06 Ohmâ��s law - Ganago
/
Ohm’s
Law
 
 


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Unformatted text preview: Ganago
/
Ohm’s
Law
 
 
 In
the
beginning
of
the
course,
we
discussed
the
current
I
through
a
circuit
element
 and
the
voltage
V
across
this
element
as
two
variables
that
hardly
related
to
each
 other.
Their
magnitudes
seemed
totally
independent;
moreover,
in
some
circuit
 elements,
the
current
entered
the
more
positive
terminal,
while
in
other
elements
 the
direction
of
current
was
opposite.
(Recall
that
this
distinction
in
current
 directions
related
to
whether
the
elements
absorbed
or
supplied
electric
power.)
 Such
lack
of
clarity
was
typical
of
the
state
of
knowledge
in
the
early
1800s,
when
 Georg
Simon
Ohm
started
his
experiments
on
electric
circuits.
Through
careful
 measurements
and
mathematical
analysis
of
data,
Georg
Ohm
found
that
the
 voltage
V
across
a
circuit
element
was
directly
proportional
to
the
current
I
 through
this
element.
His
finding,
which
we
call
Ohm’s
law,
served
as
 breakthrough
for
electric
circuit
analysis.

 
 
 Figure
2‐5.
German
physicist
Georg
Simon
Ohm
(16
March
1789
–
6
 July
1854)
formulated
the
relationship
between
electric
voltage
and
 current
based
on
his
experiments
with
circuits
done
when
he
taught
 physics
at
a
prestigious
high
school
in
1820s.

 
 The
mathematical
expression
of
Ohm’s
law
is:
 V
=
I
⋅
R

 
 
 
 [equation
2‐1]

 
 Here
the
coefficient
R
between
the
voltage
V
across
the
circuit
element
and
the
 current
I
through
this
element
is
called
electric
resistance.

 ©
2010
Alexander
Ganago

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Ohm’s
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 Ganago
/
Ohm’s
Law
 
 
 The
SI
unit
of
resistance
is
named
ohm
after
its
discoverer;
the
symbol
is
capital
 Greek
letter
Ω
(omega),
chosen
because
of
its
consonance
with
the
scientist’s
name.
 The
units
of
measure
are:
 
 1
volt
=
(1
ohm)
⋅
(1
amp),

 
 or


 1
V
=
(1
Ω)
⋅
(1
A)
[equation
2‐2]
 
 An
alternative
formulation
of
Ohm’s
law
uses
conductance
G,
which
is
reciprocal
of
 resistance
R:
 I
=
G
⋅
V,

 where

 G
=
1/R
 
 [equation
2‐3]
 
 The
unit
of
conductance
used
to
be
called
mho,
that
is
ohm
spelled
backwards;
then
 its
symbol
was
the
capital
omega
flipped
over
(put
on
its
head).
The
contemporary
 SI
unit
of
conductance
is
siemens,
abbreviated
S,
called
after
German
inventor
and
 industrialist
Ernst
Werner
von
Siemens
(see
Figure
2‐6).

 
 
 Figure
2‐6.
Ernst
Werner
von
Siemens
(13
December
1816
–
6
 December
1892)
was
a
German
inventor
and
industrialist,
founder
of
 one
of
the
largest
electro‐technological
companies
in
the
world,
which
 is
known
today
as
Siemens
AG.
The
fourth
(out
of
fourteen)
son
of
a
 tenant
farmer,
he
left
school
without
finishing
education
to
become
a
 soldier.
When
he
become
inventor
and
founded
his
own
company,
his
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2010
Alexander
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Ohm’s
Law
 
 younger
brothers
became
its
representatives
in
England
(Carl
 Wilhelm
Siemens
was
knighted
and
became
Sir
William)
and
in
Russia
 (Carl
Heinrich
was
ennobled
by
Tsar
Nicholas
II).
The
SI
unit
of
 conductance
was
named
in
his
honor.

 
 The
units
of
measure
are:

 (1
A)
=
(1
S)
⋅
(
1
V)

 
 
 
 [equation
2‐4]

 
 
 Learn
to
use
the
derivative
units,
for
example:
 1
V
=
(1
mA)
⋅
(1
kΩ)

 but

 1
mA
=
(1
V)
⋅
(1
mS)
[equation
2‐5]

 
 The
circuit
elements,
which
are
characterized
with
their
resistance,
are
called
 resistors,
often
denoted
R
and
shown
on
diagrams
with
the
zigzag
symbol,
as
in
 Figure
2‐7.

 
 Figure
2‐7.
The
circuit
symbol
for
a
resistor.
Note
that
the
reference
 marks
for
the
voltage
across
and
the
current
through
the
resistor
 indicate
that
resistors
always
absorb
electric
power.

 


 According
to
Ohm’s
law,
resistance
is
positive:
the
current
enters
the
more
positive
 terminal
of
the
resistor.
Combined
with
the
Passive
Sign
Convention
(PSC),
it
means
 that
resistors
always
absorb
electric
power.
The
power
absorbed
by
resistors
is
 converted
into
heat
thus
resistors
in
circuits
can
get
hot
and
burn
your
skin
if
you
 touch
them.

 
 There
are
two
meanings
of
Ohm’s
law.
The
first
one
may
be
called
empirical:
for
 example,
if
the
current
through
the
given
resistor
equals
2
A,
and
the
voltage
across
 this
resistor
equals
10
V,
we
can
find
from
Ohm’s
law
[equation
2‐1]
that
the
 resistance
R
equals

 (10
V)
/
(2
A)
=
5
Ω.

 
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2010
Alexander
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Law
 
 Of
course,
such
calculations
are
always
possible
and
can
often
be
justified.

 
 The
deeper
meaning
of
Ohm’s
law
is
its
predictive
power:
for
example,
if
the
current
 through
the
same
resistor
increases
to
10
A,
we
expect
that
the
voltage
across
this
 resistor
will
be
equal

 (10
A)
⋅
(5
Ω)
=
50
V.

 
 In
other
words,
Ohm’s
law
states
the
linear
dependence
between
the
voltage
and
the
 current,
as
shown
in
Figure
2‐8.

 
 
 
 Figure
2‐8.
According
to
Ohm’s
law,
the
current‐voltage
 characteristics
are
linear.

 
 The
accuracy
of
predictions
based
on
Ohm’s
law
is
limited
in
several
respects.
First
 of
all,
such
predictions
are
valid
only
for
certain
circuit
elements,
which
we
call
 resistors.
The
good
news
is
that
resistors
are
used
in
the
vast
majority
of
electronic
 circuits
and
are
known
among
the
most
reliable
components.
The
bad
news
is
that,
 in
other
classes
of
circuit
elements,
the
current‐voltage
dependence
is
non‐linear;
 such
elements
are
called
non‐ohmic
(see
next
chapter).
Although
Ohm’s
law
does
 not
apply
to
those
circuit
elements
in
its
original
form,
Electrical
Engineers
eagerly
 develop
its
modifications
in
order
to
simplify
the
analysis
of
circuits
with
non‐ohmic
 elements.
We
will
learn
about
some
of
them
in
further
chapters
of
this
book.
 Secondly,
even
good
resistors
have
linear
I‐V
characteristics
(such
as
shown
in
 Figure
2‐8)
only
under
certain,
limited
conditions;
beyond
those
limits,
their
 resistance
is
no
longer
constant
and
their
volt‐amp
characteristics
become
 nonlinear.
One
of
the
important
parameters
is
the
power
absorbed
by
the
resistor:
if
 this
power
gets
high,
resistors
heat
up,
and
their
resistances
change
due
to
the
 temperature
dependence.
In
the
extreme
cases,
the
absorbed
power
grossly
exceeds
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2010
Alexander
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Law
 
 the
manufacturer’s
specifications,
then
the
resistor
can
literally
go
in
flames,
burn
 out
and
become
useless.


