Transcript Phasor Relationships for Circuit Elements (8.4

Phasor Relationships for Circuit
Elements (8.4); Impedance and
Admittance (8.5)
Dr. Holbert
February 1, 2006
ECE201 Lect-5
1
Phasor Relationships for Circuit
Elements
• Phasors allow us to express current-voltage
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.
ECE201 Lect-5
2
I-V Relationship for a Resistor
+
i(t)
v(t)
R
v(t )  R i(t )
–
Suppose that i(t) is a sinusoid:
i(t) = IM ej(wt+q)
Find v(t)
ECE201 Lect-5
3
Computing the Voltage
v(t )  R i(t )  R I M e
v(t )  VM e
jwt  jq
jwt  jq
VI R
ECE201 Lect-5
4
Class Example
• Learning Extension E8.5
ECE201 Lect-5
5
I-V Relationship for a Capacitor
+
i(t)
v(t)
C
–
dv (t )
i (t )  C
dt
Suppose that v(t) is a sinusoid:
v(t) = VM ej(wt+q)
Find i(t)
ECE201 Lect-5
6
Computing the Current
jwt  jq
dv(t )
dVM e
i(t )  C
C
dt
dt
i(t )  jwCVM e
jwt  jq
ECE201 Lect-5
 jwCv(t )
7
Phasor Relationship
• Represent v(t) and i(t) as phasors:
V = VM  q
I = jwC V
• The derivative in the relationship between
v(t) and i(t) becomes a multiplication by jw
in the relationship between V and I.
ECE201 Lect-5
8
Example
v(t) = 120V cos(377t + 30)
C = 2mF
• What is V?
• What is I?
• What is i(t)?
ECE201 Lect-5
9
Class Example
• Learning Extension E8.7
ECE201 Lect-5
10
I-V Relationship for an Inductor
+
i(t)
v(t)
L
–
di (t )
v(t )  L
dt
V = jwL I
ECE201 Lect-5
11
Example
i(t) = 1mA cos(2p 9.15•107t + 30)
L = 1mH
• What is I?
• What is V?
• What is v(t)?
ECE201 Lect-5
12
Class Example
• Learning Extension E8.6
ECE201 Lect-5
13
Circuit Element Phasor Relations
(ELI and ICE man)
Element V/I Relation Phasor Relation
Phase
Capacitor I = C dV/dt I = j ω C V
I leads V
= ωCV 90° by 90º
Inductor V = L dI/dt V = j ω L I
V leads I
by 90º
= ωLI 90°
Resistor V = I R
V=RI
In-phase
= R I 0°
ECE201 Lect-5
14
Impedance
• AC steady-state analysis using phasors
allows us to express the relationship
between current and voltage using a
formula that looks likes Ohm’s law:
V=IZ
• Z is called impedance (units of ohms, W)
ECE201 Lect-5
15
Impedance
• Resistor:
V=IR
– The impedance is ZR = R
• Inductor:
V = I jwL
– The impedance is ZL = jwL
ECE201 Lect-5
16
Impedance
• Capacitor:
1
VI
jwC
– The impedance is ZC = 1/jwC
ECE201 Lect-5
17
Some Thoughts on Impedance
•
•
•
•
Impedance depends on the frequency, w2pf
Impedance is (often) a complex number.
Impedance is not a phasor (why?).
Impedance allows us to use the same
solution techniques for AC steady state as we
use for DC steady state.
ECE201 Lect-5
18
Impedance Example:
Single Loop Circuit
20kW
10V  0
+
–
1mF
+
–
VC
w = 377
Find VC
ECE201 Lect-5
19
Impedance Example
• How do we find VC?
• First compute impedances for resistor and
capacitor:
ZR = 20kW= 20kW  0
ZC = 1/j (377·1mF) = 2.65kW  -90
ECE201 Lect-5
20
Impedance Example
20kW  0
10V  0
+
–
+
VC
–
ECE201 Lect-5
2.65kW  -90
21
Impedance Example
Now use the voltage divider to find VC:
2.65kW - 90


VC  10V 0

 2.65kW - 90  20kW0 
VC  1.31V - 82.4
ECE201 Lect-5
22
Low Pass Filter:
A Single Node-pair Circuit
+
5mA  0
0.1mF
1kW
V
–
Find v(t) for w=2p 3000
ECE201 Lect-5
23
Find Impedances
+
-j530W
5mA  0
V
1kW
ECE201 Lect-5
–
24
Find the Equivalent Impedance
+
5mA  0
Zeq
V
–
Z eq
1000 j530)

1000 j530
ECE201 Lect-5
25
Parallel Impedances
1000 j530) 10 0  530  90


1000 j530
1132  27.9
3
Z eq
Zeq  468.2W  62.1
ECE201 Lect-5
26
Computing V
V  IZ eq  5mA0  468.2W  62.1
V  2.34 V  62.1
v(t )  2.34V cos(2p 3000t 62.1)
ECE201 Lect-5
27
Impedance Summary
Element
Impedance
Capacitor
ZC = 1 / jwC = -1/wC  90
Inductor
ZL = jwL = wL  90
Resistor
ZR = R = R  0
ECE201 Lect-5
28
Class Examples
• Learning Extension E8.8
• Learning Extension E8.9
ECE201 Lect-5
29