PRINCIPLES OF ALTERNATING CURRENT
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A
sinusoidal voltage can be described by the equation:
v (t) = VM sin (w t + f) or v (t) = VM cos (w t + f)
| where | v (t) | Instantaneous value of the voltage, in volts (V). |
| VM | Maximum or peak value of the voltage,
in volts (V) |
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| T | Period: The time taken for one cycle, in seconds |
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| f | Frequency – the number of periods in 1 second,
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| w | Angular
frequency, expressed in
radians/s w = 2*p*f or w = 2*p / T. |
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| f | Initial phase given in radians or degrees. This quantity determines the
value of the sine or cosine wave at t = 0. |
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| Note:
The amplitude of a sinusoidal voltage is sometimes expressed as VEff,
the effective or RMS value. This is related to VM according to the
relationship VM=Ö2VEff,
or approximately VEff = 0.707 VM |
Here
are a few examples to illustrate the terms above.
The
properties of the 220 V AC voltage in household electrical outlets in Europe:
Effective
value: VEff = 220 V
Peak
value: VM = 2
220 V = 311 V
Frequency:
f = 50 1/s = 50 Hz
Angular frequency:
w
= 2*p*f
= 314 1/s = 314 rad/s
Period:
T = 1/f = 20 ms
Time
function: v(t)=311 sin (314 t)
Let’s
see the time function using TINA’s Analysis/AC Analysis/Time Function command.
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here to load or save this circuit
You
can check that the period is T=20m and that VM = 311 V.
The
properties of the 120 V AC voltage in the household electrical outlet in the US:
Effective
value: VEff = 120 V
Peak
value: VM = Ö2
120 V = 169.68 V »
170 V
Frequency:
f = 60 1/s = 60 Hz
Angular frequency:
w
= 2*p*f
= 376.8 rad/s »
377 rad/s
Period:
T = 1/f = 16.7 ms
Time
function: v(t)=170 sin (377 t)
Note that in this case the time function could be given either as v(t)=311 sin (314 t+f) or v(t)=311 cos (314 t+f), since in the case of the outlet voltage we do not know the initial phase.
The initial phase plays an important role when several
voltages are present simultaneously. A good practical example is the three-phase
system, where three voltages of the same peak value, shape and frequency are
present, each of which has a 120° phase shift relative to the others. In a 60 Hz
network, the time functions are:
vA(t)=170 sin
(377 t)
vB(t)=170 sin (377 t - 120°)
vC(t)=170 sin (377 t + 120°)
The following figure made
with TINA shows the circuit with these time functions as TINA’s voltage
generators.
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The
voltage difference vAB= vA(t) - vB(t) is shown
as solved by TINA’s Analysis/AC Analysis/Time Function command.
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Note
that the peak of vAB (t) is approximately 294 V, larger than the 170
V peaks of the vA(t) or
vB(t) voltages, but also not simply the sum of their peak voltages.
This is due to the phase difference. We will discuss how to calculate the
resulting voltage (which is
Ö3
* 170 @
294 in this case) later in this chapter and also in the separate Three-phase
Systems chapter.
Characteristic
values of sinusoidal signals
Though an AC signal
continuously varies during its period, it is easy to define a few characteristic
values for comparing one wave with another: These are the peak, average and
root-mean-square (rms) values.
We have already met the peak value VM , which is simply the maximum value of the time function, the amplitude of the sinusoidal wave.
Sometimes the peak-to-peak
(p-p) value is used. For sinusoidal voltages and currents, the peak-to-peak
value is double the peak value.
The average
value of the sine wave is the arithmetic average of the values for the
positive half cycle. It is also called absolute
average since it is the same as the average of the absolute value of the
waveform. In practice, we encounter this waveform by rectifying the sine wave with a circuit called a full wave
rectifier.
It can be shown that
the absolute average of a sinusoidal wave is:
VAV = 2 / p
VM @ 0.637 VM
Note that the average of a whole cycle is zero.
The rms or effective value of a sinusoidal voltage or current corresponds to the equivalent DC value producing the same heating power. For example, a voltage with an effective value of 120 V produces the same heating and lighting power in a light bulb as does 120 V from a DC voltage source. It can be shown that the rms or effective value of a sinusoidal wave is:
Vrms = VM / Ö2 @ 0.707 VM
These values can be calculated the same way for both voltages and currents.
The
rms value is very important in practice. Unless indicated otherwise, power line
AC voltages (e.g. 110V or 220V) are given in rms values. Most AC meters are
calibrated in rms and indicate the rms level.
Example
1
Find the peak value of the sinusoidal voltage in an electrical network
with 220 V rms value.
