52. (33.1) Phasors and the complex representation of
electrical quantities.
Complex functions of time and rotating vectors (called
phasors) in the complex plane representing these numbers
are convenient representations of oscillating quantities.
a) Voltage.
The complex function V(t) such
that the voltage across the
element is
V( t ) = Im V ( t )
is called the complex voltage.
Im V
Vm
V(t)
V (t)
ωt+δ V
Example. Sinusoidal voltage is
represented by an exponential
complex function
(
)
V ( t ) = Vm e iδ V e iωt = V0 ⋅ e iωt
(because Im V ( t ) = Vm sin(ω t + δ V ) )
V0 is called the complex amplitude of the voltage. Note
that Vm = V0 .
The magnitude of the voltage phasor is equal to the
peak value Vm of the voltage. The phase of the voltage is
equal to the angle between the voltage phasor and the
x-axis; the instantaneous value of the voltage is equal to the
y-component of the phasor.
61
Re V
b) Current
The complex function
I(t) such that the current
across the element is
I( t ) = Im I ( t )
is called the complex current.
Im I
I(t)
Im
I(t)
ωt+δ I
Re I
Example.
Sinusoidal current is
represented by a complex
exponential function
I( t ) = (I m e iδ I )e iω t = I0 ⋅ e iω t
(because
Im I( t ) = I m sin (ω t + δ I ) )
I0 is called the complex amplitude of the current. Note
that I m = I0 .
The magnitude of the current phasor is equal to the
peak value Im of the current. The phase of the current is
equal to the angle between the current phasor and the
x-axis; the instantaneous value of the current is equal to the
y-component of the current phasor.
62
53. (33.5) The relation between voltage and current.
a) The coefficient Zω relating the peak values of the
voltage across the system with the peak value of the current
through the system is called the impedance of the system:
Vm = Zω ⋅ I m
(note. Vrms = Z ω ⋅ I rms )
b) The number ϕ ω relating the phase of the voltage across
the system with the phase of the current through the system
is called the phase angle between the current and the
voltage:
(ω t + δ V ) = (ω t + δ I ) + ϕ ω
c) The complex coefficient Zω, relating the complex
voltage across an element with the complex current through
this element, is called the complex impedance of this
element at frequency ω :
Vω (t ) = Iω (t ) ⋅ Zω or V0 ,ω = I0 ,ω ⋅ Zω
Complex impedance Z includes information about both the
impedance Z, relating the amplitudes of voltage and current
Z= Z
63
and the phase angle between the voltage and current
tan ϕ =
Im Z
Re Z
Explanation.
V

V = (Vmeiδ V )eiω t =  m eiϕ  ⋅ (I meiδ I )eiω t = Z ⋅ I
 Im

Z=
Vm i( δ V − δ I )
e
Im
Z=
Vm
= Z
Im
Im Z Z sin ϕ
=
= tan ϕ
Re Z Z cos ϕ
where ϕ = δ V − δ I
In summary
Z = Ze iϕ
64
Example 1. (33.2) A resistor and an alternating current.
I
a
ε
V(t ) = Va − Vb = I(t ) ⋅ R
R
For a sinusoidal current:
b
I( t ) = I m sin (ω t + δ I )
From Ohm's law the following function represents the
voltage across the resistor
V(t ) = I (t ) ⋅ R = (I m ⋅ R )sin (ω t + δ I )
Vm = Im R and δ V = δ I
The impedance of a resistor is equal to its resistance:
ZR (ω ) = R
The phase angle for a resistor is:
ϕR =
0
The average power dissipated in the resistor
Pav = I rms ⋅ Vrms ⋅ cos0
65
= I2rmsR =
2
Vrms
R
Complex analysis
Note that (because α ⋅ Im f ( t ) = Im (α ⋅ f ( t )) for real α )
Ohm's law relates also the complex voltage and the
complex current
V (t ) = I (t ) ⋅ R
Therefore the complex impedance of a resistor is also equal
to the resistance of the resistor
Z R (ω ) = R
Im
I
V
V
I
ZR
t
Re
66
Example 2. (33.3) An inductor and an alternating current.
I
V(t ) = Va − Vb = L
dI
dt
a
ε
For a sinusoidal current:
L
b
I( t ) = I m sin (ω t + δ I )
We obtain that the voltage across the inductor is

π
V( t ) = (I m ⋅ ω L) cos(ω t + δ I ) = (I m ⋅ ω L) sin  ω t + δ I + 

2
Therefore
Vm = I m ⋅ ω L
and
δV = δI +
π
2
(Coefficient X L = ω L is called inductive reactance.)
The impedance of an inductor is equal to its
(inductive) reactance:
Z L (ω ) = ω L
The phase angle for an inductor is: φ L =
π
2
The average power dissipated in the inductor
Pav = Irms ⋅ Vrms ⋅ cos π = 0
2
67
Complex analysis
Because
d
 d