 
 Besides
temperature,
electric
resistance
depends
on
many
parameters
such
as
 mechanical
stress,
light
intensity,
applied
voltage,
etc.
We
can
take
advantage
of
 such
dependences
by
turning
these
devices
into
sensors,
as
you
will
learn
later
in
 this
course.



 
 In
this
discussion,
we
will
assume
that
Ohm’s
law
holds
and
its
predictions
for
 resistors
are
valid
in
all
circuits.

 
 
 With
Ohm’s
law
in
addition
to
KCL
and
KVL,
we
can
do
power
calculations
in
several
 ways:
for
each
type
of
circuit,
we
can
choose
the
easiest
and
most
straightforward
 strategy.
We
begin
with
one
resistor
and
continue
with
two
or
more
resistors,
which
 are
connected
either
in
series
or
in
parallel.
Learning
these
techniques
is
especially
 valuable,
because
nearly
every
circuit
has
parts
that
can
be
seen
as
series/parallel
 combinations
of
resistors.


 

 
 2.4.1.
A
single
resistor.
 
 You
have
learned
that
electric
power
P
absorbed
by
a
circuit
element
is
the
product
 of
the
voltage
V
across
this
element
and
the
current
I
through
this
element;
 according
to
the
Passive
Sign
Convention
(PSC),
the
current
enters
the
more
positive
 terminal
and
the
absorbed
power
value
is
positive.
You
also
learned
that
some
 circuit
elements
absorb
power
while
others
supply
electric
power
to
the
circuits;
if
 the
power
is
supplied,
the
value
of
P
is
negative.

 
 Resistors
always
absorb
electric
power.
It
means
that,
for
a
given
voltage
across
 the
resistor,
the
current
enters
the
more
positive
terminal
as
shown
in
Figure
2‐7
 above;
if
the
polarity
of
voltage
is
swapped,
the
direction
of
current
swaps
 accordingly.
This
statement
might
sound
trivial,
but
it
is
not
necessarily
true
for
 other
circuit
elements:
for
example,
later
in
the
course
you
will
learn
that
the
 current
can
flow
into
or
out
of
the
positive
terminal
of
a
capacitor
while
the
polarity
 of
the
capacitor
voltage
remains
unchanged.

 
 <Text
box
on
the
margin>
Resistors
always
absorb
electric
power.

 
 <Text
box
on
the
margin>
The
current
always
enters
the
more
 positive
terminal
of
a
resistor.
If
the
polarity
of
voltage
across
the
 2.4
Power
absorbed
by
resistors.
 Voltage
and
current
division.
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Alexander
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Ohm’s
Law
 
 resistor
is
swapped,
the
direction
of
current
through
this
resistor
 swaps
accordingly.

 
 Using
Ohm’s
law
in
calculations
of
power,
we
can
apply
any
of
the
following
 relationships:

 P =V ⋅I V2 = V 2 ⋅G R 
 I2 P = I2 ⋅R = G P= 
 
 
 [equation
2‐7]
 Optimal
choice
of
the
particular
equation
[2‐7]
depends
on
what
you
already
know
 about
the
circuit.
Several
options
are
possible:

 (a) both
the
current
and
voltage
are
known,
but
the
resistance
is
not
known;
 (b) the
voltage
and
resistance
are
known;
 (c) the
current
and
resistance
are
known.
 
 For
practice,
decide
which
of
the
equations
[2‐7]
you
would
use
for
each
of
the
 options
(a
–
c)
listed
above.
In
the
next
section
you
will
learn
that
equations
with
 1 conductance
 G = 
lead
to
more
effective
solutions
of
circuits
where
resistors
are
 R connected
in
parallel.

 
 The
electric
power
absorbed
by
resistors
is
converted
into
heat.
The
more
power
 the
resistor
absorbs
the
hotter
it
may
become.
Each
resistor
that
you
will
use
in
the
 lab
or
any
project
has
a
certain
power
rating
in
watts;
if
the
actual
power
absorbed
 by
the
resistor
exceeds
its
rating,
consequences
may
be
dire.
For
example,
the
 resistive
layer
of
a
film
resistor,
which
absorbs
excessive
power,
gets
too
hot
and
 may
even
burst
in
flames
destroying
the
resistor.
Such
explosions
are
dangerous
not
 only
for
your
project
but
also
for
yourself
because
hot
pieces
flying
from
the
blown
 resistor
may
injure
your
face.
This
is
why
careful
calculations
of
the
power
absorbed
 by
each
resistor
are
extremely
important.

 
 Each
manufacturer
of
resistors
specifies
the
maximal
power
that
the
resistor
of
a
 certain
type
can
safely
absorb.
Usually,
film
and
composition
resistors
are
rated
for
 up
to
2
W,
while
some
wire‐wound
resistors
can
be
manufactured
for
much
higher
 powers
up
to
1
kW
and
above.

 Note
that
the
manufacturer’s
specifications
of
resistors’
power
ratings
assume
 relatively
low
ambient
temperatures
(usually,
below
50
˚C)
and
a
sufficient
flow
of
 air
to
cool
the
resistor.
If
such
conditions
are
not
met,
for
example,
in
a
tight
 enclosure
that
gets
really
hot
inside,
the
resistor’s
safe
power
limit
should
be
 reduced,
or
derated.
The
conservative
guideline
suggests
derating
by
a
factor
of
2,
 for
example,
make
sure
that
a
resistor
rated
for
1
W
does
not
absorb
more
than
½
 W.
To
follow
this
guideline,
calculate
the
power
absorbed
by
each
resistor
in
your
 ©
2010
Alexander
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Ohm’s
Law
 
 circuit
and
double
it
when
choosing
a
resistor:
for
example,
if
you
calculated
that
a
 certain
resistor
in
your
device
could
absorb
up
to
¼
W,
install
a
½
W
resistor.



 
 
 2.4.2
Resistors
connected
in
series:
Voltage
division.
Resistors
 connected
in
parallel:
Current
division.


 
 You
have
learned
that
(1)
the
same
current
flows
through
two
or
more
circuit
 elements
are
connected
in
series,
and
(2)
the
same
voltage
is
applied
across
 two
or
more
circuit
elements
are
connected
in
parallel.
Of
course,
these
general
 rules
apply
to
resistors.
The
simplest
cases
of
two
resistors
are
shown
in
Figures
2‐ 14
and
2‐15;
the
cases
of
many
resistors
are
presented
in
Figures
2‐16
and
2‐17.

