Download Review: Lecture 8

Review: Lecture 8
v(t )  Vm cos(t   )

V  Vm e j
RZR
L  Z  jL
1
CZ
j
C
(1)
Time
domain
Results in time domain
(3)
Frequency
domain
Analysis in frequency domain:
nodal/meash, theorems…
(2)
The same frequency: t
Maximum power transfer
• Minimizing power losses in the process of
transmission and distribution
• Maximize the power delivered to a load
p  i 2 RL
i
2
 VTh 
 RL
p  i 2 RL  
R

R
L 
 Th
VTh
RTh  RL
Power of the elements
p(t )  v(t )i(t )
Resistor (R):
Internal energy
Capacitor (C):
Electric potential energy
dissipated
Stored
Source:
Electric potential energy
Stored
Inductor (L):
Magnetic energy
DC steady-state:
C and L are open and short circuits, respectively.
The electric/magnetic field of the elements does not vary.
No energy being transferred to/from C and L
Electric field moves charges
dw
dq
,i 
dq
dt
dw dw dq
p

 v i
dt dq dt
v
Lecture 9 AC Circuits
AC Power Analysis
Contents
•
•
•
•
•
•
•
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Correction
5
Instantaneous power
The instantaneously power, p(t)
p(t )  v(t ) i (t )  Vm I m cos ( t   v ) cos ( t   i )
1
1
 Vm I m cos ( v   i )  Vm I m cos (2 t   v   i )
2
2
Constant power
More energy absorbed by the elements
Sinusoidal power at 2t
Power is a period
function of time
T
Less energy absorbed by the
elements
energy absorbed by
the source
2

,T
p(t) > 0: power is absorbed by the circuit;
p(t) < 0: power is absorbed by the source.
T
2
6
Average Power
The average power, P, is the average of the
instantaneous power over one period.
1
P
T
Phase difference between
v(t) and i(t)
   v   i

T
0
p (t ) dt 
1
Vm I m cos ( v   i )
2
V  RI,   0

V  jLI,  
2
1

Vj
I,   C
2
Phase domain
•
•
•
•
P is not time dependent.
When θv = θi , it is a purely
resistive load case.
When θv– θi = ±90o, it is a
purely reactive load case.
P = 0 means that the
circuit absorbs no average
power.
7
Instantaneous and Average Power
Example 1
Calculate the instantaneous power and
average power absorbed by a passive linear
network if:
v(t )  80 cos (10 t  20)
i(t )  15 sin (10 t  60)
Answer:
385.7  600 cos( 20t  10 )W, 387.5W
8
Instantaneous and Average Power
Example 2
A current I  10  30 flows through an
impedance Z  20  22Ω . Find the average
power delivered to the impedance.
Solution:
V  IZ  200e
V I
10  200
P  m m cos( ) 
cos(8o  30 o )  927.2W
2
2
j 8o
Answer: 927.2W
9
Maximum Average Power Transfer
Z TH  RTH  j X TH
Z L  RL  j X L
The other part of
the circuit
The load
The maximum average power can be
transferred to the load if
XL = – XTH and RL = RTH
*
i.e. Z L  RL  jX L  RTH  jX TH  Z TH
Pmax 
If the load is purely real, then
RL 
VTH
2
8 RTH
2
2
RTH
 X TH
 ZTH
10
Maximum Average Power Transfer
•
Current flow through the load:
•
Voltage across the load:
IL 
VL 

VL
1
VTH
Z L  ZTH
ZL
VTH
Z L  ZTH
If the load is purely real, set
XL = 0, and then follow the steps.

1
1
P  Re Vm e j v I m e  j i  Vm I m cos( v   i )
2
2
I *L
•
Average power:

1
1 
ZL
1
*

P  Re VL I *L  Re 
VTH
VTH
*
2
2  Z L  ZTH 
Z L  ZTH  


1 
RL  jX L
 Re 
VTH
2  RL  RTH 2   X L  X TH 2
1
RL
2

VTH
2
2
2 RL  RTH    X L  X TH 
•
Maximum:
2




Pmax 
P
P

0
X L RL
P
RL  X L  X TH 

X L
RL  RTH 2   X L  X TH 2


2
Z L  ZTH  RL  jX L  RTH  jX TH  RL  RTH   j  X L  X TH 
Z L  ZTH *  RL  RTH   j  X L  X TH 
VTH
2
8 RTH
VTH  0  X L   X TH
2
P 1 RL  RTH    X L  X TH   2 RL RL  RTH 
2
2
2

VTH  0  RL  RTH
  X L  X TH 
2
RL 2
RL  RTH 2   X L  X TH 2
2

2

*
Z L  RL  jX L  RTH  jX TH  ZTH
Maximum Average Power Transfer
Example 3
For the circuit shown below, find the load impedance ZL that
absorbs the maximum average power. Calculate that maximum
average power.
Answer: 3.415 – j0.7317W, 1.429W
12
Solution (Example 3)
Z2
Z3
(1) VTh
Z4
Z1
Thevenin equavilent
Z2
 Z4 


