32 Sources and power for HW 03

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/
Sources
and
Power
/
for
HW
03
 
 The
circuit
models
for
independent
voltage
sources
(Figure
6‐1)
and
independent
 current
sources
(Figure
6‐2),
although
convenient
for
solving
simple
problems,
are
 overly
simplistic
for
several
reasons.
Obviously,
they
fail
to
predict
how
much
power
 can
be
transferred
from
the
source
to
the
load.

 6.1
Overly
simplistic
models
for
independent
sources
–
 why
overly?
 
 
 Figure
6‐1.
The
circuit
model
for
independent
voltage
sources
and
its
 voltage‐current
characteristic.
 
 Figure
6‐2.
The
circuit
model
for
independent
current
sources
and
its
 voltage‐current
characteristic.

 
 ©
2010
Alexander
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03
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Sources
and
Power
/
for
HW
03
 Consider
an
independent
voltage
source
connected
to
a
load,
as
shown
in
Figure
6‐3,
 and
assume
that
the
load
resistance
is
variable.

 
 
 
 Figure
6‐3.
An
independent
voltage
source
connected
to
a
load
 resistor.
 
 For
example,
assume
that
the
source
VS
=
1.5
V
serves
as
a
model
for
AA‐type
 battery,
and
the
load
RL
=
3
Ω
represents
a
small
incandescent
lamp.
Then
the
power
 absorbed
by
the
load
 PL RL in
parallel,
their
equivalent
resistance
of
1.5
Ω
plays
the
role
of
RL
and
the
absorbed
 power
increases
to
1.5
W.
If
10
lamps
are
connected
in
parallel,
we
obtain
RL
=
0.3
 Ω,
and
the
absorbed
power
PL
=
7.5
W.
Adding
thousands
lamps
in
parallel,
we
 would
have
to
conclude
that
a
single
battery
would
supply
kilowatts
of
power,
 enough
to
illuminate
a
lecture
room,
which
is
of
course
absurd.
The
wrong
 conclusion
results
from
the
simplified
circuit
model.

 
 Similarly,
load
resistance
RL
connected
to
independent
current
source
IS
,
as
shown
 2 in
Figure
6‐4,
absorbs
the
power PL = ( I S ) ⋅ RL ,
which
endlessly
grows
as
RL
gets
 larger;
for
example,
if
more
and
more
lamps
are
connected
in
series.

Again,
an
 overly
simplified
model
leads
to
absurd
results.

 
 (VS )2 
equals
0.75
W.
If
two
identical
lamps
are
connected
 = ©
2010
Alexander
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03
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Sources
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Power
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for
HW
03
 
 
 
 Figure
6‐4.
An
independent
current
source
connected
to
a
load
 resistor.
 
 The
circuit
models
for
independent
sources
are
greatly
improved
by
addition
of
 resistances
RS
–
in
series
with
a
voltage
source
and
in
parallel
with
a
current
source.
 Think
about
these
resistances
as
internal:
they
belong
to
the
source
and
cannot
be
 disconnected
from
it.
For
example,
the
whole
complex
chemistry
of
a
battery
can
be
 approximately
represented
by
a
single
parameter
–
its
internal
resistance
RS
;
 function
generators
often
have
RS
=
50
Ω.

 
 6.2
Better
models
for
the
sources
 
 Figure
6‐5.
The
Thévenin
equivalent
circuit
includes
resistance
RS
 connected
in
series
with
the
voltage
source
VS
.

 
 ©
2010
Alexander
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03
 
 Figure
6‐6.
The
Norton
equivalent
circuit
includes
resistanceRS
 connected
in
parallel
with
the
current
source
IS
.
 
 
 Due
to
historic
reasons
(see
the
History
section
below),
the
improved
circuit
model
 for
an
independent
voltage
source
VS
with
its
resistance
RS
in
series
(Figure
6‐5)
is
 called
the
Thévenin
equivalent
circuit,
and
its
twin
model
for
an
independent
 current
source
IS
with
its
resistance
RS
in
parallel
(Figure
6‐6)
is
known
as
the
 Norton
equivalent
circuit.


