the phase angle. It should be pointed out that in phasor notation, the sine wave is always the reference, and the frequency is not represented.
Phasor algebra for sinusoidal quantities is applicable only for waveforms having the same frequency.
Example 1: Find the input voltage of the circuit of Fig. 5 if $$v_a = 50 \sin(377t + 30^\circ) \, \text{and}\\ v_b = 30 \sin(377t + 60^\circ)$$
Fig. 5: For Example 1.
Solution:
Applying Kirchhoff's voltage law, we have $$e_{in} = v_a + v_b$$ Converting from the time to the phasor domain yields $$\begin{split} v_a &= 50 \sin(377t + 30^\circ) = {50 \over \sqrt{2}} \angle{30^\circ}\\ &= 35.35V \angle{30^\circ}\\ \end{split}$$ and $$e_{in} = v_a + v_b$$ Converting from the time to the phasor domain yields $$\begin{split} v_b &= 30 \sin(377t + 60^\circ) = {30 \over \sqrt{2}} \angle{60^\circ}\\ &= 21.21V \angle{60^\circ}\\ \end{split}$$ Converting from polar to rectangular form for addition yields $$V_a = 35.35V \angle{30^\circ}= 30.61 V + j 17.68 V$$ $$V_b = 21.21 V \angle{60^\circ} = 10.61 V + j 18.37 V$$ Then $$\begin{split} E_{in} &= V_a + V_b \\ &= (30.61 V + j 17.68 V) + (10.61 V + j 18.37 V)\\ &= 41.22 V + j 36.05 V\\ \end{split}$$ Converting from rectangular to polar form, we have $$E_{in} = 41.22 V + j 36.05 V = 54.76 V \angle{41.17^\circ}$$ Converting from the phasor to the time domain, we obtain $$E_{in} = 54.76 V \angle{41.17^\circ} \Rightarrow\\ e_{in} = \sqrt{2}(54.76) \sin(377t + 41.17^\circ)\\$$ and $$e_{in} = 77.43 \sin(377t + 41.17^\circ)$$ A plot of the three waveforms is shown in Fig. 6. Note that at each instant of time, the sum of the two waveforms does in fact add up to $e_{in}$. At $t = 0 (wt = 0), e_{in}$ is the sum of the two positive values, while at a value of $wt$, almost midway between $\pi/2$ and $\pi$, the sum of the positive value of $v_a$ and the negative value of $v_b$ results in $e_{in} = 0$.
Fig. 6: Solution for Example 1.