Representation of AC Current and Voltage by Rotating Vectors: In a conventional AC supply, the voltage and current are sinusoidal in nature. By sinusoidal, we mean the sine and cosine functions. Voltage and current vary sinusoidally with respect to time. Voltage and current can be represented in the following way:
\(V\left( t \right) = {V_0}\sin \left( {\omega t} \right)\)
\(i\left( t \right) = {i_0}\sin \left( {\omega t + \phi } \right)\)

But why sinusoidal? It isn’t that we are particularly fond of the sine waveform. A sine wave is the natural consequence of the way we produce or generate electricity. Also, sinusoidal AC supply leads to some interesting behaviour of inductive and capacitive loads.

Let us, once again, consider the above two equations \(V\left( t \right) = {V_0}\sin \left( {\omega t} \right)\) and \(i\left( t \right) = {i_0}sin\left( {\omega t + \phi } \right)\).
The symbol \(\phi\) represents the phase difference between the current and voltage. You must have heard of the terms “lag,” “lead,” “in phase,” “out of phase.” All these terms refer to this phase difference \(\phi\). When the load is inductive, the current lags the voltage or \( – {90^0} \leqslant \phi < 0\). When the load is capacitive, the current leads the voltage or \(0 \leqslant \phi < {90^0}\).

Fig 1: How to Tell If the Current is Leading or Lagging?

If we look at figure 1, it isn’t readily apparent why the left red curve is leading and why the right red curve is lagging. This is where phasors come in handy. Phasors help us visualize the phase difference \(\phi\) and help us connect the two, the sinusoidal equations and their plots. But is that all phasors are good for. Let’s find out!

What is a Phasor?

Fig 2: A Voltage Phasor Having Initial Phase 0

Consider a simple sinusoidal voltage source given by
\(V\left( t \right) = {V_0}\sin \left( {\omega t} \right)\)
A phasor or phase vector is a vector that rotates about its tail in a 2D plane. To represent the above voltage source, our phasor vector will have a magnitude of \(V_0\) and will rotate anticlockwise (anticlockwise sense is positive) with an angular speed \(\omega\).  In other words, the magnitude of the phasor is equal to the peak voltage of the voltage source and the angular speed of the phasor is equal to the angular frequency of the voltage source (fig. 2).

Fig 3: Relation Between Phasor and Sinusoid

Let’s examine this phasor (fig. 3) closely. As this phasor rotates with an angular speed of \(\omega\), its angular position (angle made by the phasor with the positive \(x-\)axis) at any time \(t\) can be written as \(\omega t\). Using simple geometry, we can evaluate the \(y\) coordinate of the head of the phasor at time \(t\) as \({V_0}\sin \left( {\omega t} \right)\).

Doesn’t this look familiar? That’s correct! The height of the phasor from the \(x\) axis, or the \(y\) coordinate, is identical to \(V(t)\). In other words, as the phasor rotates about the centre with angular speed \(\omega\), at any given instant, its \(y\) coordinate gives us the instantaneous voltage of the source at that instant of time.

Fig 4: Phasor and V-t Side by Side Comparison

Let us draw the voltage phasor and the \(V – t\) curve side by side (fig. 4) so that we can observe the connection between them. The positions of the phasor have been mapped with the corresponding voltage-time curve at \(\pi / 6 \omega\) second intervals, in other words, at \(\omega t = 0,\,30^\circ,\,60^\circ,\,90^\circ,\,120^\circ,\,150^\circ,\,180^\circ,\,210^\circ,\,240^\circ,\,270^\circ,\,300^\circ,\,330^\circ\). At \(t = 0\), the phasor points along the positive \(x\) axis. At any instant of time, the \(y\) coordinate of the phasor gives us the instantaneous voltage of the source at that instant.

Have a close look at the phasor indicated by the brown coloured vector. Trace the corresponding point on the sine wave where \(\omega t = 30^\circ\). Observe that the height of the phasor is equal to the \(y\) coordinate of the sine wave at that point. When the phasor points to the right, the voltage is zero; when the phasor points upwards, the voltage reaches its positive maximum value; when the phasor points left, the voltage reaches zero again, and when the phasor points downwards, the voltage reaches its negative maximum value.  We observe that, as the phasor completes one rotation, the voltage completes one cycle of the sine wave.

Now how do we draw a phasor if the voltage is given by
\(V\left( t \right) = {V_0}\sin \left( {\omega t + \phi } \right)\)
Do we do something different? It’s quite simple, really. In this instance, our phasor simply starts at an angular position of \(\phi\) (fig. 5). In other words, at \(t = 0\), the phasor will make an angle \(\phi\) with the positive \(x\) axis. The rest is as before, the phasor will start rotating anticlockwise with angular speed \(\omega\) from \(t = 0\) and its \(y\) coordinate will give us the instantaneous voltage at any given instant of time, just like before!

Fig 5: Phasor Having Initial Phase \(\phi\) at \(t=0\)

How do Phasors or Representation of AC Current and Voltage by Rotating Vectors Help Us?

When current leads the voltage: Consider a voltage source \(V = {V_0}\sin \left( {\omega t} \right)\) which produces a current given by \(i = {i_0}\sin \;\left( {\omega t + \frac{\pi }{6}} \right)\). Here the current leads the voltage by a phase difference of \(\frac{\pi }{6}\). The corresponding phasors for the voltage and current are drawn at \(t = 0\).

Fig 6: Phasor, V-t, and i-t at \(t = 0\)

The voltage phasor is drawn in red, while the current phasor is drawn in blue (fig. 6). Recall from earlier that the current phasor will start at an angular position of \(\frac{\pi }{6}\) or \(30^\circ\) with respect to the \(x\) axis, due to the phase constant of \(\frac{\pi }{6}\) appearing in the current equation.

