Transcript Document

Engineering Circuit Analysis
CH5 AC circuit power analysis
5.1 Instantaneous Power
5.2 Average Power
5.3 Effectives values of Current ＆ Voltage
5.4 Apparent Power and Power Factor
Ch5 AC circuit power analysis
5.1 Instantaneous Power
Given instantaneous current i t  ＆ voltage v t  , the instantaneous power
delivered to any device is :
pt   it  vt 
If it is a resistor R 
vt   it  R
If it is an inductor L 
vt   L 
di t 
dt
v 2 t 
pt   i t   R 
R
2
di t 
p t   L  i t 
dt
If it is a capacitor C 
dv t 
i t   C 
dt
pt   C  vt 
dv t 
dt
Ch5 AC circuit power analysis
5.1 Instantaneous Power
The total amount of power supplied by the source = total amount of power
delivered to the circuit elements
V
 Rt
Known i t   0 1  e L 

R
Power supplied by the source
R
+
 Rt 
2
+
V0 1  e L 
LvL 
V0 
pt   V0  it   
R
V 1  e
pR t   i 2 t   R  
R
2
0
Power delivered to R:
Power delivered to L:
PL(t) = vL(t) ∙ iL(t)
 Rt
L


2
Ch5 AC circuit power analysis
5.1 Instantaneous Power
Since
vL t   L
V  Rt  R
di t 
 Rt
 L 0 e L 
 V0 e L
dt
R
L
 Rt 

V 1  e L  Rt



pL t   vL t   iL t  
e L
R
2
0
pt   pR t   pL t 
Ch5 AC circuit power analysis
5.1 Instantaneous Power
How if both i t  ＆v t  are sinusoidal signals?
vt   Vm cos wt
it   I m coswt   
 is the phase different between v(t) and i(t). E.g., in the 1st order RL circuit,
Im 
Vm
R 2  w2 L2
  arctg
wL
R
Vm I m
cos 2wt     cos  
2
V I
V I
 m m cos   m m cos 2wt   
2
2
pt   vt i t   Vm I m cos wt    cos wt 
Time invariant

T
0
twice of
w
Vm I m
cos 2 wt     0 , leading to Vm I m cos  possible to be the
2
2
averaged power?
Ch5 AC circuit power analysis
5.1 Instantaneous Power
• Example 8.1 Given a voltage source of 40 + 60μ(t) V, a 5 μF capacitor and
a 200 Ω resistor being connected in series. Find at t = 1.2ms, the powers
being absorbed by the resistor and capacitor respectively. If PS = PC + PR?
 
 
When t < 0 ,Vs = 40V, VC 0  VC 0  40V
C  5uF
)
60V
 0.3A
200
In this 1st order RC circuit   RC  5uF  200  1ms
Hence
PC
vS +
~-
When t = 0 , Vs = 100V , and VR  0  100V  40V  60V
PR
R  200
iO 
Hence , i(t )  iO (t )e

t


 0.3e
t

1ms
A
1.2ms
1ms
i(1.2ms)  0.3  e
 t90.36mA
t


vC (t )  100  (40  100)e   100  60e 1ms V
PR  i(1.2ms)  90.36mA2  200  1.633W
VC (1.2ms)  81.93V
PC (1.2ms)  vC (1.2ms)  iC (1.2ms)  7.403W
PS (1.2ms)  100V  i(1.2ms)  9.036W
PS  PC  PR
Ch5 AC circuit power analysis
5.2 Average Power
The average power (over time) for the instantaneous power pt 
1 t2
P
pt dt

t
t2  t1 1
The average power over time intervals
t1
to t 2
If pt  is periodic as pt   pt  T  , where T is the period
1
P
T

t T
t
pt dt
Validate the previous suggestion !
Ch5 AC circuit power analysis
5.2 Average Power
• Example : With vt   Vm cos wt and it   I m coswt    , the average
power for sinusoidal signals
Vm I m
Vm I m
pt  
cos  
cos 2wt   
2
2

It is a periodic signal with a period of T 
w
Vm I m
1 T Vm I m
[
cos 
cos2 wt   ]dt

0
T
2
2
V I
 m m cos
2
Phase different between
Validate the previous suggestion !
i
V
＆ v , or arctg
I
Ch5 AC circuit power analysis
5.2 Average Power
Two observations:
V I
-- power absorbed by an ideal resistor,   0 , P  m m
2
,
-- power absorbed by a purely reactive element,  

2
Example 8.2 (p.213)
Given V  400V and Z  2600 
Determine P and p(t).
w

6
or   

2
, P0
Ch5 AC circuit power analysis
5.2 Average Power
0

V
4

0
0
I  

2


60
A
0
Z 260
 t

it   2 cos wt  60 A  2 cos  600 A
6

t


v t  4 cos V
6

0




 
Vm I m
P
cos 00   60 0  4 cos 60 0 W  2W
2
pt   vt   it   4 cos
t
 t

 2 cos  600  W
6
6

 t

 2  4 cos  600  W
3

P
Ch5 AC circuit power analysis
5.2 Average Power
• Example 8.4 Determine the average power absorbed by each of the
passive elements and supplied by each of the sources.I
 j 2
j 2
1
Apply KVL, we can determine the currents are:
)
I1  5  j 10  11.18  63.43A
I 2  5  j  5  7.071  45A
The current that passes through the resistor is:
I1  I 2  5 j  5  90A
1
1
2
Hence , PR 
I m  R  5 2  2  25 W
2
2
Both L & C are reactive elements, and PL  PC  0
+
~-
200V
I1
2
I2
+
~-
100V
For the source on the left mesh
For the source on the right mesh
1
1
P