 
 
 Figure
2‐14.
The
same
current
flows
through
both
resistors
connected
 in
series
and
enters
the
more
positive
terminal
of
each
resistor.
The
 total
voltage
is
divided
between
the
resistors.
Series
connection
of
 resistors
gives
rise
to
voltage
division.

 
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2010
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Ohm’s
Law
 
 
 
 
 Figure
2‐15.
The
same
voltage
is
applied
across
the
two
resistors
 connected
in
parallel,
while
the
total
current
is
divided
between
the
 resistors.
Parallel
connection
of
resistors
leads
to
current
division.

 
 Figure
2‐16.
In
a
general
case,
many
resistors
are
connected
in
series:
 the
same
current
flows
through
all,
while
the
total
voltage
is
divided
 among
the
resistors
like
it
is
divided
between
two
resistors
(see
 Figure
2‐14).
 
 
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2010
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Law
 
 
 In
order
to
determine
how
the
voltage
is
divided
between
resistors
in
series
and
 how
the
current
is
divided
between
resistors
in
parallel,
we
will
apply
KCL,
KVL,
and
 Ohm’s
law
to
the
simpler
circuits
each
containing
two
resistors
(Figures
2‐14
and
2‐ 15);
our
results
will
be
easily
expanded
to
the
general
case
of
many
resistors
in
a
 circuit.

 
 Albeit
the
final
results
for
voltage
division
and
for
current
division
are
distinct,
their
 derivations
are
remarkably
similar
if
we
use
resistance
for
series
connection
and
 conductance
for
parallel
connection.
To
highlight
this
similarity,
we
show
the
logic
 steps
and
equations
side‐by‐side.

 
 
 
 Resistors
in
series
(Figure
2‐14)
 Resistors
in
parallel
(Figure
2‐15)
 
 
 From
KVL:
 From
KCL
at
the
top
node:
 −VTotal + V1 + V2 = 0 
 − I Total + I1 + I 2 = 0 
 thus
 thus
 V1 + V2 = VTotal 
 I1 + I 2 = I Total 
 
 
 From
KCL,
the
same
current
I
flows
 From
KVL,
the
voltage
V
across
the
 through
both
resistors.
 two
resistors
is
the
same.

 From
Ohm’s
law:

 From
Ohm’s
law:

 V1 = I ⋅ R1; V2 = I ⋅ R2 
 I 3 = V ⋅ G3; I 4 = V ⋅ G4 
 
 
 Combine
and
obtain:
 Combine
and
obtain:
 
 Figure
2‐17.
When
more
than
2
resistors
are
connected
in
parallel,
the
 same
voltage
is
applied
across
all
resistors,
while
the
total
current
is
 divided
among
them
(compare
with
Figure
2‐15).
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2010
Alexander
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Law
 
 I ⋅ ( R1 + R2 ) = VTotal 
 
 thus
 V I = Total [equation
2‐8]
 R1 + R2 
 Finally,

 R1 R2 
 V1 = VTotal ⋅ ; V2 = VTotal ⋅ R1 + R2 R1 + R2 [equation
2‐10]
 

 Thus
the
total
voltage
is
divided
in
 direct
proportionality
to
resistances.
 
 
 The
power
P
absorbed
by
each
 resistor
equals
the
product
of
current
 I
and
voltage
V1
(or
V2):
 ⎛ 1 ⎞⎛ R1 ⎞ P1 = ⎜ VTotal ⋅ ⋅ ⎜ VTotal ⋅ R1 + R2 ⎟ ⎝ R1 + R2 ⎟ ⎝ ⎠ ⎠ V ⋅ ( G3 + G4 ) = I Total 
 
 thus
 I V = Total [equation
2‐9]
 G3 + G4 
 Finally,
 G3 G4 
 I 3 = I Total ⋅ ; I 4 = I Total ⋅ G3 + G4 G3 + G4 [equation
2‐11]
 
 Thus
the
total
current
is
divided
in
 direct
proportionality
to
 conductances.
 
 The
power
P
absorbed
by
each
 resistor
equals
the
product
of
voltage
 V
and
current
I3
(or
I4):
 ⎛ 1 ⎞⎛ G3 ⎞ P3 = ⎜ I Total ⋅ ⋅ ⎜ I Total ⋅ G3 + G4 ⎟ ⎝ G3 + G4 ⎟ ⎝ ⎠ ⎠ 
 2 2 
 P1 = I ⋅ R1 P3 = V ⋅ G3 
[equation
2‐12]
 [equation
2‐13]
 
 
 Similarly,
 Similarly,

 2 P2 = I ⋅ R2 
 P4 = V 2 ⋅ G4 
 
 
 If
more
than
2
resistors
are
 If
more
than
2
resistors
are
 connected
in
series
(see
Figure
2‐16),
 connected
in
parallel
(see
Figure
2‐ the
formulas
expand
naturally:

 17),
the
formulas
expand
naturally:

 VTotal I Total 
 
 I= V= R1 + R2 + ... + RN G3 + G4 + ... + GM [equation
2‐14]
 
[equation
2‐15]
 R1 G3 
 
 V1 = VTotal ⋅ I 3 = I Total ⋅ R1 + R2 + ... + RN G3 + G4 + ... + GM [equation
2‐16]
 
[equation
2‐17]
 and
so
forth.

 and
so
on.

 
 
 In
the
case
of
series
connection,
the
voltage
is
divided
between
two
resistors
(or
 among
many
resistors)
in
direct
proportionality
with
their
resistances.
This
result
is
 ©
2010
Alexander
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law
 Ganago
/
Ohm’s
Law
 
 simple,
straightforward,
and
intuitively
clear:
if
the
same
current
I
flows
through
 different
resistors
R1
and
R2
then,
according
to
Ohm’s
law,

 V1 I ⋅ R1 R1 .

 = = V2 I ⋅ R2 R2 
 In
other
words,
you
need
to
apply
more
voltage
across
a
larger
resistance
in
order
to
 push
the
same
current
through
it.

 
 Current
division
might
seem
less
intuitive
because
equations
[2‐11,
2‐13,
and
2‐15]
 are
written
for
conductances.
However,
these
equations
can
be
rewritten
using
 resistances;
for
2
resistors
the
results
are
elegant
but
for
many
resistors
the
algebra
 becomes
cumbersome.
From
equation
[2‐11]
we
obtain:

 1 G3 R4 R3 I 3 = I Total ⋅ = I Total ⋅ = I Total ⋅ 1 1 G3 + G4 R3 + R4 + R3 R4 

 [equation
2‐18]
 1 G4 R3 R4 I 4 = I Total ⋅ = I Total ⋅ = I Total ⋅ 1 1 G3 + G4 R3 + R4 + R3 R4 
 I R Note
that
 3 = 4 ;
in
other
words,
the
current
in
the
left
branch
of
the
circuit
in
 I 4 R3 Figure
2‐15
is
proportional
to
the
resistance
in
the
right
branch
of
this
circuit.
An
 intuitive
explanation
of
this
result
is
that,
in
order
to
maintain
the
same
voltage
 across
both
resistors,
we
have
to
push
more
current
through
the
smaller
resistor.

