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].
Since a pen and paper drawing cannot be animated so easily, a
2D drawing
of a rotating vector shows the vector inscribed in the centre of a
circle
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
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
blue
vector is said to be leading the red vector or conversely the red
vector
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:
A_{x}
= 5cos(30°
+ 4cos(140°)
= 1.26595
A_{y} = 5sin(30° + 4sin(140°) = 5.07115 Thus and 


So the Pythagorean theorem and simple trigonometry produces the result
5.23sin(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
V_{R} = IR . (1)
For capacitors
V_{C} = q/C . (2)
For inductors
V_{L} = 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 I_{max}sin(ωt + φ) then V_{R} has the identical form V_{max}sin(ωt + φ) where V_{max} = I_{max}R. For capacitors and inductors it is a little more complicated. Consider the capacitor. Imagine the current entering the capacitor has the form I_{max}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 = dq/dt. Since we know the form of I simple calculus tells us that q should have the form − (I_{max}/ω)cos(ωt + φ) or (I_{max}/ω)sin(ωt + φ  90°). Thus V_{C} has the form (I_{max}/ωC)sin(ωt + φ  90°) = V_{max}sin(ωt + φ  90°). The capacitor current leads the capacitor voltage by 90°. Also note that V_{max} = I_{max}/ωC. The quantity 1/ωC is called the capacitive reactance X_{C} and has the unit of Ohms. For the inductor, we again assume that the current entering the capacitor has the form I_{max}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 ωLI_{max}cos(ωt + φ) or V_{max}sin(ωt + φ + 90°). The inductor current lags the inductor voltage. Here note that the quantity ωL is called the inductive reactance X_{L}. It also has units of Ohms.
The phase relationship of the three elements is summed up in
the following
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
V_{max} = I_{max}Z . (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
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, V_{max} and frequency ω, to be given. The unknown quantity would be the characteristics of the current leaving the power supply, I_{max} 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.
X_{C} = 1/ωC = 1/[1000 rad/s× 10 μF] = 100 Ω ,
and
X_{L} = ωL = 1000 rad/s× 40 mH = 40 Ω .
Next we assign a current to each branch of the circuit.
V_{2} = (20 Ω)I_{2} .
The branch carrying I_{1} 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 Z_{1} and
the phase angle
φ_{1} that we can use to replace the
capacitor/resistor combination.
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 I_{1} leads V_{1} by 63.4°.
Now these two branches containing the 20 Ω
resistor and Z_{1 }are in parallel, that is V_{1}
= V_{2}
= V. Since the voltage is common we draw a diagram like the following
to
find the equivalent impedance Z_{12} and phase
angle φ_{12}
.
To do the vector addition, we will treat the voltage vector as the xaxis. Then
and
Thus
Hence the equivalent impedance of the two arms together is Z_{12} = 18.319 Ω. The phase angle is
As we see from Figure 9, the current leads the voltage.
Next the current I_{12} equals I and this
current passes through
the 100 Ω resistor, the impedance Z_{12},
and X_{L}. Figure 10 shows the appropriate diagram
for determining
the total circuit's equivalent impedance Z_{eq} and
phase angle
φf.
To do the vector addition, we will treat the current vector as the xaxis. Then
and
Thus
Hence the equivalent impedance of the circuit together is Z_{eq} = 123.9 Ω. The phase angle is
As can be seen from Figure 10, the voltage leads the current. Since ε_{max} = 10 Volts, we have I_{max} = ε_{max}/Z = 10/123.875 A = 80.73 mA. The requested equation for the current is
I = (80.73 mA)sin(ωt − 17.53°) .
V_{100} = IR = (8.073 V) sin(ωt − 17.53°) .
For the inductor
V_{Lmax} = I_{max}X_{L }= 80.73 mA × 40 Ω = 3.229 Volts .
The phase relation between V_{L} and I yields
V_{L} = V_{Lmax}sin(ωt − 17.53° + 90°) = (3.229 V) sin(ωt + 72.47°) .
The maximum voltage drop across Z_{12} is
V_{max} = I_{max}Z_{12} = 80.73 mA × 18.319 Ω = 1.479 Volts.
Since the voltage lags I by φ_{12}, we find
V = V_{max}sin(ωt − 17.53°  φ_{12}) = (1.479 V) sin(ωt − 25.96°) .
From here on we reverse the steps we took to find Z_{12}
in
the first place.
By definition, the voltage drop across Z_{12} is
also the voltage
across the 20 Ω resistor. The maximum current
through the resistor will be
I_{2max} = V_{max}/R = 1.479 V / 20 Ω = 73.95 mA .
The equation for this current is
I_{2} = (73.95 mA) sin(ωt − 25.96°) .
The voltage V is also the potential drop across Z_{1}. The maximum current in this branch is
I_{1max} = V_{max}/Z_{1} = 1.479 V / 111.803 Ω = 13.23 mA .
Recalling the phase information we derived for Z_{1}, the current formula will be
I_{1} = I_{1max} 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. The maximum voltage drop over the capacitor is
V_{Cmax} = I_{2max}X_{C} = 13.23 mA × 100 &Ohms; = 1.323 Volts .
We know that V_{C} must lag I_{1} by 90°. Hence the equation for the voltage will be
V_{C} = V_{Cmax}sin(ωt + 37.48°  90°) = (1.323 V) sin(ωt − 52.52°) .
Finally the maximum voltage drop over the 50 Ω resistor will be
V_{50max} = I_{2max}R_{50} = 13.23 mA × 50 &Ohms; = 0.661 Volts.
Current and voltage are in phase for a resistor, so the equation will be
V_{50} = (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°) 
40 mH Inductor  (3.229 V) sin(ωt + 72.47°)  (80.73 mA)sin(ωt − 17.53°) 
Z_{12}  (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°) 
Z_{1}  (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? mike.coombes@kwantlen.ca