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Chapter 10
EGR 272 – Circuit Theory II
1
Read: Chapter 10 in Electric Circuits, 9th Edition by Nilsson
Chapter 10 - Power Calculations in AC Circuits
Instantaneous Power:
p(t) = v(t)i(t) = instantaneous power
Average Power:
• Instantaneous power is not commonly used.
• Average power, P, is more useful.
• P = PAVG = PDC = average or real power (in watts, W)
• P can be defined for any waveform as:
lim
P 
T
• For periodic waveforms with period T
(including sinusoids), P can be expressed as:
T
1
p(t)dt

2T -T
T
1
P   p(t)dt
T0
1
EGR 272 – Circuit Theory II
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Average power can be calculated:
1) by inspection (for simple cases)
2) by integration (for more complex cases)
Example: Find the average power absorbed to the resistor below.
i(t)
i(t) [A]
+
v(t)
10
10
4
_
t [s]
0
2
4
6
8
10
12
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Example: Find the average power absorbed to the resistor below.
v(t) [V]
i(t)
+
v(t)
_
20
10
0
t [s]
0
2
4
6
8
10
12
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RMS Voltage and Current
VRMS = root-mean-square voltage (also sometimes called Veff = effective
voltage)
VRMS = the square root of the average value of the function squared
T
VRMS 
1 2
v (t)dt

T0
T
I RMS 
1 2
i (t)dt

T0
Note how the name RMS essentially gives the definition.
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Calculating average power for a resistive load using RMS values
A key reason that RMS values are used commonly in AC circuits is that they
are used in power calculations that are very similar to those used in DC circuits.
Recall that power in DC circuits can be calculated using:
V2
P  VI 
R
 I2  R
Similarly, instantaneous power to a resistor can be calculated using
v2 (t)
p(t)  v(t)  i(t) 
 i 2 (t)  R
R
Show that power can be also be calculated using RMS values as follows:
P  VRMS  IRMS 
Development:
 VRMS 
R
2

 IRMS 
2
R
Chapter 10
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RMS values can be calculated:
1) by inspection (for simple cases)
2) by integration (for more complex cases)
Example: Find IRMS for the waveform below and use IRMS to calculate the
average power absorbed to the resistor. Recall that P was calculated earlier
using instantaneous power.
i(t)
i(t) [A]
+
v(t)
10
10
4
_
t [s]
0
2
4
6
8
10
12
EGR 272 – Circuit Theory II
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Example: Find VRMS for the waveform below and use VRMS to calculate the
average power absorbed to the resistor. Recall that P was calculated earlier
using instantaneous power.
v(t) [V]
i(t)
+
v(t)
_
20
10
0
t [s]
0
2
4
6
8
10
12
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RMS value of sinusoidal voltages and currents
Using the definition of RMS voltage and current, it can be shown that for a
sinusoidal voltage v(t) = Vpcos(wt) and a sinusoidal current i(t) = Ipcos(wt) that:
VRMS 
Vp
2
 0.707Vp
IRMS 
Ip
2
 0.707Ip
Development: Derive the expression for VRMS above.
Example: Find the power absorbed by a 10  resistor with v(t) = 20cos(377t) V.
Chapter 10
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Superposition of Power
Recall that, in general, superposition applies to voltage and current, but not to
power. i
Applying superposition to find the current i:
R
i = i1 + i2
or
+
+
V
V
2
_
_
1
i = current due to V1 + current due to V2
Show that this leads to: P  P + P
1
Development:
T
T
1
1
P   p(t)dt   i 2 Rdt 
T0
T0
2
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Power to a Complex Load


