In AC electrical theory every power source supplies a voltage
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
wave such as V(t) = Asin(ωt
is that it can be considered to be directly related to a vector of
A revolving in a circle with angular velocity ω -
in fact just the y component of the vector. The phase constant
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].
Since a pen and paper drawing cannot be animated so easily, a
of a rotating vector shows the vector inscribed in the centre of a
as indicated in Figure 2 below. The angular frequency ω
may or may not be indicated.
When two sine waves are produced on the same display, one wave
said to be leading or lagging
the other. This terminology
makes sense in the revolving vector picture as shown in Figure 3. The
vector is said to be leading the red vector or conversely the red
is lagging the blue vector.
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 ignore ω. For example consider
Asin(ωt + φ) = 5sin(ωt + 30°) + 4sin(ωt + 140°) ;
the corresponding vector addition is:
Ay = 5sin(30° + 4sin(140°) = 5.07115
So the Pythagorean theorem and simple trigonometry produces the result
5.23sin(ωt + 76.0°) .
Make a note not to forget to put ωt
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)
VC = q/C . (2)
VL = LdI/dt . (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 Imaxsin(ωt + φ) then VR has the identical form Vmaxsin(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 Imaxsin(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 = dq/dt. 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°) = Vmaxsin(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 reactance XC and has the unit of Ohms. For the inductor, we again assume that the current entering the capacitor has the form Imaxsin(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 Vmaxsin(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
diagram, 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
just an extension of Ohm's Law to AC circuits. For circuits
any combination of circuit elements, we can define a unique equivalent
impedance and phase angle that will allow us to find the current
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 supply, Imax and the phase angle φ relative to the power supply. To solve one needs only to follow the rules:
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.
XC = 1/ωC = 1/[1000 rad/s× 10 μF] = 100 Ω ,
XL = ωL = 1000 rad/s× 40 mH = 40 Ω .
Next we assign a current to each branch of the circuit.
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
We need to find the equivalent impedance Z1 and
the phase angle
φ1 that we can use to replace the
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
= V. Since the voltage is common we draw a diagram like the following
find the equivalent impedance Z12 and phase
To do the vector addition, we will treat the voltage vector as the x-axis. Then
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
the total circuit's equivalent impedance Zeq and
To do the vector addition, we will treat the current vector as the x-axis. Then
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°) .
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 = VLmaxsin(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 = Vmaxsin(ωt − 17.53° - φ12) = (1.479 V) sin(ωt − 25.96°) .
From here on we reverse the steps we took to find Z12
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 .
Recalling the phase information we derived for Z1, the current formula will be
I1 = I1max sin(ωt − 25.96° + f1 = (13.23 mA) sin(ωt + 37.48°) .
This in turn is the current through the capacitor and 50 Ω resistor. The maximum 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 = VCmaxsin(ωt + 37.48° - 90°) = (1.323 V) sin(ωt − 52.52°) .
Finally the maximum voltage drop over the 50 Ω resistor will be
V50max = I2maxR50 = 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°) .
|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°)|
|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°)||(80.73 mA)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°)|