VM = 220/0.707 = 311.17
V
Example
2
Find the peak value of the sinusoidal voltage in an electrical network with 110
V rms value.
VM = 110/0.707 = 155.58
V
Example
3 Find
the (absolute) average of the sinusoidal voltage if its rms value is 220 V.
Va
= 0.637 * VM = 0.637*311.17 = 198.26 V
Example
4 Find
the absolute average of the sinusoidal voltage if its rms value is 110 V.
The
peak of the voltage from Example 2 is155.58 V and hence:
Va
= 0.637 * VM = 0.637 * 155.58 = 99.13
V
Example
5 Find
the ratio between the absolute average (Va) and rms (V) values for
the sinusoidal waveform.
V/Va
= 0.707/0.637 = 1.11
Note
that you cannot add average values in an AC circuit because it leads to improper
results.
PHASORS
As we have already seen in
the previous section, it is often necessary in AC circuits to add sinusoidal
voltages and currents of the same frequency. Though it is possible to add the
signals numerically using TINA, or by employing trigonometric relations, it is
more convenient to use the so-called phasor
method. A phasor is a complex number
representing the amplitude and phase of a sinusoidal signal. It is important to
note that the phasor does not represent the frequency, which must be the same
for all phasors.
A phasor can be handled as
a complex number or represented graphically as a planar arrow in the complex
plane. The graphic representation is called a phasor diagram. Using phasor
diagrams, you can add or subtract phasors in a complex plane by the triangle or
parallelogram rule.
There are two forms of
complex numbers: rectangular and polar.
The rectangular representation is in the form
a + jb, where j
= Ö-1 is the imaginary unit.
The polar representation is
in the form Aejj , where A is the absolute value (amplitude) and f is the angle of the phasor from the positive real
axis, in the counterclockwise direction.
We will use bold
letters for complex quantities.
Now let’s see how to
derive the corresponding phasor from a time function.
First, assume that all the
voltages in the circuit are expressed in the form of cosine functions. (All
voltages can be converted to that form.) Then the phasor
corresponding to the voltage of v(t) = VM cos(wt+f)
is: VM = VMe jf, which is also called the complex peak value.
For example, consider the voltage: v(t) = 10 cos(wt+30°)
The corresponding phasor
is:
V
We can calculate the time function from a phasor in the
same way. First we write the phasor in polar form e.g. VM = VMe jr
and
then the corresponding time function is
v(t)=VM (cos(wt+r).
For example, consider the
phasor VM
=10 - j20 V
Bringing it to polar form:
And hence the time function is: v(t) = 22.36 cos(wt – 63.5°)
V
Phasors are often used to define the complex effective
or rms value of the voltages and currents in AC circuits. Given v(t) = VMcos(wt+r)=
10cos(wt+30°)
Numerically:
v(t)
= 10*cos (wt-30°)
The complex effective (rms) value: V
= 0.707*10* e- j30° = 7.07 e- j30°
= 6.13 – j 3.535
Vice versa: if the complex
effective value of a voltage is:
V = - 10 + j
20 = 22.36 e j 116.5°
then the complex peak
value:
and the time function:
v(t) = 31.63 cos ( wt
+ 116.5° ) V
A
short justification of the above techniques is as follows. Given a time function
VM (cos(wt+r), let’s define the complex time function as:
v(t) =VM e jr
e jwt = VMe
jwt = VM (cos(r) + j sin(r))e jwt
where
VM =VM e jr t
= VM (cos(r)
+ j sin(r)) is just the phasor introduced above.
For
example, the complex time function of v(t)=10 cos(wt+30°)
v(t) = VMe
jwt = 10 e j30 e jwt = 10e jwt (cos(30) + j
sin(30))= e jwt (8.66+j5)
By
introducing the complex time function, we have a representation with both a real
part and an imaginary part. We can always recover the original real function of
time by taking the real part of our result: v(t) = Re
{v(t)}
However the complex time function has the great
advantage that, since all the complex time functions in the AC circuits under
consideration have the same ejwt multiplier,
we can factor this out and just work with the phasors. Moreover, in practice we
do not use the ejwt part
at all--just the transformations from the time functions to the phasors and
back.
To demonstrate the
advantage of using phasors, let’s see the following example.
Example
6 Find
the sum and the difference of the voltages:
v1 = 100 cos (314*t) and v2 = 50 cos (314*t-45°)
First write the phasors of
both voltages:
V1M
= 100 V2M=
50 e – j 45° = 35.53 – j
35.35
Hence:
Vadd = V1M + V2M
= 135.35 – j 35.35 = 139.89 e- j 14.63°
Vsub = V1M – V2M
= 64.65 + j35.35 =
73.68 e j 28.67°
and then the time
functions:
vadd(t)
= 139.89 * cos(wt
– 14.63°)
vsub(t)
= 73.68 * cos(wt + 28.67°)
As this simple example
shows, the method of phasors.is an extremely powerful tool for solving AC
problems.