V
VmÐ0
If v(t) = Vm cos(wt) then i(t) = Im cos(wt - )  since I = Z  ZÐ  ImÐ   


i(t)
+
Useful identities:
v(t)
_
ZÐ°
Show that:
cos(wt - ) = cos()cos(wt) + sin()sin(wt)
cos2 (wt) = ½ + ½ cos(2wt)
sin(wt)cos(wt) = ½ sin(2wt)
V I
P  m m cos ( )  VRMS  I RMS cos ( ) 
2
Development:
T
T
1
1
P   p(t)dt   v(t)i(t)dt 
T0
T0
 VRMS 
Z
2
cos ( ) 
 I RMS 
2
 Z cos ( )
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Key definitions:
Recall that real or average power, P, was just defined as:
V I
P  m m cos ( )  VRMS  I RMS cos ( ) 
2
 VRMS 
Z
2
cos ( ) 
 I RMS 
2
 Z cos ( )
Other important terms are apparent power and power factor, as defined below:
Apparent power = (VRMS)(IRMS) measured in volt-amperes, VA
real power
power factor  p.f.  Fp 
 cos( )
apparent power
Since cos is an even function (i.e., cos() = cos(-)), calculating cos() does not
reveal the sign of the angle, so the terms leading and lagging are used with
power factor.
• The power factor is leading if  < 0, (i.e., I leads V or the load is capacitive).
• The power factor is lagging if  > 0, (i.e., I lags V or the load is inductive).
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Examples: Determine P and the p.f. for each case below:
+
10V RMS
_
Z
a) Z  4
b) Z  j3
d) Z  4  j3
e) Z  4 - j3
c) Z  -j3
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Complex Power:
If v(t)  Vm cos (wt),
then V  Vm Ð0 (phasor using the peak value of v(t))
but we can also define V RMS  VRMSÐ0
(phasor using the RMS value of v(t))
The use of phasors with RMS values is very convenient for power calculations.
IRMS
VRMS  VRMSÐ0 implies that IRMS  IRMSÐ 
+
VRMS
Z
since
_
Define
I RMS
VRMS
VRMSÐ0
=

 I RMSÐ  
ZÐ
Z
S  complex power  V RMS  I RMS (in volt-amperes, VA)
*
Chapter 10
Since
EGR 272 – Circuit Theory II
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S  complex power  V RMS  I RMS (in volt-amperes, VA)
*
Show that S  S Ð    VRMS  IRMSÐ  
 P  jQ
where P  real or average power  VRMS  IRMScos( ) (in watts, W)
Q  reactive power  VRMS  IRMSsin( ) (in volt-amperes reactive, VAR)
S  apparent power  VRMS  IRMS 
p.f.  power factor 
Development:
P
 cos( )
S
P2  Q2 (in volt-amperes reactive, VAR)
(expressed as leading if   0 or lagging if   0)
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Complex Power Diagram
A complex power diagram is sometimes used to illustrate the relationship
Im
between P, Q, |S|, and S.
S
Q
|S|
or
|S|
Q


P
Re
P
Example:
If VRMS = 100Ð0° V and IRMS = 5 Ð-30° A, find P, Q, |S|, and S and illustrate
them on a complex power diagram. Also find the power factor.
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Other useful relationships
Substitute VRMS = ZIRMS and Z = R + jX into S = VRMSIRMS* to show that:
PR 
 VRMS 
R
QL (or QC ) 
2

 IRMS 
 VRMS 
X
Development:
2

2
and PL  PC  0
R
 IRMS 
2
X
and
QR  0
-1 

where
X

wL
and
X

L
C


wC 

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Two approaches for performing power calculations:
1. Find ZTotal and IRMS  V RMS
ZTotal
Then find S  V RMS  I RMS  P  jQ
*
2. Find the voltage or current for each component (magnitude only). Then
calculate P or Q for each component and add the results for total P and Q.
Then use PT 
P
component
,
QT 
Q
component
, and
ST  PT  jQT
EGR 272 – Circuit Theory II
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Example: Find S, S , P, Q, and p.f.
A) By first finding the total circuit impedance
+
100V RMS
60 Hz
_
10
15
26.5 mH
132.6 uF
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Example: (continued) Find S, S , P, Q, and p.f.
B) By finding P or Q for each component (use mesh equations first to find the
currents for each component).
KVL, mesh 1 : - 100Ð0°  10I1  j10(I1 - I 2 )  0
+
100V RMS
60 Hz
_
10
KVL, mesh 2 : j10(I 2 - I1 )  15I 2 - j20I 2  0
15
26.5 mH
132.6 uF
10  j10 - j10   I1  100
 - j10 15 - j10     0 

