Transcript ECE 3144 Lecture 4

ECE 3144 Lecture 34
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Chapter 9 Steady-state power analysis
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Instantaneous power.
Maximum average power transfer
Effective or rms (root mean square) values
The power factor
Complex power
Single phase three-wire circuits
Safety considerations
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Instantaneous power
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The instantaneous power is defined as product of instantaneous voltage across
the device and the instantaneous current through it. Thus p(t) =v(t)i(t).
Based on passive sign convention, if the product is positive, the device absorbs
power; if the product is negative, the device is supplying power.
Consider the circuit given, the general expressions for the steady state voltage
and current can be written as
v(t) = VMcos(t+v), i(t) = IMcos(t+i)
•
Z
The instantaneous power is
P(t) = v(t)i(t) = VMIMcos(t+v)cos(t+i)
= VMIM/2[cos(v- i) + cos(2 t+v +i)]
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p(t) consists of two parts: the first term is constant and the second term is
timing varying. We will investigate more details on these two terms later.
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Average Power
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The average power can be computed as follows:
t 0 T
1
P 
T
1

T
1

T

p (t ) dt
t0
t 0 T
V
M
I M cos(t   v ) cos(t   i )dt
t0
t 0 T

t0
VM I M
[cos( v   i )  cos( 2t   v   i )]dt
2
1
 VM I M cos( v   i )
2
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Notice (v-i) is the phase angle of the circuit impedance.
– For a purely resistive impedance, P=VMIM/2 =I2MR/2=V2M/2R
– For a purely reactive impedance, P= VMIMcos(90o)/2 =0, which means purely
reactive impedances (such as a capacitor or an inductor) absorbs no average power.
So they are also called lossless elements. They operate in a mode in which they
store energy over one part of the period and release it over another.
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Maximum Average Power transfer
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We have learned that in a resistive network, maximum power transfer is achieved if the
values of the load resistor is equal to the Thevenin equivalent resistance.
We will reexamine this issue with networks given as follows:
We know that the average power at the load is
1
(1)
PL  VL I L cos( vL   iL )
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The phasor voltage and phasor current at the load are:
I
ZTH
L
VL ZL
IL 
Voc
ZTH  Z L
VL  I L Z L 
Voc Z L
ZTH  Z L
Since both ZTH and ZL are complex numbers,
ZTH=RTH + jXTH and ZL= RL+jXL
Thus the magnitude of the phasor voltage and phasor current are
IL 
Voc
( RTH  RL ) 2  ( X TH  X L ) 2
VL 
(2)
Also we know that
cos( vL   iL )  cos( Z L ) 
Voc RL2  X L2
( RTH  RL )  ( X TH  X L )
2
(3)
2
RL
RL2  X L2
(4)
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Maximum Average Power transfer
Substituting equations (2)-(4) to equation (1) yields the average power as follows:
Voc2 RL
1
1 2
PL 

I L RL
2
2
2 ( RTH  RL )  ( X TH  X L )
2
(5)
In order to get the maximum average power, two steps are performed:
Step 1: the derivative of equation (5) with respect to XL is zero => XL=-XTH
Step 2: the derivative of equation (5) with respect to RL is zero with XL=-XTH. The problem reduces to
maximizing
Voc2 RL
1
PL 
2 ( RTH  RL ) 2
This is the same format we dealt with in the purely resistive network => The maximum transfer power
happens when RL = RTH. Therefore, the maximum average power transfer to the load is achieved when
ZL = RL +jXL = RTH –jXTH = Z*TH
Then the maximum average power transfer is
2
Voc
PL max 
8 RTH
If the load impedance can only be purely resistive (XL = 0), the condition for maximum average power
can be derived from
Voc2
1
dPL
 0 => R  R 2  X 2
and PL max  4 R  R
L
TH
TH
dRL
TH
L
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Effective values of current and voltage
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The effective value of a periodic waveform, representing either voltage or current, is defined as a
constant or dc value Ieff or Veff, which as current or voltage deliver the same average power to a
resistor R.
We know the average power delivered to R as a result of Ieff is
P = I2eff R
The average power delivered to the resistor by the periodic current i(t) is
1
P
T
T
R
i
(
t
)
R
dt

0
T
2
1
T
I eff 
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T
i
2
2
(t )dt  I eff
R
0
=>
T
i
2
(t ) dt
0
Thus the effective value is obtained by performing root mean square operations.
Therefore Ieff is also denotes as Irms.
Special case: effective (rms) value of a sinusoidal waveform
– i(t) = IMcos(t+i) and T=2/
I eff 
 IM
 IM
1
T

2
T
I
cos 2 (t   i )dt
0
2 / 

4
2
M

0
1 1
[  cos( 2t  2 i )]dt
2 2
2 / 
 dt
0

IM
2
I eff 
IM
2
Veff 
VM
2
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Average power expressed in effective values
• The average power written in effective values
of voltage and current is
– P = Veff Ieff cos(v-i)
• If the load is a resistor R
– P = I2effR = V2eff / R=I2MR/2=V2M/2R
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Examples
• Examples will be provided during the
lecture.
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Homework for lecture 34
• Problems 9.2, 9.7, 9.18, 9.31
• Due April 17
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