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Phasor Diagrams for electrical circuits
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Phasor Diagrams
In AC electrical theory every power source supplies a voltage that is
either a sine wave of one particular frequency or can be considered as
a sum of sine waves of differing frequencies. The neat thing about a
sine wave such as V(t) = Asin(ωt + δ) is that it can be considered to
be directly related to a vector of length A revolving in a circle with
angular velocity ω - in fact just the y component of the vector. The phase constant δ is the starting angle at t = 0. In Figure 1, an animated
GIF shows this relation [you may need to click on the image for it to
animate].
Figure 1
Since a pen and paper drawing cannot be animated so easily, a 2Ddrawing of a rotating vector shows the vector inscribed in the centre
of a circle as indicated in Figure 2 below. The angular frequency ω
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may or may not be indicated.
Figure 2
When two sine waves are produced on the same display, one wave is
often said to be leading or lagging the other. This terminology makes
sense in the revolving vector picture as shown in Figure 3. The bluevector is said to be leading the red vector or conversely the red vector
is lagging the blue vector.
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Figure 3
Considering sine waves as vertical components of vectors has more
important properties. For instance, adding or subtracting two sine
waves directly requires a great deal of algebraic manipulation and the
use of trigonometric identities. However if we consider the sine waves
as vectors, we have a simple problem of vector addition if we ignoreω. For example consider
Asin(ωt + φ) = 5 sin(ωt + 30°) + 4 sin(ωt + 140°) ;
the corresponding vector addition is:
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Ax = 5cos(3
Ay = 5 sin(30
Thus
and
Figure 4
So the Pythagorean theorem and simple trigonometry produces the
result
5.23 sin(ωt + 76.0°) .
Make a note not to forget to put ωt back in!
Phasors and Resistors, Capacitors, and Inductors
The basic relationship in electrical circuits is between the current
through an element and the voltage across it. For resistors, the famous
Ohm's Law gives
VR = IR . (1)
For capacitors
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VC = q/C . (2)
For inductors
VL = Ld I/d t . (3)
These three equations also provide a phase relationship between the
current entering the element and the voltage over it. For the resistor,
the voltage and current will be in phase. That means if I has the form
Imax sin(ωt + φ) then VR has the identical form Vmax sin(ωt + φ) where
Vmax = ImaxR. For capacitors and inductors it is a little more
complicated. Consider the capacitor. Imagine the current entering the
capacitor has the form Imax sin(ωt + φ). The voltage, however,depends on the charge on the plates as indicated in Equation (2). The
current and charge are related by I = d q/d t. Since we know the form
of I simple calculus tells us that q should have the form −
(Imax/ω)cos(ωt + φ) or (Imax/ω) sin(ωt + φ - 90°). Thus VC has the
form (Imax/ωC) sin(ωt + φ - 90°) = Vmax sin(ωt + φ - 90°). The
capacitor current leads the capacitor voltage by 90°. Also note that
Vmax = Imax/ωC. The quantity 1/ωC is called the capacitive reactanceXC and has the unit of Ohms. For the inductor, we again assume that
the current entering the capacitor has the form Imax sin(ωt + φ). The
voltage, however, depends on the time derivative of the current as
seen in Equation (3). Since we assumed the form of I, then the
voltage over the inductor will have the form ωLImaxcos(ωt + φ) or
Vmax sin(ωt + φ + 90°). The inductor current lags the inductor voltage.
Here note that the quantity ωL is called the inductive reactance XL. It
also has units of Ohms.
The phase relationship of the three elements is summed up in the
following diagram, Figure 5.
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Figure 5
Note that in all three cases, resistor, capacitor, and inductor, the
relationship between the maximum voltage and the maximum current
was of the form
Vmax = ImaxZ . (4)
We call Z the impedance of the circuit element. Equation (4) is justan extension of Ohm's Law to AC circuits. For circuits containing
any combination of circuit elements, we can define a unique
equivalent impedance and phase angle that will allow us to find the
current leaving the battery. We show how to do so in the next
section.
Phasors and AC Circuit Problems
Phasors reduce AC Circuit problems to simple, if often tedious,
vector addition and subtraction problems and provide a nice graphical
way of thinking of the solution. In these problems, a power supply is
connected to a circuit containing some combination of resistors,
capacitors, and inductors. It is common for the characteristics of the power supply, Vmax and frequency ω, to be given. The unknown
quantity would be the characteristics of the current leaving the power
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supply, Imax and the phase angle φ relative to the power supply. To
solve one needs only to follow the rules:
Circuit elements in parallel share the same voltage.1.
Circuit elements in series share the same current.2.
Do one branch of the circuit at a time.3.Maintain the phase relationships given in Figure 5.4.
Use Ohm's Law V = IZ where Z is the equivalent impedance of
any combination of circuit elements being considered.
5.
Consider the following example.
The emf for the circuit in Figure 6 is ε = 10sin(1000t). Find the current delivered to the circuit. Find the
equivalent impedance of the circuit. Find the equation
of the current and voltage drop for each element of the
circuit.
Figure 6
Step 1. Determine the reactances and assign a current to each
branch of the circuit.
First we determine the reactances:
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XC = 1/ωC = 1/[1000 rad/s× 10 µF] = 100 Ω ,
and
XL = ωL = 1000 rad/s× 40 mH = 40 Ω .