 I 2   
I1  5.099Ð - 41.8° A
I 2  2.828Ð81.9° A
I L  I1 - I 2  7.071Ð - 61.2° A
Chapter 10
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Power Factor Correction
Power companies charge their customers based on the real power (P) used. If
the customer has a significant amount of reactive power (or a low power factor)
this has no effect on P, but results in higher current levels. Power companies
“encourage” their large customers to “correct” their power factor. If their
power factor drops too low (perhaps less than 0.9), the power company may
charge higher rates.
Large companies often have lagging power factors due to large amounts of
machinery (inductive loads) and they can correct their power factor by adding
in large parallel capacitors (or use synchronous motors). This generates
negative reactive power that cancels the positive reactive power due to the
inductive loads resulting in a power factor which is near or close to unity. The
corrected power factor results in lower current levels and thus lower line losses
(I2 R) as the power company delivers the power to the customer. Use user also
benefits from lower current levels since maximum current ratings on wire and
equipment can be reduced.
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Illustration - Power Factor Correction
I1
I2
+
I1
+
IC
C
Circuit
_
Circuit
_
Original Circuit
New Circuit (Note that |I2| < |I1| )
Im
Im
Im
Q
Re
Re
Re
P
P
QC = -Q
Original Circuit
S = P + jQ
+
Added Capacitor
S = P + jQ
=
New Circuit
S = P + jQ - jQ = P
Chapter 10
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Example: Add a capacitor in parallel with the source in order to:
A) correct the power factor to unity
Also calculate the total source current and compare it to the original current.
Note that this is the same circuit analyzed previously.
+
100V RMS
60 Hz
_
10
15
26.5 mH
132.6 uF
Chapter 10
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Example: (continued) Add a capacitor in parallel with the source in order to:
B) correct the power factor to 0.95, lagging
Also calculate the total source current and compare it to the original current.
Summary: (Parts A and B)
Circuit
Original
Part A
Part B
p.f.
1.0 (unity)
0.95, lagging
C (added)
---
current, |I|
Reduction in |I|
---
Chapter 10
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Power Factor in Systems
Just as total power can be determined by adding the power dissipated by each
component, so can the total power for a system be determined by adding the
power dissipated by each subsystem.
Then use PTotal 
P
subsystem
,
QTotal 
Q
subsystem
, and
ST  PT  jQT
Example: Find the power factor of the entire system below that is comprised
of three subsystems as shown below.
+
_
Subsystem 1:
|S| = 2000 VA
p.f. = 0.8, lagging
Subsystem 2:
P = 2000 W
p.f. = 0.9, leading
Subsystem 3:
|S| = 2000 VA
Q = 1500 VAR
EGR 272 – Circuit Theory II
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Transformers:
Our earlier study of inductors introduced the idea that current through a set of
windings creates a magnetic field and magnetic flux, , flows through the core.
If  then flows through a second set of windings another magnetic field is
created and a secondary voltage is induced across these windings. This
arrangement of two windings on a common core is called a transformer
(illustrated below).

i1
i2
+
+
V1
N1
turns
_
N2
turns
V2
_
Primary
windings
Secondary
windings
Load
Chapter 10
EGR 272 – Circuit Theory II
Recall that
 = flux linkage
N = number of turns
 = magnetic flux (in Webers, Wb)
and  = N = Li , where L = inductance in Henries, H
Also
So
d
d(N )
v

dt
dt
d
v  N
dt
Since the same magnetic flux, , flows through the windings it is also true that
the rate of change of magnetic flux, d/dt is the same, so
d
V
V
N1
V
 1  2 , so
 1
dt
N1
N2
N2
V2
a 
N1
V
 1
N2
V2
where a = turns ratio
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Chapter 10
EGR 272 – Circuit Theory II
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Key transformer relationships
We just saw that
a 
N1
V
 1
N2
V2
where a = turns ratio
Similarly
a 
N1
I
 2
N2
I1
so
V1
If Z1 
I1
and Z2
V2