 
 The
circles
on
circuit
diagrams
of
Figures
6‐5
and
6‐6
denote
the
terminals
a
and
b,
 to
which
a
load
can
be
connected.
Therefore,
the
improved
circuit
models
for
 sources
with
loads
necessarily
include
the
internal
resistances,
as
shown
in
Figures
 6‐7
and
6‐8.

 
 
 ©
2010
Alexander
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 Figure
6‐7.
The
Thévenin
equivalent
circuit
of
an
independent
voltage
 source
connected
to
the
load
includes
both
resistances
RS
and
RL
in
 series.

 
 
 Let
us
see
whether
these
models
are
indeed
better.
Consider
the
Thévenin
 equivalent
circuit
with
VS
=
1.5
V
and
RS
=
1
Ω,
and
examine
what
happens
when
the
 load
resistance
RL
is
varied.
Due
to
the
connection,
the
same
current
I
flows
through
 the
source,
the
source
resistance
RS
and
the
load
resistance
RL
thus
the
formula
for
 voltage
division
applies:

 
 VS I= RS + RL RL VL = I ⋅ RL = VS ⋅ 
 
 [equation
6‐1]
 RS + RL PL (VL )2 = RL 
 Let
us
delegate
boring
calculations
to
EXCEL
and
view
the
results
in
Table
6.1:
 
 
 Table
6.1.
 
 ©
2010
Alexander
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03
 
 As
the
load
resistance
RL
is
varied
by
6
orders
of
magnitude,
or
by
factor
of
a
million,
 from
0.001
⋅
RS
to
1000
⋅
RS
,
the
voltage
VL
across
the
load
increases
from
almost
 zero
to
nearly
the
source
voltage
VS
,
and
the
current
I
decreases
from
nearly
1.5
A
 to
practically
zero.
The
power
absorbed
by
the
load
reaches
its
peak
at
RL
=
RS
and
 decreases
to
zero
both
at
higher
and
at
lower
values
of
the
load
resistance
RL
.

 
 Equation
[6‐1]
shows
that
the
maximal
voltage
VL
across
the
load
is
reached
when
 the
load
resistance
RL
gets
much
higher
than
the
source
resistance
RS
;
this
value
is
 called
the
open‐circuit
voltage
VOC
because
the
current
practically
does
not
flow
 through
this
circuit
where
the
very
high
RL
acts
as
an
open
switch.
The
open­circuit
 voltage
VOC
equals
the
source
voltage:
 
 
 
 [equation
6‐2]
 VOC = VS 
 
 Equation
[6‐1]
also
shows
that
the
maximal
current
I
through
the
circuit
is
reached
 when
the
load
resistance
RL
gets
much
lower
than
the
source
resistance
RS
;
this
 value
is
called
the
short‐circuit
current
ISC
because
a
very
low
load
resistance
 practically
acts
as
a
closed
switch
or
a
short
circuit.
The
short­circuit
current
ISC
 equals
the
ratio
of
the
source
voltage
to
the
source
resistance:
 
 V 
 
 [equation
6‐3]
 I SC = S 
 RS 
 In
the
circuit
of
Figure
6‐7
with
the
source
voltage
VS
=
1.5
V
and
the
source
 resistance
RS
=
1
Ω,
we
obtain
the
open‐circuit
voltage
VOC
=
1.5
V
and
the
short‐ circuit
current
ISC
=
1.5
A.


 
 The
power
absorbed
by
the
load
equals
the
product
of
current
I
through
the
circuit
 and
voltage
VL
across
the
load.
As
the
load
resistance
gets
very
low,
the
current
 asymptotically
approaches
ISC
but
the
voltage
VL
across
the
load
continues
to
drop
 (compare
the
first
two
rows
of
Table
6.1),
which
causes
the
decrease
of
power.
 Conversely,
when
the
load
resistance
gets
very
high,
the
voltage
VL
across
the
load
 asymptotically
approaches
VOC
but
the
current
continues
to
drop
(compare
the
last
 two
rows
of
Table
6.1),
which
again
causes
the
decrease
of
power.



