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108
Chapter 2
Resistive Circuits
Problems
Section 2.1: Resistances in Series and Parallel
*P2.1.
Reduce
each of the networks shown in Figure
P2.1
to
a single equivalent resistance by com
bining resistances in series and parallel.
3Q
30Q
12Q
24Q
7Q
4Q
(a)
lOQ
6Q
1s
n
60Q
9Q
sn
(b)
Figure P2.1
*P2.2.
A 4Q resistance
is
in series with
the
paral
lel combination
of
a 20Q resistance and
an
unknown resistance
Rx.
The
equivalent resis
tance for the network
8
Q.
Determine
value of
Rx.
*P2.3.
Find
equivalent resistance looking into
terminals
a
and
bin
Figure P2.3.
Figure P2.3
*P2.4.
Suppose
that
we
need
a resistance of 1.5 kQ
and you have a box
1kQ resistors. Devise
a network of 1kQ resistors so the equivalent
resistance
1.5 kQ.
Repeat
for
equivalent
resistance of 2.2 kQ.
*P2.5.
equivalent resistance between ter
minals
a
Figure P2.5.
Figure P2.5
P2.6.
equivalent resistance between termi
nals
a
b
for each of the networks shown
in Figure P2.6.
b
16!2
20!2
180
n
(a)
a
b
3n
(b)
Figure P2.6
63!2
*
Denotes
answers
are
contained in the
Student
Solutions
files.
See
Appendix
F for
more
information
about
accessing the
Solutions.
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30Q
24Q
6Q
20Q
(c)
Figure P2.6
(Cont.)
P2.7.
What
resistance
in
parallel with 120
Q
results
an
equivalent
of
48
Q?
P2.8.
a.
Determine
the
between
termi
nals
a
and
b
for
network
shown in Figure
P2.8.
b.
Repeat
after
connecting c
d
with
a
short
circuit.
aoi
{)
c
bo

Figure P2.8
P2.9.
Two resistances having values
R
2R
are
parallel.
R
resis
tance
both
integers.
possible
values for
R?
P2.10.
A
connected
terminals
a
b
consists
two parallel combinations
that
in series.
Th
e first parallel combi
nation
is
compo
sed
a 16Q resistor
a
48Q resistor.
The
se
cond
parallel combina
tion
composed
a 12Q resistor
a 24Q
resistor.
Draw
determine
its
resistance.
P2.11.
Two resistances
R1
R2
in
parallel. We
know
R1 =
90
Q
current
through
R2
is
three
times
value
R1.
R2.
P2.12.
Find
infi
nite
shown
Figure P2.12(a).
Becau
se
its form, this
is called
a semiinfinite ladder.
(Hint:
If
another
Problems
1
09
section
added
to
ladder
as
in
Figure P2.12(b ),
is
same.
Thus
, working from Figure
P2.12(b ),
we
can
write
expression for
R
eq
terms
Req.
Then,
solve
Req.)
JQ
>
lQ
2Q
(a)
(b)
Figure P2.12
Ladder
(a)
• • •
P2.13.
connect
n
1000Q resistances
parallel,
what
resistance?
P2.14.
heating
element
electric
cook
top
has
two resistive elements,
R1
=
57.6
Q
R2
=
115.2
Q,
be
operated
separately, in series,
or
parallel from volt
ages
either
120 V
240
V.
For
low
est
power,
is
series with
R
2,
combination is
from 120V.
is
lowest
power?
highest power,
how
should
elements
operated?
power
results? List
more
modes
operation
resulting
for each.
P2.15.
We
designing
electric space
heater
operate
from 120 V. Two heating elements
with resistances
R1
R2
used
parallel, separately,
in
series.
highest
1280 W,
is
240
W.
values
needed
R1
R
2?
intermediate
settings
available?
P2.16.
Sometimes
, we can use symmetry consid
erations
find
a circuit
cannot
reduced
by
series
paral
lel combinations. A classic
problem
this
type is illustrated
Figure P2.16. Twelve
110
Chapter
2
Resistive Circuits
1Q resistors are arranged
on
the
edges
of
a
cube,
and
terminals
a
bare
connected
to
diagonally opposite corners
cube.
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This document was uploaded on 02/14/2012.
 Spring '09
 Mechatronics

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