Download Chapter 11

Fundamentals of
Electric Circuits
Chapter 11
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Overview
• The concept of power in an AC circuit.
• The difference between instantaneous power
and average power.
• The difference between resistive and reactive.
• Other forms of averaged measurements will
be covered
• Apparent power and complex power.
2
Instantaneous Power
• This is the power at any instant in time.
• It is the rate at which an element absorbs
power
• Consider the generalized case where the
voltage and current at the terminals of a
circuit are:
v t   Vm cos t  v  i t   I m cos t  i 
• Multiplying the two together, yields:
1
1
p  t   Vm I m cos  v  i   Vm I m cos  2t   v  i 
2
2
3
Instantaneous Power II
• Note that this has two components.
– One is constant, depending on the phase
difference between the voltage and current
– The second is sinusoidal with a frequency twice
that of the voltage and current.
• A sketch of the possible instantaneous
power is below.
1
1
p  t   Vm I m cos  v  i   Vm I m cos  2t   v  i 
2
2
4
Instantaneous Power III
• Note that the figure shows times where the
power goes negative.
• This is possible with circuit elements like
inductors or capacitors which can store and
release energy.
• Note also that instantaneous power is very
hard to measure as it is constantly changing.
• The more common power measured is
average power.
5
Average Power
• Average power is the instantaneous power
averaged over a period.
• It is given by:
T
1
P   p  t  dt
T0
• When evaluated, this returns the component
of instantaneous power that was constant.
• The time dependent part is a sinusoid and
thus averages to zero.
6
Average Power II
• In order to get the instantaneous power, you
need to work in the time domain.
• But for average power it is possible to work
in frequency domain.
• In this case, the average power is:
1
1
P  Re VI *  Vm I m cos v  i 
2
2
7
Resistive vs. Reactive
• Consider the case when θv= θi the voltage
and current are in phase and the circuit is
purely resistive:
1
1
1 2
P  Vm I m  I m2 R  I R
2
2
2
• When θv- θi = ±90°, the circuit absorbs no
power and is purely reactive
1
P  Vm I m cos 90  0
2
8
Example
• Find the average power supplied by the
source and the average power absorbed by
the resistor.
9
Example
• Calculate the average power absorbed by
each of the five elements.
10
Maximum Average Power
Transfer
• How to maximize power
delivered to a resistive
load.
• The maximum power
was transferred when
the load resistance
equaled the Thevenin
resistance of the supply
circuit.
11
Maximum Average Power
• In rectangular form, the Thevenin impedance
and load impedance are:
ZTh  RTh  jX Th
Z L  RL  jX L
• The current through the load is:
VTh
VTh
I

ZTh  Z L  RTh  jX Th    RL  jX L 
• The average power delivered to the load is:
2
P
VTh RL / 2
1 2
I RL 
2
2
2
 RTh  jX Th    RL  jX L 
12
Maximum Average Power II
• We want to find the peak in this function,
which means finding the derivative and
identifying where it goes to zero.
• This must be done for both RL and XL.
• For ∂P/∂XL
VTh RL  X Th  X L 
P

2
X L
 RTh  jX Th 2   RL  jX L 2 


2
• For ∂P/∂RL
V
P
  Th
RL
2
 RTh  RL    X Th  X L   2 RL  RTh  RL 
2
 RTh  jX Th 2   RL  jX L 2 


2
2
13
Maximum Average Power III
• Combining these one finds that XL= -XTh and
RL=RTh satisfy the requirements:
*
Z L  RL  jX L  RTh  jX Th  ZTh
• The load impedance must be equal to the
complex conjugate of the Thevenin
impedance.
• The maximum average power will be:
Pmax 
VTh
2
8RTh
14
Example
• Determine the load impedance Z L that
maximizes the average power. What is the
maximum power.
15
Effective Value
• When a time varying source is delivering
power to a resistive load, we often want to
know the effectiveness of that source on
delivering power.
• This value is the DC current that delivers the
same average power to a resistor as the
periodic current
• For a periodic current, the average power
absorbed is:
T
T
1
R
P   i 2 Rdt   i 2 dt
To
T o
16
Average Power
• For a DC current, the average power
absorbed is:
P  ieff2 R
• Equating these two and solving for the
effective DC current yields:
T
I eff
1 2

i dt
T 0
• The effective voltage is found in a similar
manner:
T
1 2
Veff 
v dt

T0
17
RMS
• The effective values for both current and
voltage take the form of the square root of
the average of the square of the periodic
signal.
• This is typically referred to as the root mean
square, or RMS value for short.
• This can be extended to any periodic
function:
T
X rms
1 2

x dt

T0
18
RMS II
• If the parameter of interest is a constant, the
RMS value will be that constant.
• The RMS value is applicable to any periodic
function, regardless of its shape.
• However, for a sinusoidal waveform, the RNS
value is related to the amplitude as follows:
Vrms 
Vm
2
• RMS power can be determined from either
RMS current or voltage:
PI
2
rms
2
Vrms
Vm2
R