If
 more
than
2
resistors
are
connected
in
parallel,
the
current
through
each
of
them
is
 elegantly
expressed
through
conductances
[equation
2‐17]
but
the
expression
with
 resistances
becomes
awkward
.

 
 Now
we
can
explore
the
question
of
which
resistor
–
the
larger
or
the
smaller
one
–
 absorbs
more
power.
Interestingly,
the
answer
depends
on
the
connection
between
 resistors,
not
on
the
resistances
themselves!
If
resistors
are
connected
in
series,
 the
power
absorbed
by
each
resistor
is
directly
proportional
to
its
resistance
 (see
equation
[2‐12])
but
if
resistors
are
connected
in
parallel,
the
power
 absorbed
by
each
resistor
is
directly
proportional
to
its
conductance,
or
 inversely
proportional
to
its
resistance
(see
equation
[2‐13]).

 
 <Text
box
on
the
margin>
When
resistors
are
connected
in
series,
the
 larger
resistor
absorbs
more
power.
Conversely,
when
resistors
are
 connected
in
parallel,
the
smaller
resistor
absorbs
more
power!

 
 ©
2010
Alexander
Ganago

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law
 Ganago
/
Ohm’s
Law
 
 Note
that
the
same
rules
apply
to
the
general
cases,
in
which
(a)
more
than
2
 resistors
are
connected
in
series,
and
(b)
more
than
2
resistors
are
connected
in
 parallel.

 
 Examples
and
Practice
Problems
 
 We
begin
with
Examples
and
Practice
Problems,
which
I
like
to
call
the
“Fireworks
 #1
and
#2”
because
they
focus
on
the
safe
power
limits,
and
you
already
know
that
 if
these
limits
are
surpassed,
unwanted
fireworks
may
happen
on
the
circuit
board.
 In
both
cases,
we
work
with
circuits
built
of
the
same
two
resistors;
the
difference
is
 in
their
connection.

 
 
 Example
2‐5
“Fireworks
#1”
 
 In
“Fireworks
#1”
we
examine
a
voltage
divider
built
of
two
resistors
(Figure
2‐14),
 in
which
both
resistances
are
given
along
with
the
power
ratings
of
resistors,
and
 the
applied
voltage
VTotal
is
variable.
Suppose
that
VTotal
is
gradually
increased
so
that
 the
power
absorbed
by
each
resistor
grows.
The
question
is,
at
what
voltage
level
 the
power
absorbed
by
one
of
the
resistors
reaches
its
safety
limit,
while
the
other
 resistor
still
absorbs
the
power
below
its
limit.

 
 For
simplicity,
let
us
assume
that
the
power
limit
for
both
resistors
is
the
same
and
 equal
to
PMAX
=
¼
W.
For
simplicity,
we
neglect
derating
and
allow
the
maximal
 absorbed
power
to
reach
exactly
¼
W.
Suppose
that
the
resistances
are
R1
=
100
Ω
 and
R2
=
200
Ω.
From
the
discussion
above,
we
already
know
that
in
a
series
 connection
the
larger
resistor
absorbs
higher
power
(see
equation
[2‐12]),
thus
our
 calculations
are
focused
on
the
larger
resistor
R2.
At
a
certain
voltage
VTotal,
the
 current
in
the
circuit
equals:


 V I = Total 
 R1 + R2 
 and
the
power
absorbed
by
R2
can
be
found
as:
 ⎛V ⎞ I ⋅ R2 = ⎜ Total ⎟ ⋅ R2 
 ⎝ R1 + R2 ⎠ 
 Assuming
that
this
power
equals
PMAX,
we
obtain
VTotal,
MAX
:
 2 ⎛ VTotal , MAX ⎞ ⎜ R + R ⎟ ⋅ R2 = PMAX ⎝1 2⎠ 

 VTotal , MAX PMAX PMAX = thus VTotal , MAX = ( R1 + R2 ) ⋅ R1 + R2 R2 R2 
 2 2 ©
2010
Alexander
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Ohm’s
Law
 
 Substitute
and
obtain
VTotal,
MAX
=
10.6
V.

 
 Thus,
in
order
to
operate
this
circuit
within
its
safety
limits,
one
should
never
apply
 VTotal
above
10.6
V.
Note
that
the
total
voltage
of
10.6
V
is
divided
between
the
two
 resistors:
7.07
V
is
dropped
across
the
larger,
200‐Ω
resistor
and
3.53
V
across
the
 smaller,
100‐Ω
resistor.
Also
note
that,
at
this
voltage,
the
100‐Ω
resistor
absorbs

 ( 3.53 V )2 = 0.1246 W = 124.6 mW 
 P1 = 100 Ω 
 which
is
well
below
its
power
rating
of
¼
W.
 
 We
can
also
ask
at
what
current
I
the
power
limit
is
reached.
This
problem
is
much
 simpler,
because
we
already
know
that
it
is
the
larger
resistor
that
reaches
its
power
 limit
first,
and
all
that
remains
to
be
done
is
to
apply
equation
[2‐12]
to
the
 condition,
at
which
the
power
limit
is
reached:
 2 PMAX = ( I MAX ) ⋅ R2 
 
 Therefore,

 PMAX 
 IMAX = R2 
 Substitute
the
given
numerical
values
and
obtain:

 0.25 W I MAX = = 0.0354 A = 35.4 mA 
 200 Ω 
 It
is
a
good
habit
to
check
the
consistency
of
two
solutions
of
the
same
circuit:
 according
to
KCL,
KVL,
and
Ohm’s
law,
the
product
of
IMAX
and
the
sum
of
resistances
 should
equal
VMAXwe
calculated
earlier.
Do
the
calculation
yourself
and
verify
that
 the
two
solutions
agree.

 
 If
we
follow
the
practical
guideline,
which
demands
power
derating
by
a
factor
of
2,
 1 and
allow
each
resistor
to
absorb
no
more
than
 W = 0.125 W = 125 mW ,
the
 8 maximal
voltage
applied
to
the
voltage
divider
should
be
limited
at
 PMAX VMAX , practical = ( R1 + R2 ) ⋅ = 7.5 V .
 2 ⋅ R2 
 As
an
exercise,
calculate
the
maximal
current
through
this
circuit
according
to
this
 practical
guideline,
and
verify
the
consistency
of
two
solutions.

 
 
 ©
2010
Alexander
Ganago

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law
 Ganago
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Ohm’s
Law
 
 
Practice
problem
2‐5
(Moderate)
 
 In
the
circuit
of
Figure
2‐14,
the
resistances
are:
R1=
100
Ω
and
R2
=
270
Ω.
The
total
 voltage
applied
to
the
circuit
is
increased
from
1
V
in
1‐V
steps.
Determine
the
 maximal
voltage,
at
which
the
power
absorbed
by
each
resistor
remains
below
¼
W.

