Complex Impedance
Electric Circuits
Dr. Ayman Yousef
43
Complex Impedance
Ø The Impedance Z of a circuit is the ratio of phasor
voltage V to the phasor current I (Ohm’s Law) .which
is usually represented by complex number.
V
Z=
I
or V =ZI
(Ohm’s Law)
Ø This Impedance Z is usually a complex number and not
a sine wave.
Electric Circuits
Dr. Ayman Yousef
44
Complex Impedance
in case of Resistace ZR
Ø
In the Resistance circuit, the phasor current I is in phase with
phasor voltage V.
(current in phase with voltage)
Appling Ohm’s law:
Electric Circuits
Dr. Ayman Yousef
45
Complex Impedance
in case of Inductive reactance ZL
Ø
In the Inductance circuit, the phasor current I is LAGs voltage V
by 90o.
(current lags voltage by 90o)
Appling Ohm’s law:
Electric Circuits
Dr. Ayman Yousef
46
Complex Impedance
in case of Capacitive reactance Zc
Ø
In the Capacitive circuit, the phasor current I is LEADs
voltage V by 90o.
(current leads voltage by 90o)
Appling Ohm’s law:
Electric Circuits
Dr. Ayman Yousef
47
Impedance of Joint Elements
Ø
The Impedance Z represents the opposition of the circuit to the
flow of sinusoidal current.
V
Z = = R + jX =
I
=Resistance + j × Reactance
= Z ∠θ
Z = R +X
2
R = Z cos θ
2
X
θ = tan
R
X = Z sin θ
−1
Ø The Reactance is Inductive if X is positive and it is Capacitive
if X is negative.
Electric Circuits
Dr. Ayman Yousef
48
Series and Parallel
Circuits
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Dr. Ayman Yousef
49
Series Circuits
Ø
The Kirchoff”s Voltage Law (KVL) holds in the frequency domain. For
series connected impedances:
Electric Circuits
Dr. Ayman Yousef
50
Voltage Divider rule
Ø
The Voltage Division for two elements in series is:
Z1
V1 =
V
Z1 + Z2
Electric Circuits
Dr. Ayman Yousef
Z2
V2 =
V
Z1 + Z2
51
Admittance of Joint Elements
Ø
The Admittance Y represents the admittance of the circuit to
the flow of sinusoidal current. The admittance is measured in
Siemens (s)
Electric Circuits
Dr. Ayman Yousef
52
Ex.5: consider the circuit,
(a) Calculate the sinusoidal voltages v1 and v2
using phasors and voltage divider rule
(b) Sketch the phasor diagram showing E, V 1
and V2.
Solution
(a) the sinusoidal voltages v1 and v2 using phasors


40 − j80
(70.71∠0o ) = 78.4∠ − 33.69 o V
V1 = 
 (40 − j80) + (30 + j 40) 


30 + j 40
(70.71∠0o ) = 43.9∠82.87o V
V2 = 
 (40 − j80) + (30 + j 40) 
V1 =
Z1
V
Z1 + Z2
V2 =
Z2
V
Z1 + Z2
(b) Phasor diagram
Electric Circuits
Dr. Ayman Yousef
53
Parallel Circuits
Ø
The Kirchoff”s Current Law (KCL) holds in the frequency domain. For
Parallel connected impedances:
In case of two resistors
Electric Circuits
Dr. Ayman Yousef
54
Current Divider rule
Ø
The Current Division for two elements is:
Ø
The Current Division for more elements is:
Electric Circuits
Dr. Ayman Yousef
55
Ex. 6: Refer to the circuit;
(a) Find the total impedance, ZT.
(b) Determine the supply current, IT.
(c) Calculate I1, I2, using current divider rule.
(d) Verify Kirchhoff’s current law at node a.
Solution
a.
1
1
1
1
1
=
+ +
=
ZT 20 j1 − j1 20
b. I T =
ZT = 20∠0 kΩ
o
V
5∠0
=
= 250 x10 −6 ∠0 o A = 250∠0 o µA
3
ZT 20 x10 ∠0
ZR = 20∠0o kΩ
Z L = 1∠90o kΩ
ZC = 1∠ − 90o kΩ
56
Electric Circuits
Dr. Ayman Yousef
Delta to Star and
Star to Delta
Conversions
Electric Circuits
Dr. Ayman Yousef
57
Y-Δ and Δ -Y Equivalent Circuits
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Dr. Ayman Yousef
58
Y-Δ and Δ -Y Equivalent Circuits
ØY-Δ and Δ -Y type equivalent conversions will be useful when considering
Three Phase circuits.
Y − ∆ Conversion
Z1Z2 + Z2 Z3 + Z3 Z1
Za =
Z1
∆ − Y Conversion
Zb Zc
Z1 =
Za + Zb + Zc
Z1Z2 + Z2 Z3 + Z3 Z1
Zb =
Z2
Z2 =
Z1Z2 + Z2 Z3 + Z3 Z1
Zc =
Z3
Za Zc
Za + Zb + Zc
Z3 =
Za Zb
Za + Zb + Z59c
Electric Circuits
Dr. Ayman Yousef
Balanced Y-Δ and Δ -Y Equivalent Circuits
ØA delta (Δ ) or Y (wye) circuit is balanced if it has equal impedances
in all three branches.
ØY-Δ and Δ -Y conversions is very simple for balanced circuits.
Electric Circuits
Dr. Ayman Yousef
60
Ex. 7: Find the current I in the circuit.
