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].
This, however, can be a long and tedious process with limited accuracy. A shorter method uses the rotating radius vector shown in [Fig. 2],
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)].
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
In other words, if we convert $v_1$ and $v_2$ to the phasor form using
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)]. [Fig. 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.
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
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.
Fig. 1: Adding two sinusoidal waveforms on a point-by-point basis.
Fig. 2: Generating a sinusoidal waveform through the vertical projection of a rotating vector
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.
(b) finding the sum of two sinusoidal waveforms of v1 and v2.
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
$$1 V \angle{0^\circ} + 2 V \angle{90^\circ} = 2.236 V \angle{63.43^\circ}$$
$$v = V_m \sin(wt + \theta) = V_m \angle{\pm \theta}$$
Fig. 4: Adding two sinusoidal currents with phase angles other than 90 degrees.
In general, for all of the analyses to follow, the phasor form of a sinusoidal voltage or current will be
$$\text{V} =V \angle {\theta} \,\text{and}\, \text{I} =I \angle {\theta}$$
Example 1: Find the input voltage of the circuit of [Fig. 5] if
Solution:
Applying Kirchhoff's voltage law, we have
Converting from the time to the phasor domain yields
and
Converting from the time to the phasor domain yields
Converting from polar to rectangular form for addition yields
Then
Converting from rectangular to polar form, we have
Converting from the phasor to the time domain, we obtain
and
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 $e_{in} = 0$.
$$v_a = 50 \sin(377t + 30^\circ) \, \text{and}\\ v_b = 30 \sin(377t + 60^\circ)$$
Fig. 5: For Example 1.
Applying Kirchhoff's voltage law, we have
$$e_{in} = v_a + v_b$$
$$\begin{split}
v_a &= 50 \sin(377t + 30^\circ) = {50 \over \sqrt{2}} \angle{30^\circ}\\
&= 35.35V \angle{30^\circ}\\
\end{split}$$
$$e_{in} = v_a + v_b$$
$$\begin{split}
v_b &= 30 \sin(377t + 60^\circ) = {30 \over \sqrt{2}} \angle{60^\circ}\\
&= 21.21V \angle{60^\circ}\\
\end{split}$$
$$V_a = 35.35V \angle{30^\circ}= 30.61 V + j 17.68 V$$
$$V_b = 21.21 V \angle{60^\circ} = 10.61 V + j 18.37 V$$
$$\begin{split}
E_{in} &= V_a + V_b \\
&= (30.61 V + j 17.68 V) + (10.61 V + j 18.37 V)\\
&= 41.22 V + j 36.05 V\\
\end{split}$$
$$E_{in} = 41.22 V + j 36.05 V = 54.76 V \angle{41.17^\circ}$$
$$E_{in} = 54.76 V \angle{41.17^\circ} \Rightarrow\\
e_{in} = \sqrt{2}(54.76) \sin(377t + 41.17^\circ)\\
$$
$$e_{in} = 77.43 \sin(377t + 41.17^\circ)$$
Fig. 6: Solution for Example 1.
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