 
 The
power
PL
absorbed
by
the
load
reaches
its
maximum
when
the
load
 resistance
equals
the
source
resistance:
RL
=
RS
.
To
prove
this
statement
 dP analytically,
take
the
derivative
 L of
the
expression
in
[6‐1],
find
where
this
first
 dRL d 2 PL derivative
equals
zero,
and
prove
that
the
second
derivative
 
is
negative
at
this
 2 dRL value
of
RL
.

For
circuit
analysis
we
need
only
the
final
result,
that
is
RL
=
RS
.
 According
to
equation
[6‐1],
the
maximal
power
equals:

 ©
2010
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HW
03
 
 PL , MAX (VS )2 = (VOC ) ⋅ ( I SC ) 

 = 4 ⋅ RS 4 
 [equation
6‐4]

 
 Note
that
the
maximal
power
is
obtained
when
the
voltage
across
the
load
equals
 half
the
source
voltage
and
the
current
through
the
circuit
equals
½
ISC
,
as
seen
 from
both
equation
[6‐4]
and
Table
6.1.
In
other
words,
neither
the
highest
current
I
 by
itself
nor
the
highest
voltage
VL
by
itself
can
ensure
the
highest
power:
their
 product
is
maximal
“in
the
middle.”
Observe
the
remarkable
symmetry
of
voltage,
 current,
and
power
values
in
Table
6.1.

 
 Using
our
example
of
a
1.5‐V
battery,
to
which
several
lamps
can
be
connected,
we
 conclude
from
the
equations
above
and
Table
6.1
that
the
maximal
power
is
 absorbed
by
the
load
when
RL
=
RS
=
1
Ω,
which
corresponds
to
3
lamps
(3
Ω
each)
 in
parallel.
Neither
more
or
fewer
lamps
in
parallel
nor
lamps
in
series
ensure
 higher
values
of
the
total
power
absorbed
by
the
load
in
the
circuit
of
Figure
6‐7.

 
 Noteworthy,
in
the
improved
circuit
model
for
the
independent
voltage
source,
the
 power
absorbed
by
the
load
resistor
always
remains
finite:
no
circuit
condition
 leads
to
the
absurd
result
of
infinite
power
transferred
from
the
source
to
the
load.
 This
is
a
great
advantage
of
the
improved
model
(Figure
6‐5)
over
the
simplistic
one
 (Figure
6‐1).


 
 
 Now
let
us
examine
the
Norton
equivalent
circuit
for
a
current
source
connected
to
 the
load,
as
shown
in
Figure
6‐8.

 
 
 
 ©
2010
Alexander
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03
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03
 Figures
6‐8.
The
Norton
equivalent
circuit
of
an
independent
current
 source
connected
to
the
load
includes
both
resistances
RSand
RL
in
 parallel.
 
 
 All
elements
of
this
circuit
are
connected
in
parallel
thus
the
same
voltage
V
is
 applied
to
them.
The
source
current
IS
gets
divided
between
the
source
resistance
RS
 and
the
load
resistor
RL
according
to
the
familiar
equations:
 
 RS I L = IS ⋅ RS + RL 

 
 
 [equation
6‐5]
 2 PL = ( I L ) ⋅ RL 
 
 The
short‐circuit
current
in
this
case
equals
the
source
current:
 
 
 
 
 [equation
6‐6]
 I SC = I S 
 
 
 The
open‐circuit
voltage
can
be
found
from
Ohm’s
law:
 
 
 
 [equation
6‐7]
 VOC = I S ⋅ RS 
 
 
 
 The
analytic
solution
for
maximal
power
absorbed
by
the
load
is
very
similar
to
that
 for
the
voltage
source
of
Figure
6‐7,
which
was
given
by
equation
[6‐4]:
the
power
 absorbed
by
the
load
reaches
its
maximum
when
the
load
resistance
equals
 the
source
resistance
RL
=
RS.
The
maximal
power
value
is:
 
 ( I ) ⋅ (VOC ) 
 
 1 2 PL , MAX = ⋅ ( I S ) ⋅ RS = SC 
[equation
6‐8]
 4 4 
 The
numerical
solution
is
also
surprisingly
similar
to
that
given
above.
If
you
use
the
 same
value
for
the
source
resistance
RS
=
1
Ω
along
with
the
source
current
IS
=
1.5
A
 and
run
calculations
for
the
same
load
resistance
values
as
in
Table
6.1,
you
will
 obtain
exactly
the
same
numerical
values
for
the
power
absorbed
by
the
load
as
in
 Table
6.1.
As
you
will
learn
in
the
next
section,
it
is
not
a
coincidence.