Solution
j 4( 2 − j 4)
16 + j8
=
= 1.6 + j 0.8 Ω
j4 + 2 − j4 + 8
10
j 4(8)
8(2 − j 4)
=
= j 3 .2 Ω
Zcn =
= 1 .6 − j 3 .2 Ω
10
10
Zan =
Z bn
Electric Circuits
Dr. Ayman Yousef
61
Ex. 7: Find the current I in the circuit.
Electric Circuits
Dr. Ayman Yousef
62
AC Bridges
Electric Circuits
Dr. Ayman Yousef
63
AC Bridges
Ø
The AC bridge is Balanced when no current flows through the meter.
AC bridges are used in measuring inductance and capacitance values.
At balance case: Imeter = 0
AC bridge circuit
Electric Circuits
Dr. Ayman Yousef
64
AC Bridges
ØUnknown capacitance and inductances Cx and Lx are measured in terms
of the known standard values C s and Ls
AC Bridge for measuring L
AC Bridge for measuring C.
Electric Circuits
Dr. Ayman Yousef
65
Ex.8: Determine the series components that make up Zx
for balancing the bridge. Assume Z1 = 4.8 kΩ resistor, Z2
is 10 Ω in series with 0.25 μH, Z3 = 12 kΩ and f = 6 Mhz.
Solution
The series components are
25 Ω with a 0.625 µH
66
Resonance
in
AC Circuits
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Dr. Ayman Yousef
67
Resonance in AC Circuits
l
Resonance is a condition in an RLC circuit in
which the capacitive and reactive reactance are
equal in magnitude, the result is a purely resistive
impedance.
l
Resonance circuits are useful for constructing
filters and used in many application as signal
processing and communications systems .
Electric Circuits
Dr. Ayman Yousef
68
Series Resonant Circuit
l
l
l
l
At resonance, the impedance consists only resistive
component R.
The value of current will be maximum since the total
impedance is minimum.
The voltage and current are in phase.
Maximum power occurs at resonance since the power
factor is unity.
Electric Circuits
Dr. Ayman Yousef
69
Series Resonant Circuit
Total impedance of series RLC Circuit is
Z Total = R + jX L - jX C
ZTotal = R + j(X L - X C )
At Resonance:
The impedance now reduce to
The highest power is at ω o
The current at resonance
Im =
Vs
V
= m
ZTotal
R
2
V
P (ω o ) = I 2 R =
R
Electric Circuits
Dr. Ayman Yousef
70
Series Resonance
Resonance Frequency
Resonance frequency is the frequency where the condition of
resonance occur.
Im(z) = 0
1
ωL =
ωC
XL = XC
1
ωL −
=0
ωC
1
ωo L =
ωoC
rad/sec
Hz
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Dr. Ayman Yousef
71
Series Resonance
Half-power Frequency
ω
1
& ω
2
Half-power frequencies is the frequency when the magnitude of the output
voltage or current is decrease by the factor of 1 / √2 from its maximum value.
Also known as cutoff frequencies.
I
= 0.707 I
2
• The half-power frequencies ω1 and ω2
can be obtained by setting,
I=
Response curve
Electric Circuits
Dr. Ayman Yousef
72
Series Resonance
Maximum Power Dissipated
The maximum current at resonance where:
I=
Vm
R
Thus maximum power dissipated is:
The average power dissipated by the RLC circuit is:
Electric Circuits
Dr. Ayman Yousef
73
Series Resonance
Bandwidth of resonant circuit, β
Bandwidth, β is define as the difference between the two half power
frequencies. The width of the response curve is determine by the
bandwidth.
Resonance frequency can be obtained from
the half-power frequencies.
Response curve
Electric Circuits
Dr. Ayman Yousef
74
Quality Factor (Q-Factor)
Quality factor is used to measure the “sharpness”of response curve
Quality factor is the ratio of resonance frequency
to the bandwidth
Selectivity defines how well a resonant
circuit responds to certain frequencies.
l
l
Higher value of Q, smaller the bandwidth.
(Higher the selectivity)
Lower value of Q, larger the bandwidth.
Electric Circuits
(Lower the selectivity)
Dr. Ayman Yousef
75
High-Q
It is to be a high-Q circuit when its quality factor is equal or
greater than 10.
For a high-Q circuit (Q ≥ 10), the half-power frequencies
are, for all practical purposes, symmetrical around the
resonant frequency and can be approximated as
Electric Circuits
Dr. Ayman Yousef
76
Ex.9: For the circuit shown:
(a) Find the resonant frequency and the half
power frequencies
(b) Calculate the quality factor and bandwidth
(c) Determine the amplitude of the current
at ω o, ω 1 and ω 2
Solution
(a)
(b)
Amplitude
(c)
77
Parallel Resonant Circuit
The total admittance:
YTotal = Y1 + Y2 + Y3
YTotal
1
1
1
= +
+
R (jωL) (-j/ωC)
YTotal =
1 -j
+
+ jω C
R ωL
1
YTotal = + j(ω C − 1/ω L)
R
Resonance occur when:
l
l
l
At resonance, the impedance consists only conductance G.
The value of current will be minimum since the total admittance is minimum.
The voltage and current are in phase.
78
Electric Circuits
Dr. Ayman Yousef
Parameters in Parallel Circuit
Parallel resonant circuit has same parameters as the series
resonant circuit.
Resonance frequency
Half-power frequencies
Bandwidth
Q - Factor
79
Ex.10: For the circuit shown:
(a) Calculate ω o, Q and B
(b) Find ω 1 and ω 2
(c) Determine the power dissipated
at ω o, ω 1 and ω 2
Solution
(a)
(b)
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Dr. Ayman Yousef
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Electric Circuits
Dr. Ayman Yousef
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