 
 Answer:
11
V.

 
 
 Example
2‐6
“Fireworks
#2”
 
 Now
we
can
proceed
to
“Fireworks
#2”,
which
is
the
current
divider
built
of
two
 resistors
as
shown
in
Figure
2‐15.
Again,
both
resistors
are
given
along
with
their
 power
ratings,
and
the
applied
voltage
V
is
variable.
For
a
straightforward
 comparison
between
voltage
division
and
current
division,
let
us
use
the
same
 resistance
values
R1
=
100
Ω
and
R2
=
200
Ω,
and
the
same
power
rating
of
¼
W
for
 both
resistors.

 
 Suppose
that
V
is
gradually
increased
from
zero
so
that
the
power
absorbed
by
each
 resistor
grows.
The
question
is,
at
what
voltage
level
the
power
absorbed
by
one
of
 the
resistors
reaches
its
safety
limit,
while
the
other
resistor
still
absorbs
the
power
 below
its
limit.
Again,
for
the
sake
of
simplicity,
we
neglect
power
derating.
 
 From
our
earlier
discussion,
we
know
that
when
resistors
are
connected
in
parallel,
 the
smaller
one
absorbs
more
power
(see
equation
[2‐13]).
Thus
the
focus
of
our
 calculation
is
on
R1
=
100
Ω.
When
the
power
limit
is
reached,
we
obtain:
 PMAX (VMAX )2 
 = R1 
 Substitute
the
numerical
values
and
solve
for
VMAX:

 VMAX = PMAX ⋅ R1 = ( 0.25 W ) ⋅ (100 Ω ) = 5 V 
 
 Note
that
the
value
of
VMAX
that
we
just
found
is
much
lower
than
VTotal,
MAX
,
which
 we
obtained
in
the
case
of
voltage
division.
Of
course,
in
the
parallel
circuit
the
 whole
voltage
V
is
applied
across
each
resistor,
while
in
the
series
circuit
the
total
 voltage
is
divided
between
the
two
resistors!
Note
that,
at
this
voltage
applied
to
 the
parallel
circuit,
the
200‐Ω
resistor
absorbs

 ( 5 V )2 = 0.125 W = 125 mW 
 P2 = 200 Ω 
 which
is
well
below
its
power
rating.

 
 ©
2010
Alexander
Ganago

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law
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/
Ohm’s
Law
 
 We
can
continue
our
comparison
between
voltage
and
current
division
by
slightly
 changing
the
condition
for
the
parallel
circuit:
suppose
that
the
total
current
ITotal
is
 gradually
increased
and
determine
its
value,
at
which
one
of
the
resistors
reaches
its
 power
rating
while
the
other
absorbs
lesspower.
Again,
we
know
that
the
smaller,
 100‐Ω
resistor
will
absorb
more
power
in
the
parallel
circuit;
specifically,
at
the
 maximal
safe
power
level:

 
 PMAX = ( I1, MAX ) 2 ⎛ R2 ⎞ ⋅ R1 = ⎜ I Total , MAX ⋅ ⋅ R1 
 R1 + R2 ⎟ ⎝ ⎠ 2 
 Solve
for
ITotal,
MAX
,
substitute
the
numerical
values
and
obtain:
 ⎛ R⎞ P ⎛ 100 Ω ⎞ 0.25 W I Total , MAX = ⎜ 1 + 1 ⎟ ⋅ MAX = ⎜ 1 + = 0.075 A = 75 mA 
 ⎟⋅ ⎝ R2 ⎠ R1 200 Ω ⎠ 100 Ω ⎝ 
 Note
that
this
current
is
more
than
2‐fold
above
the
maximal
current
in
the
series
 circuit,
which
we
calculated
in
Example
1.
Naturally,
in
the
parallel
circuit
the
 total
current
is
divided
between
the
two
resistors!
 
 
 
 Practice
problem
2‐6
(Moderate)

 
 Consider
the
circuit
of
Figure
2‐15
built
of
two
resistors
R1=
270
Ω
and
R2
=
390
Ω,
 each
rated
at
1
W.
Use
the
power
derating
by
a
factor
of
2
and
determine
the
 maximal
total
current,
at
which
the
resistors
do
not
overheat.

 
 Answer:
72.8
mA.
 
 
 
 Application
2‐4.
What
exactly
is
a
“short”
circuit?

 
 You
have
probably
heard
of
the
so‐called
“short
circuit”,
a
cause
of
failure
in
electric
 and
electronic
devices.
For
example,
the
“short
circuit”
condition
can
be
 inadvertently
created
when
a
piece
of
wire
gets
connected
in
parallel
with
a
resistor
 as
shown
in
Figure
2‐19.

This
circuit
diagram
is
a
twin
of
Figure
2‐15
thus
the
rules
 for
current
division
apply.

 
 ©
2010
Alexander
Ganago

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law
 Ganago
/
Ohm’s
Law
 
 
 
 Figure
2‐19.
If
a
piece
of
wire
is
connected
in
parallel
to
a
resistor
R,
 and
the
wire
resistance
RW
is
very
small
RW
<<
R,
then
nearly
all
 current
flows
through
the
wire.
If
the
voltage
V
is
fixed,
for
example
by
 the
power
supply,
the
current
ITotal
could
get
dangerously
high.

 
 In
particular,
if
the
wire
resistance
RW
is
very
small
compared
to
the
resistance
R
of
 the
resistor,
that
is
RW
<<
R,
then
equations
[2‐11
and
2‐18]
manifest
that
practically
 no
current
flows
through
the
resistor,
that
is

 IR
<<
ITotal
,

because
nearly
all
ITotal
bypasses
the
resistor
and
flows
through
the
wire;
 as
a
result,
practically
no
power
is
absorbed
by
the
resistor.
This
is
what
Electrical
 Engineers
mean
by
saying
that
the
resistor
is
“shorted
out”
or
“short‐circuited.”

 
 A
“short
circuit”
is
created
whenever
there
happens
to
be
a
low‐resistance
bypass
 for
current
around
a
component
or
entire
circuit:
what
matters
is
the
ratio
of
 R resistances
 W thus
it
is
not
necessarily
a
piece
of
wire
that
creates
a
“short
circuit”.
 R For
example,
a
1‐MΩ
resistor
can
be
shorted
by
a
100‐Ω
resistor
connected
in
 parallel,
because
the
100‐Ω
resistor
will
get
99.99%
of
the
current
and
of
the
power.
 In
other
words,
any
creation
of
a
low‐resistance
bypass
around
a
component
or
 entire
circuit
leads
to
the
“short
circuit”
condition.
 