 
 The
bottom
line
is
that
the
improved
model
for
an
independent
current
source
does
 not
produce
the
absurd
results
of
infinitely
high
power
transferred
from
the
source
 to
the
load.

 
 
 ©
2010
Alexander
Ganago

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03
 
 Voltage‐Current,
or
Volt‐Amp
characteristics
are
important
and
convenient
tools
for
 circuit
analysis:
they
show
the
relationship
between
the
voltage
across
two
 terminals
of
a
circuit
element
and
the
current
through
these
terminals.

 
 We
are
already
familiar
with
3
such
characteristics:

 
 1. For
a
resistor,
the
voltage
across
is
directly
proportional
to
the
current
 through;
the
coefficient
of
proportionality
is
called
resistance.

 
 2. For
an
independent
voltage
source,
the
voltage
across
its
terminals
is
fixed
 but
the
current
through
them
can
have
any
magnitude
and
direction
(Figure
 6‐1).

 
 3. For
an
independent
current
source,
the
current
through
its
terminals
is
fixed,
 while
the
voltage
across
them
can
have
any
magnitude
and
polarity
(Figure
 6‐2).

 
 We
would
like
to
obtain
Volt‐Amp
characteristics
for
our
better
models
for
sources
–
 the
Thevenin
and
Norton
equivalent
circuits
(Figures
6‐7
and
6‐8).
 
 Before
starting
the
derivations,
let
us
clarify
our
goal,
compare
our
expectations
to
 the
list
above,
and
review
our
analytic
tools
along
with
the
main
parameters
of
the
 circuits:


 
 A. Our
goal
is
to
obtain
the
relationship
between
the
current
IL
that
flows
 between
the
terminals
a
and
b
(the
reference
direction
is
from
a
to
b)
and
the
 voltage
VL
across
these
terminals
(the
reference
polarity
is
positive
at
a).
 Recall
thatVL
is
the
voltage
across
the
load,
and
IL
is
the
current
through
the
 load
connected
to
these
terminals;
the
values
of

VL
and
IL
vary,
depending
on
 the
specific
load.
 
 B. For
a
particular
case
of
a
Thevenin
equivalent
circuit
that
we
examined
 above,
we
found
that
an
increase
of
VL
leads
to
a
decrease
of
IL
.
This
finding
 suggests
a
shape
of
this
Volt‐Amp
characteristic
distinct
from
all
the
three
 listed
above.

 
 C. Our
analytic
tools
include
KCL,
KVL,
and
Ohm’s
law.
The
circuit
parameters
 include:
voltage
VS
of
the
voltage
source;
current
IS
of
the
current
source;
the
 source’s
internal
resistance,
which
we
denote
RS
for
both
cases;
the
load
 resistance
RL
(again,
the
same
label
for
both
circuits);
the
current
IL
that
 flows
through
RL
,
and
the
voltage
VL
across
RL
(Figures
6‐7
and
6‐8).



 
 ©
2010
Alexander
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Sources
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03
 6.3
Volt‐Amp
characteristics
of
the
better
models
are
very
 much
alike!
 Ganago
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Sources
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HW
03
 The
logic
steps
needed
for
analysis
of
both
circuits
are
quite
similar;
therefore
we
 lay
out
our
derivations
below
side‐by‐side,
in
two
columns:
the
left
for
the
Thevenin
 equivalent
circuit
of
Figure
6‐7
and
the
right
for
the
Norton
equivalent
circuit
of
 Figure
6‐8.