Thus, the phase difference of \(\frac{\pi }{6}\) between current and the voltage appears as an angle difference in their corresponding phasors. After \(t = 0\), the two phasors rotate with the same angular speed \(\omega\) in the anticlockwise direction, thereby maintaining the angle difference of \(\frac{\pi }{6}\) or \(30^\circ\) between them.

Fig 7: Phasors, V-t, and i-t at \(t = π/3ω\)

The current leading the voltage can be understood by observing the phasors for current and voltage. As the current phasor is positioned anticlockwise with respect to the voltage phasor, the current phasor will reach any position sooner or before the voltage phasor. For example, the current phasor reaches the vertically upward position (fig 7.) sooner than the voltage phasor. Putting it another way, the current will reach its crest sooner than the voltage reaches its crest. Hence, the current leads the voltage. Such circuits are capacitive.

When current lags the voltage: Consider a voltage source \(V = {V_0}\sin \left( {\omega t} \right)\) which produces a current given by \(i = {i_0}\sin \;\left( {\omega – \frac{\pi }{6}} \right)\). Here the current lags the voltage by a phase difference of \(\frac{\pi }{6}\). The corresponding phasors for the voltage and current are drawn at \(t = 0\).

Fig 8: Phasors, V-t and i-t at \(t = 0\)

The voltage phasor is drawn in red, while the current phasor is drawn in blue (fig. 8). In this case, the current phasor will start at an angular position of \(-\pi/6\) or \(- 30^\circ\) with respect to the \(x\) axis, due to the phase constant of \(-\pi/6\) appearing in the current equation.

The phase difference of \(-\pi/6\)  between current and the voltage appears as an angle difference in their corresponding phasors.  Once again, both phasors will rotate anticlockwise with the same angular speed \(\omega\) and maintain the angular separation of \(\pi/6\) between them. This time, it is the voltage phasor that is ahead by \(\pi/6\).

Fig 9: Phasors, V-t and i-t at \(t = π/2ω\)

The current lagging the voltage can once more, be understood by observing their respective phasors. As the voltage phasor lies anticlockwise with respect to the current phasor, any position will be reached by the voltage phasor sooner than the current phasor. For example, the voltage phasor will reach the vertically upward position (fig. 9) sooner than the current phasor. Putting it another way, the voltage will reach its crest sooner than the current reaches its crest. Hence, the current lags the voltage. Such circuits are inductive.

Where Do We Apply Phasors?

We will examine one simple application of phasors. When two DC sources are connected in series, the resultant voltage is simply the scalar addition of the two voltages. When two AC sources having the same frequency(coherent) are connected in series, the calculation isn’t so straightforward, especially if the two sources have a phase difference between them.

With phasors, our problem becomes rather easy! Consider two voltage sources connected in series. Their individual voltages are given as
\({V_1}\left( t \right) = {V_1}\sin \left( {\omega t} \right)\)
\({V_2}\left( t \right) = {V_2}\sin \left( {\omega t + \phi } \right)\)
Here, we draw the phasors for the two sources (fig. 10). The phasor for \(V_1 (t)\) is drawn along the \(x\) axis and has a magnitude equal to the peak value \(V_1\). As \(V_2 (t)\) is ahead of \(V_1 (t)\) by a phase difference of \(\phi\) we will draw its phasor at an angle \(\phi\), with respect to the first phasor. Also, as \(V_2 (t)\) is leading, its phasor is anticlockwise with respect to the phasor of \(V_1 (t)\).

Fig 10: Resultant Phasor From Vector Addition

The resultant phasor that we get from the vector addition of the two phasors gives us the resultant voltage. The resultant voltage can be written as
\({V_R}\left( t \right) = {V_R}\sin \left( {\omega t + \varepsilon } \right)\)
where \(V_R\) and \(\varepsilon\) evaluated from the vector addition.
There are a lot more ways in which phasors make the study of AC circuits easier. Feel to explore them!

Summary

Phasors are very useful for visualising sinusoidally time-variant quantities. AC current and voltage being sinusoidal in nature can therefore be represented by phasors. A phasor is a rotating vector whose length is equal to the maximum value of the sinusoidally varying physical quantity and whose angular speed is numerically equal to the angular frequency of the sinusoidally varying function.

Phasors, in addition to being useful in visualizing AC current and voltage, also helps us visualize the phase difference between current and voltage. They also help simplify some circuit problems, such as the addition of phase differentiated voltages. It is, therefore a very useful tool to have in your arsenal.

FAQs on Representation of AC Current and Voltage by Rotating Vectors

Q.1. What is a phasor in AC?
Ans: A phasor is a rotating vector that represents a sinusoidally varying quantity such as AC current or voltage.

Q.2. Why are phasors useful?
Ans: Phasors help us visualize the sinusoidal quantities through rotating vectors. It is especially useful in visualizing phase differences.

Q.3. Why are phasors used in AC?
Ans: Phasors help simplify several AC circuit problems; one such problem has been explored in this article.

Q.4. Why do we need phasors in AC calculation?
Ans: In DC, most calculations involve scalar operations. In AC, however, due to the phase difference in various circuit parameters, calculations aren’t straightforward. Phasors make it easier to address such problems.

Q.5. How does phase difference appear in the phasor diagram?
Ans: In a phasor diagram, the phase difference appears as an angle difference between the phasors.

We hope this detailed article on Representation of AC Current and Voltage by Rotating Vectors helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!