 Short
circuit
conditions
are
very
dangerous
not
only
because
a
particular
resistor
 (or
part
of
the
circuit)
does
not
get
any
power
from
the
source:
more
serious
is
the
 high
current
that
can
develop
in
the
circuit
due
to
the
unexpectedly
low
resistance.
 V According
to
Ohm’s
law,
the
current
through
the
wire
 IW = 
can
get
extremely
 RW high
if
 RW → 0 and
the
voltage
V
is
fixed
by
the
source;
any
bypass
that
has
a
low
 resistance
can
cause
a
similar
increase
of
current.
Then
the
total
current
drawn
 ©
2010
Alexander
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law
 Ganago
/
Ohm’s
Law
 
 from
the
source
may
exceed
safety
levels,
the
bypass
may
absorb
monstrous
power,
 and
something
might
blow
up:
in
the
best‐case
scenario,
it
will
be
a
fuse;
otherwise,
 the
bypass,
the
power
supply,
etc.
Good
power
supplies
are
protected
from
 dangerously
high
current
by
their
internal
circuitry,
but
it
always
remains
the
 responsibility
of
the
user
to
ensure
that
the
power
connections
do
not
incur
short
 circuits.

 
 

 
 
 2.5.
What
is
the
equivalent
resistance?
 Equivalent
resistance
is
one
of
the
most
powerful
and
commonly
used
concepts
in
 circuit
analysis.
In
this
section
we
will
explain
this
concept
and
show
its
applications
 to
several
typical
problems.

 
 
 2.5.1.
The
concept
of
equivalent
resistance
 
 Application
of
Ohm’s
law
to
an
entire
circuit
or
a
part
of
the
circuit
leads
to
the
very
 powerful
concept
of
equivalent
resistance
 Req 
and
equivalent
conductance
 Geq = 1 ,
illustrated
in
Figure
2‐22.
The
equivalent
resistance
is
defined
as
the
 Req ratio
of
the
total
voltage
applied
to
the
circuit
or
its
part
to
the
total
current
 through
this
circuit;
the
equivalent
conductance
is
its
reciprocal:
 V 1 I 
 
 
 [equation
2‐19]
 Req = Total ; Geq = = Total 
 I Total Req VTotal 
 
 ©
2010
Alexander
Ganago

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law
 Ganago
/
Ohm’s
Law
 
 
 Figure
2‐22.
The
concept
of
equivalent
resistance
and
conductance
 stems
from
the
application
of
Ohm’s
law
to
the
entire
circuit
or
a
part
 of
the
circuit.


 <Text
box
on
the
margin>
Equivalent
resistance
is
defined
as
the
ratio
 of
voltage
applied
to
the
circuit
(or
its
part)
to
the
current
through
 this
circuit
(or
the
part
of
it).
 
 Here,
the
circuit
can
be
as
simple
as
a
pair
of
resistors
–
or
as
complex
as
an
entire
 laptop
or
cell
phone.
Reducing
a
whole,
complex
circuit
to
a
single
resistor
Req
 makes
sense
for
an
application,
in
which
you
focus
only
on
the
overall
voltage
and
 current
but
not
on
the
particulars
of
the
circuitry:
for
example,
when
you
test
a
 battery
for
your
laptop,
you
probably
do
not
fret
about
what
exactly
happens
in
its
 central
processing
unit
(CPU);
of
course,
in
such
applications,
Req
may
be
an
average
 value
because
the
laptop
consumes
less
current
in
the
sleep
mode
and
more
current
 when
you
play
a
video
game.
For
a
simple
circuit,
such
as
you
study
in
this
course,
 the
concept
of
equivalent
resistance
helps
you
find
all
circuit
parameters
in
the
most
 effective
way.

 
 
 2.5.2.
How
to
apply
the
concept
of
equivalent
resistance
to
circuit
 analysis?
 
 Let
us
begin
with
the
simplest
series
and
parallel
connections
of
resistors,
which
we
 learned
as
voltage
and
current
dividers.
If
two
resistors
are
connected
in
series,
as
 shown
in
Figure
2‐23,
voltage
division
takes
place,
and
we
can
easily
derive
that
the
 equivalent
resistance
equals
the
sum
of
both
resistances:

 V VV [equation
2‐20]
 Req = Total = 1 + 2 = R1 + R2 for resistors in series 
 I Total I I 
 
 ©
2010
Alexander
Ganago

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law
 Ganago
/
Ohm’s
Law
 
 
 Figure
2‐23.
The
equivalent
resistance
equals
the
sum
of
resistances
 in
a
circuit
with
series
connection.

 
 This
result
can
be
extended
to
the
case
of
many
resistors
in
series,
as
shown
in
 Figure
2‐24.

 
 
 
 Figure
2‐24.
For
any
number
of
resistors
connected
in
series,
their
 equivalent
resistance
equals
the
sum
of
resistances.

 
 Req = R1 + R2 + ... + RN for resistors in series 
 [equation
2‐21]
 
 <Text
box
on
the
margin>
Resistances
in
series
add.
 
 
 ©
2010
Alexander
Ganago

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law
 Ganago
/
Ohm’s
Law
 
 In
particular,
if
all
resistances
are
equal

 R1 = R2 = ... = RN = R 

 
 then
the
equivalent
resistance
in
the
series
circuit
is:

 Req = N ⋅ R 
 
 
 
 
 
 [equation
2‐22]

 
 Quite
often,
we
see
2
identical
resistors
R
in
series,
then
Req
=
2
⋅
R.
 
 
 If
you
choose
to
memorize
only
a
few
statements
and
equations
from
Electrical
 Engineering,
do
not
skip
Resistances
in
series
add
and
equation
[2‐21]
because
 they
really
help
in
many
practical
applications.
For
example,
if
you
add
longer
wires
 between
the
speaker
and
the
amplifier
in
your
home
theater,
the
equivalent
 resistance
increases.

 
 Another
practical
consequence
is
that
in
a
series
circuit
the
equivalent
resistance
 is
larger
than
the
largest
resistor.
This
might
sound
trivial
but
only
until
we
 consider
the
parallel
connection.

 
 <Text
box
on
the
margin>
In
a
series
circuit,
the
equivalent
resistance
 is
larger
than
the
largest
resistor.
 
 If
two
resistors
are
connected
in
parallel,
as
shown
in
Figure
2‐25,
current
division
 takes
place,
and
conductances
add:

 I II Geq = Total = 1 + 2 = G1 + G1 V VV 1 1 1 

 
 [equation
2‐23]
 =+ Req R1 R2 Req = 
 R1 ⋅ R2 for two resistors in parallel R1 + R2 ©
2010
Alexander
Ganago

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2010‐01‐11
Ohm’s
law
 Ganago
/
Ohm’s
Law
 
 
 If
more
than
2
resistors
are
connected
in
parallel,
as
shown
in
Figure
2‐26,
their
 conductances
add

 Geq = G1 + G2 + ... + GN for resistors in parallel 
 
 [equation
2‐24]
 but
the
equation
for
resistances
is
awkward:


 1 1 1 1 =+ + ... + for resistors in parallel 
 
 Req R1 R2 RN 
 [equation
2‐25]
 
 Figure
2‐25.
If
two
resistors
are
connected
in
parallel,
the
 conductances
add;
the
equivalent
resistance
is
found
from
equation
 [2‐25]
 
 Figure
2‐26.
If
many
resistors
are
connected
in
parallel,
the
 conductances
add;
the
equivalent
resistance
is
found
from
equation
 [2‐25]
 ©
2010
Alexander
Ganago

 Page
21
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29

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2010‐01‐11
Ohm’s
law
 Ganago
/
Ohm’s
Law
 
 
 
 In
particular,
if
all
resistances
are
equal

 R1 = R2 = ... = RN = R 

 
 then
the
equivalent
resistance
in
the
parallel
circuit
is:

 R 
 
 
 
 
 
 [equation
2‐26]
 Req = 
 N 
 Most
often,
we
observe
2
identical
resistors
R
in
parallel,
then
Req
=
R/2.
 