Let’s solve the problem
using the tools in TINA’s interpreter.
| {calculation of v1+v2} v1:=100 v2:=50*exp(-pi/4*j) v2=[35.3553-35.3553*j] v1add:=v1+v2 v1add=[135.3553-35.3553*j] abs(v1add)=[139.8966] radtodeg(arc(v1add))=[-14.6388] {calculation of v1-v2} v1sub:=v1-v2 v1sub=[64.6447+35.3553*j] abs(v1sub)=[73.6813] radtodeg(arc(v1sub))=[28.6751] |
The amplitude and phase
results confirm the hand calculations.
Now lets check the result
using TINA’s AC analysis.
Before performing the
analysis, let’s make sure that the Base function for AC ia set to cosine
in the Editor Options dialog box
from the View/Option menu. We will explain the role of this parameter at Example
8.
The circuits and the
results:
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Again the result is the
same. Here are the time function graphs:
v1 = 100 sin (314*t) and v2 = 50 cos (314*t-45°)
This example brings up a new question. So far we have
required that all time functions be given as cosine functions. What shall we do
with a time function given as a sine? The solution is to transform the sine
function to a cosine function. Using the trigonometric relation sin(x)=cos(x-p/2)=cos(x-90°),
our example can be rephrased as follows:
v1 = 100 cos(314t - 90°) and v2 = 50 cos (314*t - 45°)
Now the phasors of the
voltages are:
V1M
= 100 e – j 90° = -100 j
V2M= 50 e – j
45°
= 35.53 – j 35.35
Hence:
Vadd = V1M + V2M
= 35.53
– j 135.35
Vsub = V1M – V2M
= - 35.53
– j 64.47
and then the time
functions:
vadd(t)
= 139.8966 cos(wt-75.36°)
vsub(t)
= 73.68 cos(wt-118.68°)
Let’s solve the problem
using the tools in TINA’s interpreter.
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{calculation of v1+v2}
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Let’s check the result
with TINA’s AC Analysis
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Example
8 Find
the sum and the difference of the voltages:
v1 = 100 sin (314*t ) and v2 = 50 sin (314*t-45°)
This example brings up one
more issue. What if all voltages are given as sine waves and we also wish to see
the result as a sine wave?. We
could of course convert both voltages to cosine functions, compute the answer,
and than convert the result back to a sine function--but this isn't necessary.
We can create phasors from the sine waves in the same way that we did from
cosine waves and then simply use their amplitude and phases as amplitude and
phase of sine waves in the result.
This will obviously give the same result as
transforming the sine waves to cosine waves. As we could see in the previous
example, this is equivalent to multiplying by –j and then using the cos(x) = sin (x-90°) relation to transform it back to a sine wave. This is
equivalent to multiplying by j. In
other words, since –j × j = 1, we could use the phasors derived directly
from the amplitudes and phases of sine waves to represent the function and then
return to them directly. Also, reasoning in the same manner about the complex
time functions, we could consider sine waves as the imaginary parts of the
complex time functions and supplement them with the cosine function to create
the full complex time function.
Let's see the solution to
this example using the sine functions as the base of the phasors (transforming
sin(wt) to the real unit phasor (1) ).
V1M
= 100 V2M=
50 e – j 45° = 35.53 - j
35.35
Hence:
Vadd = V1M + V2M
= 135.53
– j 35.35
Vsub = V1M – V2M
= 64.47+
j 35.35
Note that the phasors are
exactly the same as in Example 6 but not the time functions:
v3(t)
= 139.9sin(wt
- 14.64°)
v4(t)
= 73.68sin(wt + 28.68°)
As you can see, it is very easy to obtain the result using sine functions, especially when our initial data are sine waves. Many textbooks prefer to use the sine wave as the base function of phasors. In practice, you can use either method, but don't confuse them.
When you create the phasors, it is very important that all time functions are first converted either to sine or cosine. If you started from sine functions, your solutions should be represented with sine functions when returning from phasors to time functions. The same is true if you start with cosine functions.
Let’s solve the same problem using TINA’s interactive mode. Since we want to
use sine functions as the base for creating the phasors, make sure that the Base
function for AC is set to sine in
the Editor Options dialog box from
theView/Option menu.
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The circuits for making the sum and difference of the waveforms and the result:
and the time funtions:
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