R
2R
19
Example
• Determine the rms value of the current waveform in the
figure. If the current is passed through a
resistor,
2
find the average power absorbed by the resistor.
20
Apparent Power
v (t )  Vm cos(t   v ), i (t )  I m cos(t  i ), V  Vm  v , I  I m i
1
p  Vm I m cos( v  i )  Vrms I rms cos( v  i )  S cos( v  i )
2
p
S  Vrms I rms the apparent power, pf  cos( v  i ) 
S
• The product of RMS voltage and
current will be called apparent power.
• Note that this is modulated by the
phase difference:
cos v  i 
• This will be referred to as the power
factor
.
21
Apparent Power II
• Apparent power is measured in VA, to
distinguish it from the average or real power.
• The angle θv-θi is called the power factor
angle.
• This is equal to the angle of the load
impedance if V is the voltage across the load
and I is the current through it.
22
Power Factor
• The power factor can range from zero to
unity.
• Leading power factor means current leads
voltage, which implies a capacitive load.
Lagging power factor means current lags
voltage, which implies a inductive load.
• Purely reactive loads will have a power factor
of zero.
• Power factors affect the way utilities bill for
electricity.
23
Example
• Calculate the average power absorbed by
each of the five elements.
24
Complex Power
• The complex power, S, absorbed by an AC
load is:
1
S  VI *
2
• This may also be expressed in terms of the
RMS values and load impedance as:
2
1 * V I
V
*
2
*
rms
S  VI 
 VrmsIrms  I rms Z  *  Vrms I rms
2
Z
2 2
25
Complex Power II
• The complex power expressed in rectangular
form is:
2
S  I rms
 R  jX   P  jQ
• Where:
2
P  Re  S   I rms
R
2
Q  Im  S   I rms
X
• P is the average or real power
• Q depends on the load’s reactance and is
called reactive (or quadrature) power
26
Real and Reactive Power
• The real power is the only useful power. It is
measured in watts.
• The reactive power is a measure of the
energy exchange between the source and the
reactive load.
• It is measured in units of volt-ampere
reactive (VAR)
27
Summarizing Power
Complex Power  S  P  jQ  Vrms  I rms  *
 Vrms I rms   v  i 
Apparent Power  S  S  Vrms I rms  P 2  Q 2
Real Power  P  Re  S   S cos  v  i 
Reactive Power  Q  Im  S   S sin  v  i 
P
Power Factor   cos  v  i 
S
28
Conservation of AC Power
• Regardless of how circuit elements are
connected, the total complex power
delivered is equal to the total complex power
absorbed by the elements.
S  S1  S2  S3 
 SN
• The same is true of real and reactive power,
but not of apparent power.
29
Power Factor Correction
• Most domestic and industrial loads, such as
washing machines, air conditioners, and
induction motors are inductive.
• They have a low, lagging power factor.
• The load cannot be changed, but the power
factor can be increased without altering the
voltage or current to the original load.
• This is referred to as power factor correction.
30
Adding a Capacitor
• To mitigate the inductive aspect of
the load, a capacitor is added in
parallel with the load.
• Looking at the phasor diagram,
showing before and after adding
the capacitor, the power factor has
improved.
31
Adding a Capacitor II
• With the same supplied voltage, the current
draw is less by adding the capacitor.
• Since power companies charge more for
larger currents because it leads to larger
power losses.
• Overall, the power factor correction benefits
the power company and the consumer.
• By choosing a suitable size for the capacitor,
the power factor can be made to be unity.
32
Adding a Capacitor III
• The capacitor needed in order to shift the
power factor angle from θ1 to θ2 is:
C
P  tan 1  tan  2 
QC

2
2
Vrms
Vrms
• Note that the real power dissipated in the
load is not affected by the shunt capacitor.
• Although it is not as common, if a load is
capacitive in nature, the same treatment with
an inductor can be used.
33
Example
• When connected to a 120-V(rms), 60-Hz
power line, a load absorbs 4KWat a lagging
power factor of 0.8. Find the value of
capacitance nexessary to raise the pf to 0.95.
pf  0.8  cos 1  0.8  1  36.87 o
S1 
p
4000

 5000VA
cos 1
0.8
Q1  S1 sin 1  5000  sin 36.87o  3000VAR
pf  0.8  0.95   2  18.19o
S2 
p
4000

 4210.5VA
cos  2
0.95
Q2  S 2 sin  2  1314.4VAR
QC  Q1  Q2  3000  1314.4  1685.6
C 
QC
1685.6

 310.5 F
2
Vrms
2  60  1202
34
Wattmeter
• Power consumption in a AC system
can be measured using a
Wattmeter.
• The meter consists of two coils; the
current and voltage coils.
• The current coil is designed with
low impedance and is connected in
series with the load.
• The voltage coil is designed with
very large impedance and is
connected in parallel with the load.
35