Im f (t ) = Im f (t ) , the complex voltage is related to
dt
 dt

the complex current:
dI
V (t ) = L
dt
For AC the complex current is represented by the following
complex function of time:
I( t ) = I0 e iω t
The complex voltage is therefore
V ( t ) = iω L ⋅ I0 eiω t = iω L ⋅ I( t )
The complex impedance of an inductor is therefore
Z L (ω ) = iω L
Note that using this complex impedance we can quickly
find the impedance and the phase angle:
Im Z L
ωL π
Z L = Z L = ω L ; ϕ = arctan
= arctan
=
Re Z L
0
2
Im
V
V
ZL I
I
t
Re
68
I
Example 3. (33.4) A capacitor and
an alternating current.
a
I( t ) =
dQ
dV
= C⋅
dt
dt
Q
ε
C
-Q
b
For a sinusoidal current:
V(t ) = Vm sin (ω t + δ V )
From the above we can find the function representing
current

I( t ) = ω CVm cos (ω t + δ V ) = ω CVm cos ω t + δ V +

π

2
Therefore
1
π
and δ V = δ I −
ωC
2
1
The coefficient XC =
is called capacitive reactance
ωC
Vm = I m ⋅
The impedance of a capacitor is equal to its
(capacitive) reactance:
1
ZC (ω ) =
ωC
The phase angle for a capacitor is: ϕ C = −
π
2
The average power dissipated in the capacitor
Pav = I rms ⋅ Vrms ⋅ cos
69
π
2
=
0
Complex analysis.
The complex voltage is related to the complex current
according to the same expression and the voltage and
current.
The complex voltage is an exponential function of
time
V (t ) = V0 eiω t
therefore
dQ
d
I( t ) =
= C⋅
V0 eiω t = ω C ⋅ V0 eiω t = iω C ⋅ V ( t )
dt
dt
1
or V (t ) =
⋅ I (t )
iω C
The complex impedance of a capacitor is therefore
1
Z C (ω ) =
iω C
Note that using this complex impedance we can quickly
find the impedance and the phase angle:
1
−
Im Z C
1
π
ZC = ZC =
;
ϕ = arctan
= arctan ω C = −
ωC
Re ZC
0
2
Im
V
I
t
Re
ZC
I
V
70
54. Complex impedance of a system of elements
a) Parallel connection
Because Im(Z 1 + Z 2 + .. + Z n ) = Im Z 1 + Im Z 2 + ... + Im Z n ,
in a parallel connection, the total complex current through a
system of elements is equal to the sum of the complex
currents through each element.
 1
V
V
V
1
1


I = I1 + I3 + ... + In = +
+ ... +
= V
+
+ ... +
Z1 Z 2
Zn
Z n 
 Z1 Z2
Therefore the inverse of the equivalent impedance is equal
to the sum of the inverses of impedance of individual
elements.
1 1 1
1
=
+
+ ...+

Z eq  Z1 Z 2
Zn 
b) Series connection
Again, using the same property of complex numbers
for series connection, we can express the equivalent
impedance of the system in terms of the impedance of
individual elements.
V = V1 + V2 + ...+ Vn = (Z 1 + Z2 + ...+ Z n )I
The equivalent impedance is equal to the sum of the
impedance of individual elements.
Z eq = Z1 + Z2 + ...+ Z n
71
Example 1. In a parallel LC circuit relate the currents to the
(sinusoidal) voltage.
I
a
IC
IL
Q
ε
C
1
L
2
-Q
b
For a sinusoidal AC:
From Kirchhoff’s rules
1
I
iω C C
1
I
iω C C
− IC
=
-V
− iω LIL
=
0
− IL
=
0
−
I
V ( t ) = V0 ⋅ e iω t
We can use Cramer’s method to solve for the complex
currents
0
D= 0
1
0
DC = 0
1
1
iω C
1
iω C
−1
−
−
V
0
0
−
0
− iω L =
−1
1 0
⋅
iω C 1
0
− iω L =
−1
V⋅
0
1
− iω L
− iω L
−1
−1
=
1
L
⋅ − iω L =
iω C
C
V ⋅ iω L
−
DV = 0
0
0
=
1
iω C
1
iω C
−1
V
DL
=
0
1
72
1
iω C
1
iω C
−1
−
0
− iω L = −
−1
−
1
V ⋅ iω C
1
− iω L
=
−1
V
0
0
=−
1
V ⋅ 0 iω C
1 −1
=
V⋅ 1
iω C
V ⋅ 
1
 iω C

+ iω L

The complex currents are
I=
DV 
1 
=  ωC −
 i⋅ V
D 
ω L
IC =
DC
= iω C ⋅ V
D
IL =
DL
i
= −
⋅ V
D
ωL
Therefore the current through the LC system is
1  iω t 