Next we assign a current to each branch of the circuit.
Figure 7
Step 2. Use the phase relationships to determine the impedance of
the circuit one piece at a time.
The branch with current I2 is easy. Since current and voltage drop arein phase for resistors, we have
V2 = (20 Ω)I2 .
The branch carrying I1 needs more work. Since the current is
common we draw a diagram that indicates the appropriate phase
relationships. We need to find the equivalent impedance Z1 and the phase angle φ1 that we can use to replace the capacitor/resistor
combination.
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Figure 8
Using the Pythagorean Theorem and trigonometry, we find the
impedance of the branch
and the phase angle
As we see from Figure 8, the current I1 leads V1 by 63.4°.
Now these two branches containing the 20 Ω resistor and Z1 are in
parallel, that is V1 = V2 = V. Since the voltage is common we draw a
diagram like the following to find the equivalent impedance Z12 and
phase angle φ12 .
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Figure 9
To do the vector addition, we will treat the voltage vector as the
x-axis. Then
and
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Thus
Hence the equivalent impedance of the two arms together is Z12 =
18.319 Ω. The phase angle is
As we see from Figure 9, the current leads the voltage.
Next the current I12 equals I and this current passes through the 100
Ω resistor, the impedance Z12, and XL. Figure 10 shows the
appropriate diagram for determining the total circuit's equivalent
impedance Zeq and phase angle φf.
Figure 10
To do the vector addition, we will treat the current vector as the
x-axis. Then
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and
Thus
Hence the equivalent impedance of the circuit together is Zeq = 123.9
Ω. The phase angle is
As can be seen from Figure 10, the voltage leads the current. Since
εmax = 10 Volts, we have Imax = εmax/Z = 10/123.875 A = 80.73 mA.The requested equation for the current is
I = (80.73 mA) sin(ωt − 17.53°) .
Step 3. Find the current or voltage for each piece using the phaserelationship and Ohm's Law.
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The battery current passes through the 100 Ω resistor and the
inductor. Using Ohm's Law, the voltage drop across the 100 Ω resistor
is
V100 = IR = (8.073 V) sin(ωt − 17.53°) .
For the inductor
VLmax = ImaxXL = 80.73 mA × 40 Ω = 3.229 Volts .
The phase relation between VL and I yields
VL = VLmax sin(ωt − 17.53° + 90°) = (3.229 V) sin(ωt + 72.47°) .
The maximum voltage drop across Z12 is
Vmax = ImaxZ12 = 80.73 mA × 18.319 Ω = 1.479 Volts.
Since the voltage lags I by φ12, we find
V = Vmax sin(ωt − 17.53° - φ12) = (1.479 V) sin(ωt − 25.96°) .
From here on we reverse the steps we took to find Z12 in the first
place.
By definition, the voltage drop across Z12 is also the voltage across
the 20 Ω resistor. The maximum current through the resistor will be
I2max = Vmax/R = 1.479 V / 20 Ω = 73.95 mA .
The equation for this current is
I2 = (73.95 mA) sin(ωt − 25.96°) .
The voltage V is also the potential drop across Z1. The maximum
current in this branch is
I1max = Vmax/Z1 = 1.479 V / 111.803 Ω = 13.23 mA .
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Recalling the phase information we derived for Z1, the current
formula will be
I1 = I1max sin(ωt − 25.96° + f 1 = (13.23 mA) sin(ωt + 37.48°) .
This in turn is the current through the capacitor and 50 Ω resistor. Themaximum voltage drop over the capacitor is
VCmax = I2maxXC = 13.23 mA × 100 &Ohms; = 1.323 Volts .
We know that VC must lag I1 by 90°. Hence the equation for the
voltage will be
VC = VCmax sin(ωt + 37.48° - 90°) = (1.323 V) sin(ωt − 52.52°) .
Finally the maximum voltage drop over the 50 Ω resistor will be
V50max = I2maxR 50 = 13.23 mA × 50 &Ohms; = 0.661 Volts.
Current and voltage are in phase for a resistor, so the equation will be
V50 = (0.661 V) sin(ωt + 37.48°) .
Summarizing
Element Voltage Current
Power Supply (10 V) sin(ωt) (80.73
mA) sin(ωt −
17.53°)
100 Ω Resistor (8.073 V) sin(ωt −
17.53°)
(80.73
mA) sin(ωt −17.53°)
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40 mH Inductor (3.229 V) sin(ωt +
72.47°)
(80.73
mA) sin(ωt −
17.53°)
Z12 (1.479 V) sin(ωt −25.96°)
(80.73mA) sin(ωt −
17.53°)
20 Ω Resistor (1.479 V) sin(ωt −
25.96°)
(73.94 mA)
sin(ωt −
25.96°)
Z1 (1.479 V) sin(ωt −
25.96°)
(13.23 mA)
sin(ωt +
37.48°)
10 µF Capacitor (1.323 V) sin(ωt −
52.52°)
(13.23 mA)
sin(ωt +
37.48°)
50 Ω Resistor (0.661 V) sin(ωt +
37.48°)
(13.23 mA)
sin(ωt +
37.48°)
Questions? [email protected]
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