then show that
I2
a 
N1
V
I
 1  2
N2
V2
I1
a2 
Z1
Z2
Chapter 10
EGR 272 – Circuit Theory II
Transformer symbols
General Transformer
Iron-core Transformer
28
Chapter 10
EGR 272 – Circuit Theory II
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Examples of transformers (reference: www.allelectronics.com)
Primary: 120V
Secondary: 28V, 1.5A
Primary: 110V
Secondary: 15V, 0.4A
Primary: 120V
Secondary: 40VCT, 0.25A
Utility pole
transformer
Chapter 10
EGR 272 – Circuit Theory II
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Examples of transformers (reference: www.allelectronics.com)
PC Mount Transformer
Primary: 120V
Secondary: 16VCT, 0.8A or 8V, 1.6A
Up/Down Transformer
(110V to 220V) or (220V to 110V)
Toroidal Transformer
Primary: 120V
Secondary: 8.5V and 9.4V
Variac (Variable Transformer)
Input: 110V
Output: 0 to 130V
Chapter 10
EGR 272 – Circuit Theory II
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Dot Convention
The direction of the windings in the secondary with respect to the direction of
the windings in the primary determines the polarity of the secondary voltage.
However, rather than showing the direction of the windings, dots are often
placed at one end of each winding. Then the following convention applies:
Dot convention: “A positive voltage at one dotted terminal
will produce a positive voltage at the other dotted terminal.”
Note that the relationships a  N1  V1  I2
N2
V2
I1
Imply that:
• the positive terminals of V1 and V2 are at the dotted terminals
• I1 enters the dotted terminal and I2 leaves the dotted terminal
a:1
N1:N2
i1
These voltage polarities and
current directions are indicated
to the right.
i2
+
+
V1
V2
_
_
Chapter 10
EGR 272 – Circuit Theory II
Four Common Transformer Uses
• Change voltage levels
• Change current levels
• Change impedance levels (impedance matching)
• Isolation (e.g., to isolate the secondary load from the ground in the primary)
Examples: Show simple examples of the four transformer uses listed above.
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EGR 272 – Circuit Theory II
Chapter 10
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Example: Determine the V1 and i2 in the circuit below. Find these values by:
A) Using KVL equations around each loop
1:2
+
20 mF 5
+
2
V1
5 cos(10t)
_
_
20 mH
i2
EGR 272 – Circuit Theory II
Chapter 10
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Example: Determine the V1 and i2 in the circuit below. Find these values by:
B) Reflecting the impedance of the secondary to the primary.
1:2
+
20 mF 5
+
2
V1
5 cos(10t)
_
_
20 mH
i2
EGR 272 – Circuit Theory II
Chapter 10
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Example: Determine the value of Zab below.
1:2
a
10
Zab
3:1
10
1:2
10
10
b
Chapter 10
EGR 272 – Circuit Theory II
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Example: Recall the example presented in EGR 271 where a transformer was
used to insure that maximum power was delivered to a 4 ohm speaker from the
output of an amplifier with an output resistance of 100 ohms. Illustrate and
discuss.
Chapter 10
EGR 272 – Circuit Theory II
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Example: Determine the inductance, L, and the turns ratio, a, required to
achieve maximum power transfer to the secondary load.
a:1
+
62.5 pF
80 
320
L
100cos(105t) V
_
Chapter 10
EGR 272 – Circuit Theory II
Transformer Models:
Transformers considered in this
course have been considered to
be ideal. Although transformer
behavior is often close to ideal,
there are cases where non-ideal
behavior might be examined.
The figures to the left
(reference: Electric Power &
Machinery by Matsch) show
models of a transformer used to
approximate non-ideal behavior.
The resistances, R1 and R2,
represent the resistance of the
windings and the inductive
reactances, Xl1 and Xl2 represent
leakage magnetic flux.
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