Circuits II
EE221
Unit 3
Instructor: Kevin D. Donohue
Instantaneous, Average, RMS, and
Apparent
pp
Power, and, Maximum Power
Transfer, and Power Factors
Power Definitions/Units:



Work is in units of newton
newton-meters
meters or joules
same as energy.
1 joule
j l is
i work
k done
d
f
for 1 ampere passed
d
through 1 ohm for 1 second, or work done
by
y a force of 1 Newton applied
pp
over 1
meter.
Power is a measure of the rate at which
work is done and is in units ofjoules per
seconds, 746 Watts in one horsepower.
Power Conversion


Electric motors and generators
enerators convert
electrical power to mechanical power and
vise versa.
versa
Other devices exist that convert electric
power to light, heat, sound, …. and vise
versa
Instantaneous and Average Power

The instantaneous power is the power absorbed by an
element at an instance of time. In an electric circuit
this is given by:
P (t )  v(t )i (t )
where i(t) is the current through the element and v(t) is
the voltage drop over the element.

Average Power over some time interval T is given by:
T
PAV 
1
v(t )i (t )dt
T 0

If the i and v product is periodic, then PAV can be
reduced to the integration over a single period. For
random/non-periodic
d /
i di signals
i
l T goes to
t iinfinity.
fi it
Instantaneous Power


Instantaneous power is simply computed by the
product of 2 functions. For sinusoidal problems,
trig identities or phasors can be used to simplify
the products for easier average power formulae.
formulae
Example:
p
Write a Matlab script
p that plots
p
the
instantaneous power of a cosine voltage across an
impedance load. Write the script so the voltage
p
can be easily
y changed
g
waveform and impedance
and instantaneous power replotted.
Matlab Scripts


A script is a series of Matlab command line
instructions typed in a text file and stored
with an *.m extension. This is referred to
as an mfile.
mfile
To run these commands change your
current directory associated with the
Matlab work space to the one containing
the mfile.
mfile Then type the base name of the
file in the Matlab command line. Or on the
editor menu there is a green “play” arrow
that will also execute the program.
program
Matlab Scripts


For this example comments are included in
the script, and are identified by text
following a “%”
% character.
character
The functions plot and cos are used in this
script. Type help plot or help cos to get a
full description
f
p
of
f these f
functions in the
Matlab workspace.
Matlab Scripts
% This script
p will pplot the instantaneous ppower absorbed
% by an impedance load with a sinusoidal voltage over it.
% Set the parameters of the Voltage signal
q
y of signal
g in Hz
f = 1000;; % Frequency
ph = 30; % Phase of signal in degrees
A=2;
% amplitude of signal
%S impedance
%Set
p
value
zm = 8; % Impedance magnitude
zp = -40; % Impedance phase in degrees
% Create a time axis of 2 periods over which to plot the power
tp = 1/f; % Determine period
t = 2*tp*[0:5000]/5000; % Make a 5001 point time axis (row vector) over 2 periods
Matlab Scripts
% Create a vector of p
points for voltage
g
v = A*cos(2*pi*f*t + pi*ph/180);
% Create a vector of points for the current (adjust magnitude and phase
% based on impedance
i = ((A/zm)*cos(2*pi*f*t
)
( p
+ pi*(ph-zp)/180);
p (p p)
);
% Take an element by element product to get power
p = v.*i;
% Plot it
p ( ,p)
plot(t,p)
title(['Instantaneous Power '] )
xlabel(['Seconds'])
ylabel(['Watts'])
Result
Instantaneous Power
0.6
0.5
0.4
Watts
0.3
0.2
0.1
0
-0.1
0
02
0.2
04
0.4
06
0.6
08
0.8
1
12
1.2
Seconds
14
1.4
16
1.6
18
1.8
2
-3
x 10
What is the meaning of negative instantaneous power?
Average Power in Periodic Signals
Given a sinusoidal voltage and current in a device:
v(t )  Av cos(t   v )
i (t )  Ai cos(t   i )
Show:
PAV 
1
Av Ai cos( v   i )
2
Average Power in Periodic Signals
Given a phasor representation of a voltage and
current in a device:
Vˆ  Av  v
Iˆ  A 
i
i
Show:
PAV
 
