Phasors From a Differential Equations Point of View
by Prof. Townsend
MTH 352 Fall 2005
Original Differential Equation (2nd order, linear, homogeneous) for a series RLC circuit with
voltage source v ( t ) :
v ( t ) = vL ( t ) + vC ( t ) + vR ( t )
1
Q = v (t )
C
(2a)
1
i ( t ) dt = v ( t )
C∫
(2b)
LQ ''+ RQ '+
This equation is also written as
Assume the voltage is
Li '+ Ri +
(1)
v ( t ) = Vm sin ωt
(3)
The resulting current is assumed to be out of phase with the voltage so is written*
i ( t ) = I m sin (ωt − φ ) (4)
To avoid the mess resulting from using trig functions, recall
Euler’s identity
e jθ = cos θ + j sin θ
(5)
sin (θ ) = Im e jθ
(6)
v ( t ) = Vm e jωt
(7)
i ( t ) = I m e j (ωt −φ )
(8)
⎛
1 ⎞ − jφ
Substitute equations (7) and (8) into equation (2b): Vm = ⎜ jω L + R +
⎟ Ime
jωC ⎠
⎝
(9)
So
The voltage and current are then rewritten as
Define the impedance, Z:
So that
( )
⎛
1 ⎞
Z = e− jφ ⎜ jω L + R +
⎟
jωC ⎠
⎝
Vm = ZI m
(10)
(11)
Since Vm and I m are the maximum values of the voltage and current, respectively, they are real
hence Z must be real. First, we define the inductive and capacitive reactances so the formulas
are easier to handle:
1
(12a,b)
X L = ωL
XC =
ωC
This statement allows us to find the phase angle, φ:
Z = e − jφ ( R + j ( X L − X C ) )
(13)
The magnitude of Z is then
Z = R2 + ( X L − X C )
(14a)
The phase, φ, is found from
2
R + j ( X L − X C ) = e jφ = cos φ + j sin φ which gives
X L − XC
R
tan φ =
(14b)
Now let’s look at the individual element voltages:
Z = e − jφ ( jX L + R − jX C ) = e − jφ e jπ 2 X L + R + e − jπ 2 X C
(
)
So, in exponential form we have
vL = I m e j (ωt −φ ) ( jX L ) = I m e j (ωt −φ ) ( e jπ 2 X L ) = e j (ωt −φ +π 2) I m X L (15a)
vC = I m e j (ωt −φ ) ( − jX C ) = e j (ωt −φ −π 2) I m X C
(15b)
j ( ωt −φ )
vR = e
Im R
To return to reality, we take the Im of each of the above three equations.
vL = Im e j (ωt −φ +π 2) I m X L = I m X L sin ωt − φ + 900
vC
vR
And recall that
(
= Im ( e (
= Im ( e (
j ωt −φ −π 2 )
j ωt −φ )
Im X C
)
)
)=I
m
XC
(
)
sin (ωt − φ − 90 )
0
I m R = I m R sin (ωt − φ )
(15c)
(16a)
(16b)
(16c)
v = Vm sin (ωt )
(16d)
In other words the voltage drop across the inductor leads the current by 90º (π/2) and the
capacitor lags by 90º (π/2). The current and voltage drop across the resistor are in phase.
Note that all four equations (16) have sines in them. If you think of each of these equations as
representing the y component of a vector, then we can construct a vector diagram. This diagram
is called a phasor diagram. Note that, due to the ωt , the entire phasor diagram rotates around the
origin counterclockwise with frequency f =
vector form.
G G G G
v = vL + vC + vR
ω
. Also note that we can write the voltage law in
2π
vL
v
ωt
φ
i
vR
O
vC
_____________________________________________________________________________
*Cathay and Nasar, Basic Electrical Engineering, Schaum’s Outline Series, McGraw-Hill, 1984, p54.