 Note
that,
in
a
parallel
circuit,
the
equivalent
resistance
is
smaller
than
the
smallest
 resistor.
 
 <Text
box
on
the
margin>
In
a
parallel
circuit,
the
equivalent
 resistance
is
smaller
than
the
smallest
resistor.
 
 Recall
our
discussion
of
the
short‐circuit
condition:
when
a
resistor
is
in
parallel
 with
a
bypass,
their
equivalent
resistance
becomes
less
than
the
bypass’
resistance.

 
 In
many
circuits,
we
can
identify
the
parts
reducible
to
parallel
and
series
 combinations.
Step‐by‐step,
such
circuits
can
be
reduced
to
their
equivalent
 resistance.
Figure
2‐27
presents
a
straightforward
example.

 
 
 
 

 Figure
2‐27.
This
simple
circuit
can
be
reduced
to
a
combination
of
 series
and
parallel
connections.

 ©
2010
Alexander
Ganago

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law
 Ganago
/
Ohm’s
Law
 
 The
circuit
in
Figure
2‐27
has
resistor
R1
in
series
with
(R2||R3);
here
we
use
the
 symbol
||
to
denote
resistors
connected
in
parallel.
We
already
found
that
the
 resistance
of
R2
and
R3
in
parallel
can
be
expressed
as:
 R ⋅R ( R2 || R3 ) = 2 3 
 R2 + R3 
 and
we
know
that
the
equivalent
resistance
is
the
sum
of
two
resistances
in
series:

 RR R R + R2 R3 + R1 R3 
 Req = R1 + ( R2 || R3 ) = R1 + 2 3 = 1 2 R2 + R3 R2 + R3 
 The
entire
circuit
can
be
seen
as
a
voltage
divider
formed
by
R1
and
(R2||R3);
the
 total
current
equals
:
 VTotal R2 + R3 
 I Total = = VTotal ⋅ R1 + ( R2 || R3 ) R1 R2 + R2 R3 + R1 R3 
 From
Ohm’s
law,
we
obtain
the
voltage
across
resistor
R1
:
 R1 R1 R2 + R1 R3 
 V1 = VTotal ⋅ = VTotal ⋅ R1 + ( R2 || R3 ) R1 R2 + R2 R3 + R1 R3 
 According
to
KVL,
the
voltages
across
resistors
R2
and
R3
are
equal;
they
can
be
 calculated
from
Ohm’s
law:

 
 ( R2 || R3 ) = V ⋅ R2 R3 
 V2 = V3 = VTotal ⋅ Total R1 + ( R2 || R3 ) R1 R2 + R2 R3 + R1 R3 The
currents
through
individual
resistors
are
obtained
from
current
division:

 R3 R2 
 I 2 = VTotal ⋅ and I 3 = VTotal ⋅ R1 R2 + R2 R3 + R1 R3 R1 R2 + R2 R3 + R1 R3 
 From
these
equations,
we
can
find
the
power
absorbed
by
each
resistor.

 
 
 Problems
and
MCQs:
Examples
and
Practice
 


 Example
2‐9
“Fireworks
#3”
 
 Here
is
the
“Fireworks
#3”:
in
the
circuit
of
Figure
2‐27,
the
resistances
are:
R1
=
5
 Ω,
R2
=
10
Ω,
and
R3
=
15
Ω;
the
voltage
VTotal
is
gradually
increased
from
zero.
All
 resistors
are
rated
at
¼
W.
At
a
certain
voltage
VTotal,
MAX
one
of
the
resistors
absorbs
 250
mW,
while
the
others
absorb
less
than
that.
Determine
which
resistor
absorbs
 more
power
at
any
voltage
VTotal;
calculate
VTotal,
MAX
,
and
find
the
power
absorbed
 by
every
resistor
at
this
voltage.

 
 ©
2010
Alexander
Ganago

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law
 Ganago
/
Ohm’s
Law
 
 
 Solution
 
 The
total
current
flows
through
R1
and
then
is
divided
between
R2
and
R3
so
that

 15 Ω I 2 = I Total ⋅ = 0.6 ⋅ I Total (10 Ω ) + (15 Ω ) 
 10 Ω I 3 = I Total ⋅ = 0.4 ⋅ I Total (10 Ω ) + (15 Ω ) 
 Thus
the
power
absorbed
by
each
resistor
equals,
in
terms
of
ITotal:
 P1 = ( I Total ) ⋅ ( 5 Ω ) = 5 ⋅ ( I Total ) ⋅ (1 Ω ) 2 2 2 2 P2 = ( 0.6 ⋅ I Total ) ⋅ (10 Ω ) = 3.6 ⋅ ( I Total ) ⋅ (1 Ω ) 
 2 P3 = ( 0.4 ⋅ I Total ) ⋅ (15 Ω ) = 2.4 ⋅ ( I Total ) ⋅ (1 Ω ) 
 It
means
that,
at
any
given
current
ITotal
,
the
5‐Ω
resistor
absorbs
more
power
than
 the
10‐Ω
or
15‐Ω
resistors.
Next,
we
can
find
at
what
value
ITotal,
MAX

the
10‐Ω
resistor
 absorbs
exactly
250
mW:
 0.25
W
=
5
·
(
ITotal,
MAX
)2
·
(
1
Ω
)
 0.25 I Total , MAX = A = 0.224 A 
 5 
 We
can
obtain
VTotal,
MAX
from
Ohm’s
law
as
 VTotal,
MAX
=
ITotal,
MAX
·
Req
 
 where
 Req
=
R1
+
(R2||
R3)
 (10 Ω ) ⋅ (15 Ω ) = 5 Ω + 6 Ω = 11 Ω 
 Req = 5 Ω + (10 Ω ) + (15 Ω ) 
 Thus
 VTotal,
MAX
=
(0.224
A)
·(11
Ω)
=
2.464
V
 
 To
determine
the
power
absorbed
by
each
resistor,
we
can
use
our
results
 (expressed
in
terms
of
ITotal):
 P1=
5
·
(0.224
A)2
·(1
Ω)
=
0.25
W,
as
expected
 P2=
3.6
·
(0.224
A)2
·(1
Ω)
=
0.181
W
 P3=
2.4
·
(0.224
A)2
·(1
Ω)
=
0.120
W
 
 Answer:

At
any
voltage
VTotal,
the
5‐Ω
resistor
absorbs
more
power
than
any
of
the
 other
two
resistors.
 VTotal,
MAX
=
2.464
V
 2 ©
2010
Alexander
Ganago

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Ohm’s
law
 Ganago
/
Ohm’s
Law
 
 At
this
voltage,
the
5‐Ω
resistor
absorbs
250
mW,
 the
10‐Ω
resistor
absorbs
181
mW,
 and
the
15‐Ω
resistor
absorbs
120
mW.
 