Z1  Z 2
Z 4  Z 3 
  2
VTH  
Z

Z
Z

Z

Z

Z
3 
2
4
3 
 4
 1
108  j 4

8  j 4  5  j10
 3.90 - j 4.88
(2) ZTh
Z TH  Z 4 //( Z1  Z 2  Z 3 )  5 //(8  j 6)
Z3
 3.41 + j0.732
(3) Pmax
Z1
Z4
Pmax 
VTH
2
8RTH
3.9 2  4.882

 1.43 W
8  3.41
*
When Z L  ZTH
 3.41  j 0.732
Effective or RMS Value – voltage/current
The average power dissipated by R is given by:
1
P
T

T
0
R T 2
2
i Rdt   i dt  I eff R
T 0
2
AC
I eff 
1
T
DC equivalent

T
0
i 2 dt  I rms
root-mean-square
The rms value is a constant itself which
depending on the shape of the function i(t).
The effective of a periodic current is the dc current that delivers the same
average power to a resistor as the periodic current.
14
RMS: sinusoids and power
The rms value of a sinusoid i(t) = Imcos(wt) is
given by:
Relationship
between effective
and maximum values
I rms
I
 m
2
1
T
I rms 
 Im
1
T

T
0

T
I m cos 2 (t) dt
2
0
1  cos2t  dt 
2
Im
2
The average power can be written in terms of the
rms values:
1
Pave  Vm I m cos (θv  θi )  Vrms I rms cos (θv  θi )
2
sinusoids
Average power in
terms of ‘effective’
current and voltage
Note: If you express amplitude of a phasor source(s) in rms, then all the answer
as a result of this phasor source(s) must also be in rms value.
15
Apparent power and Power factor
• Apparent Power, S, is the product of the r.m.s. values of voltage and
current.
• It is measured in volt-amperes or VA to distinguish it from the average
or real power which is measured in watts.
Apparent Power, S
P  Vrms I rms cos (θv  θi )  S cos (θv  θi )
Real power
Effective
Power Factor, pf
• Power factor is the cosine of the phase difference between the
voltage and current. It is also the cosine of the angle of the load
impedance.
16
Apparent Power and Power Factor
Purely resistive
load (R)
Purely reactive
load (L or C)
Resistive and
reactive load
(R and L/C)
θv– θi = 0
pf = 1
θv– θi = ±90o
pf = 0
θv– θi > 0
θv– θi < 0
P/S = 1, all power are
consumed
P = 0, no real power
consumption
• Lagging - inductive
load
• Leading - capacitive
load
Current leads voltage
17
Complex Power
Complex power S is the product of the voltage and the complex
conjugate of the current:
P  Vrms I rms cos (θv  θi )  S cos (θv  θi )
V  Vm θv  2Vrms θv
1
S  V I   Vrms I rms  θv  θi 
2
I  I m θi  2 I rms θi
 Vrms I rms cos (θv  θi )  jVrms I rms sin (θv  θi )
P
•
•
•
•
Q
1
1
S  VI  I Z
Apparent power: S  S  P 2  Q 2 , in VA
2
2
P
R  jX 

I
Z

I
pf 
 cos( v   i )
Power factor:
S
 I R  jI X
Average power: P  Re(S)  S cos( v   i ) , in watts
delivered to a load, the only useful power.
Reactive power: Q  Im( S)  S sin( v  i ) , in VAR
exchange between the source and the reactive part
2
rms
2
rms
*
2
2
rms
2
rms
 Q = 0 for resistive loads (unity pf).
 Q < 0 for capacitive loads (leading pf).
 Q > 0 for inductive loads (lagging pf).
18
Complex power: power triangle
S  Vrms I rms cos (θv  θi )  jVrms I rms sin (θv  θi )
Q
P
1st quadrant:
Inductive load
Complex power
Reactive power
Current lags voltage
Power factor angle
  v  i
Current leads voltage
Real power/average power
Power Triangle
Impedance Triangle
Power Factor
4th quadrant:
capacitive load
in 2nd/3rd quadrants?
Possible, if RL<0: active circuits.
19
Conservation of AC Power
The complex, real, and reactive powers of the sources equal
the respective sums of the complex, real, and reactive
powers of the individual loads.
For parallel connection:
S
1
1
1
1
*
V I *  V ( I1  I 2* )  V I1*  V I 2*  S1  S2
2
2
2
2
The same results can be obtained for a series connection.
20
Application: power factor correction
Can we increase the power factor without altering the
voltage or current to the original load? And why?
Power factor correction is necessary for economic reason.
21
Power Factor Correction
Capacitor: i leads v by /2
Same supplied voltage
2<1: Increasing the
power factor
Qc  Q1 – Q2
|IL|> |I|: keep IL the same, reduce
the current drawn from the source.
2
 P( tan 1- tan  2 )  CV rms
Q1  S1 sin 1  P tan 1
C
P  S1 cos 1
without altering the real power
delivered to the load
Q1  S 2 sin  2  P tan  2
Qc
P( tan θ1  tan θ2 )

2
2
ωVrms
ω Vrms
22
Lecture 10 AC Circuits
Magnetically-Coupled Circuits