But when presented mathematically in this way it is sometimes difficult to visualise this angular or phasor difference between two or more sinusoidal waveforms. One way to overcome this problem is to represent the sinusoids graphically within the spacial or phasor-domain form by using Phasor Diagrams, and this is achieved by the rotating vector method. Basically a rotating vector, simply called a “Phasor” is a scaled line whose length represents an AC quantity that has both magnitude (“peak amplitude”) and direction (“phase”) which is “frozen” at some point in time.
A phasor is a vector that has an arrow head at one end which signifies partly the maximum value of the vector quantity ( V or I ) and partly the end of the vector that rotates. Generally, vectors are assumed to pivot at one end around a fixed zero point known as the “point of origin” while the arrowed end representing the quantity, freely rotates in an anti-clockwise direction at an angular velocity, ( ω ) of one full revolution for every cycle. This anti-clockwise rotation of the vector is considered to be a positive rotation. Likewise, a clockwise rotation is considered to be a negative rotation. Although the both the terms vectors and phasors are used to describe a rotating line that itself has both magnitude and direction, the main difference between the two is that a vectors magnitude is the “peak value” of the sinusoid while a phasors magnitude is the “rms value” of the sinusoid. In both cases the phase angle and direction remains the same.
Phase Difference of a Sinusoidal Waveform
The generalised mathematical expression to define these two sinusoidal quantities will be written as:
The current, i is lagging the voltage, v by angle Φ and in our example above this is 30o. So the difference between the two phasors representing the two sinusoidal quantities is angle Φ and the resulting phasor diagram will be.
Phasor Diagram of a Sinusoidal Waveform
The phasor diagram is drawn corresponding to time zero ( t = 0 ) on the horizontal axis. The lengths of the phasors are proportional to the values of the voltage, ( V ) and the current, ( I ) at the instant in time that the phasor diagram is drawn. The current phasor lags the voltage phasor by the angle,Φ, as the two phasors rotate in an anticlockwise direction as stated earlier, therefore the angle, Φ is also measured in the same anticlockwise direction.
If however, the waveforms are frozen at time t = 30o, the corresponding phasor diagram would look like the one shown on the right. Once again the current phasor lags behind the voltage phasor as the two waveforms are of the same frequency.
However, as the current waveform is now crossing the horizontal zero axis line at this instant in time we can use the current phasor as our new reference and correctly say that the voltage phasor is “leading” the current phasor by angle, Φ. Either way, one phasor is designated as the reference phasor and all the other phasors will be either leading or lagging with respect to this reference.
Phasor Addition
Sometimes it is necessary when studying sinusoids to add together two alternating waveforms, for example in an AC series circuit, that are not in-phase with each other. If they are in-phase that is, there is no phase shift then they can be added together in the same way as DC values to find the algebraic sum of the two vectors. For example, if two voltages of say 50 volts and 25 volts respectively are together “in-phase”, they will add or sum together to form one voltage of 75 volts.
If however, they are not in-phase that is, they do not have identical directions or starting point then the phase angle between them needs to be taken into account so they are added together using phasor diagrams to determine their Resultant Phasor or Vector Sum by using the parallelogram law.
Consider two AC voltages, V1 having a peak voltage of 20 volts, and V2 having a peak voltage of 30 volts where V1 leads V2 by 60o. The total voltage, VT of the two voltages can be found by firstly drawing a phasor diagram representing the two vectors and then constructing a parallelogram in which two of the sides are the voltages, V1 and V2 as shown below.
Phasor Addition of two Phasors
The source is shown as a magnitude and phase (the phase is 0 in this case). The resistor's impedance is the same as the resistance. The capacitor's impedance is 1/jwC. Note that 1/j = -j. The frequency is 2krad/s and C is 1uF.
2. Solve for the desired value in the phasor domain. In this case, we can use a voltage divider to find the voltage over the cap:
Vc = -j.5k/(1k - j.5k) * 10<0
Note that the voltage divider uses a fraction of the source to find the desired voltage, where the denominator has the total series impedance and the numerator has the impedance of the element for which we are finding the voltage.
Now clean it up a bit by multiplying top and bottom by the conjugate of the denominator (conjugate of a+jb is a-jb). Why? Because multiplying by the conjugate produces a real number. Once the denominator is all real, it is easier to write the whole thing in rectangular form.
Vc = -j.5k/(1k - j.5k) * (1k + .j5k)/(1k + .j5k) * 10<0
= (-j.5M + .25M)/(1M +.25M) * 10<0 (notice how the denominator becomes a real number)
= (.25M - j.5M)/1.25M * 10<0
= 2- j4 (rectangular form)
Let's convert to polar form. Magnitude is the square root of the real part squared plus the imaginary part squared. Phase is the inverse tangent of the imaginary part divided by the real part.
Vc = SQRT(2^2 + -4^2) < arctan(-4/2)
Vc = 4.47 < -63.4
3. Convert answer to time-domain.
We already converted to polar form, which is the easiest form to convert back to time-domain. The time-domain solution has four parts. Two of them are from the solution we computed (magnitude and phase), and two of them we just copy from the original source (sin/cos and frequency). Thus:
Vc(t) = 4.47 sin (2kt - 63.4)
With Matildo