I( t ) = Im I = Im  V0  ω L −
i⋅ e  =
ωC
 

1
π


V
⋅
ω
L
−
sin
ω
t
+
δ
+


V
 m


ωC
2
= 
1
π
 Vm ⋅ ω L −
sin ω t + δ V − 


ωC
2
for
XL > XC
for
X L < XC
The current "through" the capacitor is

π
I C ( t ) = Im IC ( t ) = Im( iω C ⋅ V0 e iω t ) = ( Vm ⋅ ω C ) sin  ω t + δ V + 

2
The current through the inductor is
Vm
−i
π

iω t 
I L ( t) = Im I L ( t ) = Im
⋅ V0 e  =
⋅ sin ω t + δ V − 
 ωL


ωL
2
73
The complex impedance of the LC system satisfies
1
1
1 

=
+ iω C = i ω C −


Z eq iω L
ω L
The impedance of the LC system is therefore:
1
1
Z=
−
XL XC
The phase angle for the system is:
−1
 − π2
ϕ =  π
 2
for XC < X L
for XC > X L
Note that at angular frequency
1
LC
the impedance of the LC system is infinite. The current
flows only in the LC loop!!
ω =
74
Example 2. (33.8) High-pass filter
C
OUT
IN
ε
Z=∞
R
Using the definition of complex impedance, we can relate
the output voltage with the input voltage
Vout = I ⋅ R =
Vin
1
+R
iω C
⋅R =
1
1
1−
iω CR
⋅
Vin
For a constant input (ω = 0 ) the output has a zero
value. As the frequency increases, the peak value of the
output voltage approaches the peak value of the input
voltage. For ω = ∞ , Vout = Vin .
Notice that the output phase is slightly different than
the input phase
1
tan α =
ω CR
75
Example 2. (33.8) Low-pass filter
R
OUT
IN
ε
C
Z=∞
Using the definition of complex impedance, we can relate
the output voltage with the input voltage
Vout = I ⋅
1
Vin
1
1
=
⋅
=
⋅ Vin
1
iω C R + ⋅
iω C 1 + iω CR
iω C
For a constant input (ω = 0 ) the output voltage is
equal to the input voltage. As the frequency increases, the
peak value of the output voltage decreases. For ω = ∞ ,
Vout = 0 V.
Notice that the output phase is slightly different than
the input phase.
tan α = − ω CR
76
55. (32.5, 32.6) Oscillations in an RLC circuit
VR
VL
VC
-Q
Q
ε
I
When the switch is closed, the charge Q(t) on the capacitor
must satisfy the following differential equation resulting
from the loop rule
d 2Q
dQ Q
L 2 +R
+
= 0
dt C
dt
We solve this kind of complex differential equation by
assuming that the unknown (function) is an exponential
function
Q ( t ) = Q0 e α t ,
which leads to a polynomial equation (called the
characteristic equation) for α ,
1
Lα 2 + Rα + = 0 .
C
77
The solutions to the characteristic equation are
−R±
α± =
R2 − 4
2L
L
2
C = − R ±  R  − 1 .
 2 L
2L
LC
First consider a situation that R = 0. In this case the
solution to the characteristic equation is an imaginary
number
α = iω 0 .
where ω 0 = 1 , is called the resonant angular frequency
LC
of the circuit. The charge on the capacitor will oscillate
with this angular frequency
Q( t ) = Im Q( t ) = Im( Q0 e iω 0 t ) = Q m sin (ω 0 t + δ Q )
Im
Q
Q
α
(t
Re
α
78
ωt
 R
< 

 2L
2
1
.
LC
Without loss of generality, we can limit ourselves to one
solution only (+). The solution to the characteristic equation
is a complex number
Consider now a situation such that 0
α = −
<
R
+ iω d
2L
2
1  R
where ω d =
−
 , is called the angular frequency
LC  2 L
of the damped oscillator. In this case the "peak value" of
the oscillating charge decreases
Q( t ) = Im Q( t )
R
R

 
−
t
−
t
i
ω
t
2
L
d
2
L
 =  Qm e
 sin ω t + δ Q
= Im  Q 0 e
⋅e
 


 

(
)
Q
ImQ
ωt
ReQ
79
 R


 2L
2
1
, the solution to the
LC
characteristic equation is real.
Finally, when
≥
Q (t ) = Q 0 ⋅ ( A ⋅ e α + t + B ⋅ e α − t )
where α - < α + < 0 and A and B are constants dependent on
the initial conditions. The damping process is faster than the
oscillations. We say that the system is overdamped. The
smallest resistance at which overdampling occurs is called
the critical resistance.
ImQ
Q
ωt
Re
ImQ
Q
ωt
Re
80