 Note
that
in
this
circuit,
which
combines
voltage
and
current
division,
it
is
neither
 the
smallest
nor
the
largest
resistor
that
absorbs
the
highest
power.
This
is
not
a
 rule:
each
circuit,
except
for
the
simplest
ones,
for
which
you
already
know
the
 answers,
deserves
careful
analysis
and
calculations.

 
 
 Practice
problem
2‐9
(Moderate)

 
 The
circuit
of
Figure
2‐27
is
built
with
resistors
 R1
=
100
Ω,

R2
=
200
Ω,

R3
=
600
Ω.
 The
total
voltage
applied
to
the
circuit
is
increased
from
zero
in
1‐V
steps.
 At
a
certain
voltage
VTotal,
M
one
of
the
resistors
absorbs
≥
250
mW,
while
each
of
the
 others
absorbs
less
power.
 
 Questions:
 1. Which
of
the
resistors
reaches
the
250
mW
level
before
the
others?
 2. At
the
voltage
VTotal,
M

does
it
happen?
 3. What
is
the
total
current
through
the
circuit
at
this
voltage?
 4. What
was
the
total
power
dissipated
in
the
circuit
at
this
voltage?
 
 Answers:

 1. R2
reaches
the
250
mW
power
level
before
the
others.
 2. VTotal,
M
=
12
V.
 12 V 3. I
=
 
=
0.048
A
=
48
mA.
 250 Ω 4. PTotal
=
576
mW.
 
 
 Example
2‐10
 
 Figure
2‐28
shows
the
circuit
diagram
where
each
resistance
equals
R;
nodes
B
and
 C
are
connected
with
a
wire
whose
resistance
is
zero.
The
challenge
is
to
find
the
 equivalent
resistors
between
nodes
A
and
B.
 ©
2010
Alexander
Ganago

 Page
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2010‐01‐11
Ohm’s
law
 Ganago
/
Ohm’s
Law
 
 
 
 <Figure
Legend>
Figure
2‐28.
Determine
the
equivalent
resistance
 between
nodes
A
and
B
in
this
circuit
builtof
identical
resistors.
 
 Solution
 
 First
of
all,
notice
that
the
resistor
between
nodes
B
and
C
is
in
parallel
with
a
wire
 thus
it
gets
short‐circuited,
and
no
current
flows
through
it.
Therefore,
we
do
not
 show
this
resistor
on
the
following
circuit
diagrams.

 
 Secondly,
a
key
step
in
the
solution
is
to
identify
the
nodes
in
the
circuit
taking
into
 account
that
–
due
to
the
wire
–
nodes
B
and
C
become
one.
Then
we
can
produce
 step‐by‐step,
redrawing
the
equivalent
circuit
at
each
step.
Note
that
redrawing
the
 diagram
is
one
of
the
most
effective
problem‐solving
skills
not
only
in
this
problem
 as
well
as
in
many
others.

 
 Figure
2‐29
shows
the
first
step:
here
we
omit
the
shorted‐out
resistor
and
label
the
 central
node
D.

 ©
2010
Alexander
Ganago

 Page
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2010‐01‐11
Ohm’s
law
 Ganago
/
Ohm’s
Law
 
 
 <Figure
Legend>
Figure
2‐29.
This
is
step
1
of
our
solution:
we
omit
 the
shorted‐out
resistor
and
label
the
central
node
D.

 
 As
mentioned
above,
due
to
the
wire,
nodes
B
and
C
become
one;
on
the
following
 diagrams,
we
label
this
combined
node
as
B,
C.
The
equivalent
circuit
contains
only
3
 nodes
–
A,
D,
and
the
combined
node
B,
C.
Now
we
do
the
inventory:
examine
each
 resistor
and
identify
between
which
nodes
it
is
connected.
The
result
is
shown
in
 Figure
2‐30.

 

 
 <Figure
Legend>
Figure
2‐30.
This
is
step
2
of
our
solution:
we
 identify
between
which
nodes
each
resistor
is
connected.

 
 So
far,
we
did
not
use
any
equations;
now
the
time
has
come.
Apply
the
equation
for
 parallel
connection
[equation
2‐23
or
2‐26]
to
two
pairs
of
resistors
and
replace
 each
pair
with
its
equivalent
0.5
R,
as
shown
in
Figure
2‐31.

 
 ©
2010
Alexander
Ganago

 Page
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Ohm’s
law
 Ganago
/
Ohm’s
Law
 
 

 <Figure
Legend>
Figure
2‐31.
In
step
3
of
our
solution,
we
replace
two
 pairs
of
identical
resistors
with
their
equivalents
of
0.5
R
each.

 
 From
the
diagram
in
Figure
2‐31,
it
is
clear
that
between
nodes
A
and
B
there
is
a
 0.5
R
resistance
in
parallel
with
(R
+
0.5
R).
Thus
the
equivalent
resistance
equals:

 RAB
=
(R
+
0.5R)||(0.5
R)
=
0.375
R.

 
 Note
that
the
circuit
with
the
wire
is
reducible
to
a
series/parallel
combination
of
 resistors,
while
the
circuit
without
the
wire
is
not.
You
will
learn
how
to
calculate
 the
equivalent
resistance
of
more
complicated
circuits
in
Chapter
6
(?).
 
 ©
2010
Alexander
Ganago

 Page
28
of
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Ohm’s
law
 Ganago
/
Ohm’s
Law
 
 <D>
Practice
problem
2‐10
 
 
 The
integrated
circuit
(IC)
shown
in
Figure
2‐32
has
four
terminals
labeled
A
–
D.
All
 resistances
are
in
ohms.
Terminals
A
and
B
are
connected
with
a
wire
(zero
 resistance).
Determine
the
equivalent
resistance
in
ohms
between
terminals
C
and
 D.

 
 <Figure
Legend>
Figure
2‐32.
The
integrated
circuit
(IC)
has
four
 terminals
labeled
A
–
D.
All
resistances
are
in
ohms.
Terminals
A
and
B
 are
connected
with
a
wire
(zero
resistance).
Calculate
the
equivalent
 resistance
in
ohms
between
terminals
C
and
D.


 
 The
equivalent
resistance
in
ohms
between
terminals
C
and
D
equals...

 A. 13.75

 B. 17.50
 C. 19.375
 D. 21.25
 E. 25.00
 
 Hint:
Redraw
the
circuit
for
each
step
of
solution;
first
of
all,
get
rid
of
the
IC
casing
 (the
square
drawn
with
thick
line).
 
 Answer:
B

 
 
 
 
 ©
2010
Alexander
Ganago

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of
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Ohm’s
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 ...
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This note was uploaded on 12/06/2010 for the course EECS 314 taught by Professor Ganago during the Spring '07 term at University of Michigan.

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