 
 KVL:

 KCL:
 −VS + VRS + VL = 0 

 − I S + I RS + I L = 0 

 
 
 Ohm’s
law:
 Ohm’s
law:
 VRS = I ⋅ RS 
 V I RS = L 
 
 RS Thus

 
 I ⋅ RS = I L ⋅ RS = VS − VL 
 Thus

 VS VL V 
















[equation
6‐9]
 IL = − I L = I S − L 













[equation
6‐10]
 RS RS RS 
 
 In
particular,
when
VL
=
0,
we
obtain:
 In
particular,
when
VL
=
0,
we
obtain:
 VS I L = I SC = I S 
 
 I L = I SC = RS 
 When
IL
=
0,
we
obtain:
 When
IL
=
0,
we
obtain:
 VL = VOC = I S ⋅ RS 
 VL = VOC = VS 
 
 
 Our
main
result
is:
the
Volt­Amp
characteristics
for
both
Thevenin
and
Norton
 equivalent
circuits,
which
show
the
relationships
between
the
voltage
VL
 across
the
load
and
the
current
IL
through
the
load,
are
linear.
The
negative
sign
 in
equations
[6‐9]
and
[6‐10]
indicates
that
as
the
voltage
VL
increases,
the
current

 IL
decreases,
in
agreement
with
our
expectation
B
above.

 
 The
linear
Volt‐Amp
characteristics
can
be
plotted
in
several
ways.
Figure
6‐9
shows
 them
highlighting
the
crossing
points
with
the
axes
–
the
open‐circuit
voltage
VOC
 and
the
short‐circuit
current
ISC.

 
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2010
Alexander
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for
HW
03
 
 
 Figure
6‐9.
Volt‐Amp
characteristic
of
the
Thevenin
and
Norton
 equivalent
circuits.
This
plot
highlights
the
two
crossing
points
with
 the
axes
‐
the
open‐circuit
voltage
VOC
and
the
short‐circuit
current
ISC.
 
 Note
that
the
same
characteristic
can
be
plotted
with
the
opposite
choice
of
axes,
as
 shown
in
Figure
6‐10.

 
 
 
 Figure
6‐10.
Volt‐Amp
characteristic
of
the
Thevenin
and
Norton
 equivalent
circuits.
Note
that
the
characteristic
itself
is
the
same
as
in
 Figure
6‐9
but
the
choice
of
axes
is
opposite.

 
 For
the
discussion
below,
it
is
more
convenient
to
use
the
plot
of
Figure
6‐9,
where
 the
current
IL
is
on
the
horizontal
axis
and
voltage
VL
on
the
vertical.

 ©
2010
Alexander
Ganago

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03
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03
 
 As
we
already
derived,
the
open‐circuit
voltage
VOC
can
be
expressed
either
as
the
 voltage
VS
of
the
voltage
source
[equation
6‐2]
or
as
the
product
of
the
source
 resistance
RS
and
the
current
IS
of
the
current
source
[equation
6‐7].
Similarly,
the
 short‐circuit
current
ISC
can
be
expressed
in
two
ways:
as
the
ratio
of
the
source
 voltage
VS
to
the
source
resistance
RS
[equation
6‐3]
or
as
the
source
current
IS
 [equation
6‐6].
It
means
that
the
same
Volt­Amp
characteristic
belongs
to
both
 the
Thevenin
circuit
with
VS
and
RS
and
the
Norton
circuit
with
IS
and
RS,
 provided
that
two
conditions
are
met:

 
 VS = I S ⋅ RS 
 
 [equation
6‐11]
 and RS is the same in both circuits 
 This
is
a
very
important
result.

 
 Until
now,
we
could
see
the
independent
current
sources
and
the
independent
 voltage
sources
as
species
from
different
worlds:
the
distinction
between
their
Volt‐ Amp
characteristics
in
Figures
6‐1
and
6‐2
confirmed
this
opinion.
Equation
[6‐11]
 proves
that
the
sources
possess
the
same
Volt‐Amp
characteristics
if
their
 parameters
match,
satisfying
the
two
conditions
of
equation
[6‐11].
Figure
6‐11
 illustrates
this
important
conclusion.
 
 
 
 Figure
6‐11.
The
voltage
source
represented
by
its
Thevenin
circuit
is
 equivalent
to
the
current
source
represented
by
its
Norton
circuit,
 provided
that
both
conditions
of
equation
[6‐11]
are
met.
The
 equivalent
sources
possess
the
same
Volt‐Amp
characteristic.