V2 m  I 2 m cos( 0  (45))
PSL  V1m  I1m cos(0  (63.43))
SR
2
2
1
1

 10  7.071  cos 45
  20 11.18 cos63.43
2
2
 25 W
 50W
The power of the left supplies 50W power and it is absorbed by the resistor and the
source on the right
Ch5 AC circuit power analysis
5.2 Average Power
Remark: superposition is applicable for evaluating the average power of a
non periodic function which can be decomposed into two periodic functions
with different periods, i.e.,
f t   f1 t   f 2 t 
1 T1
1 T2
P  P1  P2  0 f1 t dt  0 f 2 t dt
T1
T2
Ch5 AC circuit power analysis
5.2 Average Power
• Example 8.6 (p218)
Determine the average power delivered to the 4 Ω resistor by the current i1
= 2 cos10t – 3 cos20t A.
The two different frequency components of the current can be considered
separately.
i1 = 2 cos10t – 3 cos20t A
I mA
I mB
1
1
2
2
P  I mA   4  I mB   4
2
2
1 2
1 2
  2  4   3  4  8  18  26W
2
2
Ch5 AC circuit power analysis
5.3 Effectives values of Current ＆ Voltage
i t 
I eff
v t 
Veff
v 2 (t )
For a resistor, p (t )  i (t ) R 
R
2
PR 
I eff 
1
T

T
0
2
i 2 t   Rdt PR  I eff
R
1 T 2
i t dt

0
T
2
Veff
1 T v 2 t 
PR  0
dt PR 
T
R
R
Veff 
1 T 2
v t dt

0
T
- Represent the actual effect of a periodic function (current/voltage)
in term of delivering / absorbing power
- Transforms an AC circuit into an equivalent DC circuit
- Square root of the mean of the square (RMS)
Ch5 AC circuit power analysis
5.3 Effectives values of Current ＆ Voltage
,
RMS of the Sinusoidal wave.
Given it   I m coswt   
I eff
T
1
1

f w
2
1 T 2
2

I
cos
( wt   )dt
m
∫
0
T
2
1
w[
0
2
 Im
w
2
 Im
w 1 2 I m
 

2 2 w
2

1
 cos( 2 wt  2 )]dt
2
1
For a sinusoidal function, its RMS value is
of its magnitude
2
I
V
I eff  m
Veff  m
2
2
I V
1
Peff  Ieff  Veff  m  m  I mVm  P
2 2 2
Ch5 AC circuit power analysis
5.3 Effectives values of Current ＆ Voltage
Effective value of multiple-frequency circuit
i1 t  of 50Hz
i2 t  of 100Hz
Power can be determined using superposition


Peff  I12eff  R  I 22eff  R  I12eff  I 22eff  R
I eff  I12eff  I 22eff
In general, given a circuit with N frequencies,
2
I eff  I12eff  I 22eff   I Neff
Ch5 AC circuit power analysis
5.3 Effectives values of Current ＆ Voltage
•
Example
Given two sinusoidal currents with the frequencies
of 60Hz and 120Hz, respectively,
2
P2  ? I eff  ?
5A rms(60HZ)
3A rms(120HZ)
P2  (5A)2  2  (3A)2  2  68W
I eff 
5  3  5.831A
2
2
Ch5 AC circuit power analysis
5.4 Apparent Power and Power Factor
Given vt   Vm cost   
it   I m cost   
One can determine the average power delivered to the network is
1
P  I mVm cos    
2
or
P  I effVeff cos -  
However, the effective power delivered to the network would be
Papp  Ieff  Veff
Papp is defined as the apparent power –
P  Papp
Note that the unit of the Papp is VA.
The maximal value that the
average power can take.
Ch5 AC circuit power analysis
5.4 Apparent Power and Power Factor
Power factor ( PF )
average power
P
PF 

apparent power I eff  Veff
In the above case,
PF  cos   
If passive elements are all resistive,     0 PF  1
0
If passive elements are all reactive,     90 PF  0
,
If passive elements are neither purely resistive, nor purely reactive
 900      900
 900      00, capacitive load, since V lags behind I
00      900 , inductive load, since V leads I
,
a leading PF of
“—current is leading”
0  PF  1
a lagging PF of “—current is lagging”
Ch5 AC circuit power analysis
5.4 Apparent Power and Power Factor
Example 8.8(P222)
Determine the apparent power supplied by the source3, and the power factor
of the combined loads.
2  j1
+
-
Papp  Veff  I eff
1  j5
6000Vrms
V 600V

I

 12  53.13Arms
Z
3  j 4
Papp  Veff  I eff  60V 12A  720W
P  Veff  I eff cos(0  (53.13))  432W
PF 
+
-
6000Vrms
3  j 4
P 432

 0.6
Papp 720
(a lagging PF)
Note that for the impedance 3+j4Ω,only the
resistive component will absorb power
delivered by the source. ad
Pasb  I eff  3  122  3  432W
2