1
1
*
ˆ
ˆ
 Re
R VI  Av Ai cos(( v   i )
2
2
Conservation of Power
In a given circuit the average power absorb (denoted
by positive values) equals the power delivered
(denoted by negative values).
For a circuit with N elements the sum of all power is
zero:
0
N
P
i
i 1
Note: These are all real
or average power values
values.
Passive Sign Convention
It is assumed that positive charge
entering the positive terminal of
an element implies power
absorbed by the element.
Th
Therefore,
f
charge
h
leaving
l
the
h
positive terminal of an element
implies power supplied or
d l
delivered
db
by the
h element.
l
If the words absorbed or supplied
are not g
given w
with a p
power
w value,
u ,
power absorbed will be assumed.
(A negative sign will imply power
suppl ed).
supplied).
I
V
I
V
Example Average Power
Find the average power each of the
elements given is(t) = 3cos(1000t)A
10ia
ia
25Ω
10mH
is
P25Ω=34.61W
=34 61W
PL=PC=0
P5Ω=21.13W
PCCVS=-5.54W
5 54W
Pis=-50.19W
80µF
5Ω
Maximum Power Transfer
Given a Thévenin circuit with load ZL:
Zthh
Vth
ZL
Show that for a maximum power transfer to the load:
Zˆ L  Zˆ th*
And the maximum power transfer is:
Pmax 
Vˆth
2
8 Re[ Zˆ th* ]
Example Maximum Power
Find the impedance of the load Z to result in the
maximum power transfer,
transfer and find the resulting
power. Assume vs(t) = 110cos(377t) V
40 F
40µF
8Ω
vs
Z=2.87-75.88
Z=2
87-75 88
Ith=1.6583.12A
Vth = 4.71 159V
Pz=3.98W
3 98W
12Ω
7.5mH
Z
Root Mean Square
q
((RMS)) Values

The RMS value of a
periodic current or
voltage is its DC
equivalent value for
delivering average
power to a resistor.
PAV
1

T
2
i
(t )
2
Ri 2 (t )dt  R
dt  RI rms
T
T T

I rms 
PAV
1

T

1
T

T
i 2 (t )dt
v 2 (t )
1
dt 
T
R
R

Vrms 
1
T

T
2
Vrms
v 2 (t )
dt 
T
T
R

v 2 (t )dt
RMS Formula for Sinusoids
The RMS value for a
sinusoid
i
id of
f any
frequency or phase is
its amplitude
p
divided
by square root of 2.
I rms
I rms 
1
T
I rms 
A2
T
A
T
2
cos 2 (t   )dt
2
 exp( j (t   ))  exp( j (t   )) 
 dt
T 
2



I rms
A2

4T
 exp(( j 2(t   ))  2 exp((0)  exp(( j 2(t   )) dtd
I rms
A2

4T
A2
2dt 
T
4T
A2
A

2T  0 
4T
2
T

 exp( j 2(t   ))  exp( j 2(t   ))dt
T
Apparent
pp
Power and Power Factor
Apparent power (S) for
sinusoidal waveforms is the
product of the RMS voltage and
current magnitudes without
regard to their phase offsets.
It is iin units
its of
fV
Volt-Amps
lt A s (VA)
(VA).
The cosine of their phase
difference is the power factor
(pf).
(pf)
1
1
PAV 
2
 
Re VˆIˆ* 
S  Vrms I rms
2
v(t )  Av cos(t   v )
i (t )  Ai cos(t   i )
Vˆ  Av  v
Iˆ  A 
i
Av Ai cos( v   i )  Vrms I rms cos( v   i )
((apparent
pp
ppower))
PAV  S cos( v   i ) (real power)
Q  S sin( v   i )
i
(reactive power)
pf  cos( v   i ) (power
(po er factor)
Sˆ  Vrms I rms exp j ( v   i )  (complex power)
Leading and Lagging
The sign
g of phase
p
difference between the
voltage and current is related to the angle
of the impedance of the load. The terms
l di and
leading
d lagging
l
i are used
d to describe
d
ib
this property:
 v   i  0  pf leading (Capacitive Load)
 v   i  0  pff lagging
l i (Inductive
(I d i Load)
L d)
 v   i  0  (Only Real Load)
Example
An industrial load consumes 10 kW at a
power factor of .95 leading. The voltage
across the load is 220 Vrms (assume zero
phase).
h
) Determine
D
i the
h current drawn
d
by
b
the load (including phase).
Result
I  477.5818.2 Armss
Example
Find complex
p
power
p
supplied
pp
by
y source
(both real and reactive), and power factor
at the source.
15 
VS
-j10 
5kW
.9
Lagging
6kVA
.8
Leading
+
1200 Vrms
-
Result
Vs  1.59  24.88 kVrms
pf  .85 Leading
S s  111 kW  68.6 kVAR