 
 ©
2010
Alexander
Ganago

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12
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15

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03
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Sources
and
Power
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for
HW
03
 For
example,
consider
the
Thevenin
circuit
with
VS
=
12
V
and
RS
=
3
Ω
and
the
 Norton
circuit
with
IS
=
4
A
and
RS
=
3
Ω:
for
their
parameters,
both
conditions
of
 equation
[6‐11]
are
met,
therefore,
these
two
circuits
have
the
same
Volt‐Amp
 characteristic,
as
shown
in
Figure
6‐12.

 
 
 
 


 Figure
6‐12.
An
example
of
equivalent
sources.
The
voltage
source
 (Thevenin
circuit)
and
the
current
source
(Norton
circuit)
shown
here
 are
equivalent,
because
both
conditions
of
equation
[6‐11]
are
met.
 These
sources
possess
the
same
Volt‐Amp
characteristic.


 
 The
concept
of
equivalent
sources
leads
to
a
very
convenient
tool
for
circuit
analysis
 –
the
Source
Transformation.
In
any
circuit,
a
Thevenin
equivalent
can
be
replaced
 with
its
Norton
equivalent,
provided
that
both
conditions
of
equation
[6‐11]
are
 met,
and
the
rest
of
the
circuit
would
not
notice
any
difference!

 
 Note
that
equivalent
does
not
mean
identical:
it
is
only
to
the
load
(to
the
right
of
 terminals
a
and
b
on
our
diagrams)
that
there
is
no
difference
what
source
is
used
 (to
the
left
of
terminals
a
and
b).
The
sources
themselves
are
not
exactly
the
same!
 For
example,
if
no
load
is
connected
to
terminals
a
and
b
(open
circuit
condition),
 the
voltage
source
does
not
supply
any
power,
but
its
equivalent
current
source
 2 wastes
 ( I S ) ⋅ RS 
every
second!
 
 Let
us
return
to
the
more
general
issue
of
defining
the
Volt‐Amp
characteristic
of
a
 Thevenin
or
Norton
circuit.
We
can
reach
this
goal
in
several
ways,
suggested
by
 equations
[6‐9]
and
[6‐10].
Recall
that,
geometrically,
a
straight
line
can
be
 determined
by
2
points,
or
by
one
point
and
the
slope.
Thus,
as
an
alternative
to
 defining
the
Volt‐Amp
characteristic
by
its
crossing
points
with
the
axes
as
shown
in
 Figure
6‐9,
we
can
use
the
slope
of
the
line,
which
equals
the
source
resistance
RS
,
 ©
2010
Alexander
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03
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HW
03
 and
either
one
of
the
crossing
points
with
the
axes,
as
shown
in
Figures
6‐13
and
6‐ 14.

 
 
 
 Figure
6‐13.
The
Volt‐Amp
characteristic
of
a
Thevenin
and
Norton
 circuit
can
be
determined
by
the
open‐circuit
voltage
VOC
and
the
 source
resistance
RS.

 
 
 
 Figure
6‐14.
The
Volt‐Amp
characteristic
of
a
Thevenin
and
Norton
 circuit
can
be
determined
by
the
short‐circuit
current
ISC
and
the
 source
resistance
RS.
 
 
 ©
2010
Alexander
Ganago

 Page
14
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15

File:
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03
 Ganago
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Sources
and
Power
/
for
HW
03
 Eventually,
we
know
from
geometry
that
any
two
points
on
a
straight
line
uniquely
 define
it,
not
necessarily
the
crossing
points
with
axes;
therefore,
we
can
determine
 the
Volt‐Amp
characteristic
by
two
pairs
of
voltage
and
current
measured
or
 calculated
for
two
different
loads,
which
are
labeled
{VL,1
;
IL,
1}
and
{VL,2
;
IL,
2}
in
 Figure
6‐15.

 
 

 Figure
6‐15.
The
Volt‐Amp
characteristic
of
a
Thevenin
and
Norton
 circuit
is
uniquely
defined
by
two
pairs
of
voltage
across
the
load
and
 current
through
the
load,
which
are
measured/calculated
for
two
 distinct
loads
and
labeled
here
as
{VL,1
;
IL,
1}
and
{VL,2
;
IL,
2}.


 
 
 
 
 
 ©
2010
Alexander
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