Phasor Relationships;
Impedance
Dr. Holbert
April 2, 2008
Lect17
EEE 202
1
Introduction
• Any steady-state voltage or current in a
linear circuit with a sinusoidal source is
also a sinusoid
– This is a consequence of the nature of
particular solutions for sinusoidal forcing
functions
– All steady-state voltages and currents
have the same frequency as the source
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Introduction (cont.)
• In order to find a steady-state voltage or
current, all we need to know is its
magnitude and its phase relative to the
source (we already know its frequency)
• Usually, an AC steady-state voltage or
current is given by the particular solution
to a differential equation
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The Good News!
• We do not have to find this differential
equation from the circuit, nor do we have
to solve it
• Instead, we use the concepts of phasors
and complex impedances
• Phasors and complex impedances convert
problems involving differential equations
into simple circuit analysis problems
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Phasors
• Recall that a phasor is a complex number
that represents the magnitude and phase
of a sinusoidal voltage or current
x(t) = XM cos(ωt+θ) ↔ X = XM θ
Time domain
Frequency Domain
• For AC steady-state analysis, this is all we
need---we already know the frequency of
any voltage or current
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Impedance
• AC steady-state analysis using phasors
allows us to express the relationship
between current and voltage using a
formula that looks like Ohm’s law:
V=IZ
• Z is called impedance (units of ohms, W)
• Impedance is (often) a complex number,
but is not technically a phasor
• Impedance depends on frequency, ω
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EEE 202
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Phasor Relationships for Circuit
Elements
• Phasors allow us to express currentvoltage relationships for inductors and
capacitors much like we express the
current-voltage relationship for a resistor
• A complex exponential is the mathematical
tool needed to obtain this relationship
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EEE 202
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I-V Relationship for a Resistor
v(t )  R i (t )
+
i(t)
v(t)
R
–
VI R
V  I Z  ZR  R
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I-V Relationship for a Capacitor
+
i(t)
v(t)
C
–
V  I Z  ZC 
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EEE 202
dv(t )
i (t )  C
dt
1
VI
jC
1
j C
9
I-V Relationship for an Inductor
+
i(t)
v(t)
L
–
di (t )
v(t )  L
dt
V  j L I
V  I Z  Z L  j L
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Impedance Summary
Lect17
Element
Impedance
Capacitor
ZC = 1 / jC = –1/C  90
Inductor
ZL = jL = L  90
Resistor
ZR = R = R  0
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Class Examples
• Drill Problems P8-4, P8-7, P8-5 (and P8-1,
if time permits)
• Remember: sin(ωt) = cos(ωt–90°)
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