CHAPTER 21: CIRCUITS, BIOELECTRICITY, AND DC ...

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Oct 7, 2013 (4 years and 1 month ago)

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College  Physics
 
Student  Solutions  Manual
 
Chapter  21
 
157
 
 
CHAPTER  21:  
CIRCUITS,  BIOELECTRI
CITY,  
AND  DC  INSTRUMENTS
 
21.1  
RESISTORS  IN  SERIES  
AND  PARALLEL
 
1.
 
(a)  What  is  the  resistance  of  ten  
Ω
-
275
 
resistors  connected  in  series?  (b)  In  parallel?  
 
Solution
 
(a)  
From  t
he  equation  
...
R
R
R
R








3
2
1
s




 
we  
kn
ow  that  resistors  in  series  add
:
 




Ω

Ω






k

75
.
2


10


275










10
3
2
1
s
R
........
R
R
R
R
 
(b)
 
From  
the  equation  
....
R
R
R



2
1
p
1
1
1
,  we  know  th
at  resistors  in  series  add  like
:
 


Ω
×

⎟
⎠
⎞
⎜
⎝
⎛
Ω






2
10
2
1
p
10
64
.
3

275
1

10
1
1
1
1
R
.........
R
R
R
 
So  that
 
Ω

Ω
⎟
⎠
⎞
⎜
⎝
⎛
×



5
.
27

10
64
.
3
1
2
p
R
 
7.
 
Referring  to  the  example  combining  series  and  parallel  circuits  and  
Figure  21.6
,  
calculate  
3
I
 
in  the  following  two  different  ways:  (a)  from  the  known  values  of  
I
 
and  
2
I
;  (b)  using  Ohm’s  law  for  
3
R
.  In  both  parts  explicitly  show  how  you  follow  the  steps  
in  the  
Problem
-­‐
Solving  Strategies  for  Series  and  Parallel  Resistors
.
 
Solution
 
Step  1:
 
The  circuit  diagram  is  drawn  in  
Figure  21.6
.
 
Step  2:
 
Find  
3
I
.
 
Step  3
:
 
Resistors  
2
R
and  
3
R
 
are  in  parallel.  Then,  resistor  
1
R
 
is  in  series  with  the  
combination  of  
2
R
and  
3
R
.
 
Step  4:
 
College  Physics
 
Student  Solutions  Manual
 
Chapter  21
 
158
 
 
(
a)  Looking  at  the  point  where  the  wire  comes  into  the  parallel  combination  of  
2
R
 
and    
3
R
 
,  we  
see  that  the  current  coming  in
 
I
 
is  equal  to  the  current    going  out  
2
I
 
and  
3
I
 
,  so  that  
,
I
I
I
3
2


or  
A

74
.
0
A

61
.
1
A

35
.
2
2
3





I
I
 
(
b)
 
Using  Ohm’s  law  for  
3
R
,  and  voltage  for  the  combination  of  
2
R
 
and  
3
R
 
,  found  in  
Example  21.3
,  we  can  determine  the  current:  
A

742
.
0

0
.
13
V

65
.
9
3
p
3

Ω


R
V
I
 
Step  5:
 
The  result  is  reasonable  because  it  is  smaller  than  the  incoming  current,  
I
,  
and  both  methods  produce  the  same  answer.
 
21.2  
ELECTROMOTIVE  FORCE:
 
TERMINAL  VOLTAGE
 
15.
 
Carbon
-­‐
zinc  dry  cells  (sometimes  referred  to  as  non
-­‐
alkaline  cells)  have  an  emf  of  1.54  
V,  and  they  are  produced  as  single  cells  or  in  various  combinations  to  form  other  
voltages.  (a)  How  many  1.54
-­‐
V  cells  are  needed  to  make  the  common  9
-­‐
V  battery  
used  in  
many  small  electronic  devices?  (b)  What  is  the  actual  emf  of  the  
approximately  9
-­‐
V  battery?  (c)  Discuss  how  internal  resistance  in  the  series  connection  
of  cells  will  affect  the  terminal  voltage  of  this  approximately  9
-­‐
V  battery.
 
Solution
 
(
a)  To  determine
 
the  number  simply  divide  the  9
-­‐
V  by  the  emf  of  each  cell:
 
6
84
.
5
V

54
.
1
V

9



 
(
b)
 
If  six  dry  cells  are  pu
t  in  series  ,  the  actual  emf  is  
V

24
.
9
6
V

54
.
1

×
 
(c)  Internal  resistance  will  decrease  the  terminal  voltage  because  there  will  be  voltage  
dro
ps  across  the  internal  resistance  that  will  not  be  useful  in  the  operation  of  the  
9
-­‐
V  battery.
 
30.
 
Unreasonable  Results
 
(a)  What  is  the  internal  resistance  of  a  1.54
-­‐
V  dry  cell  that  
supplies  1.00  W  of  power  to  a  
Ω
-
0
.
15
 
bulb?  (b)  What  is  unreasonable  about  this  
result?  (c)  Which  assumptions  are  unreasonable  or  inconsistent?
 
College  Physics
 
Student  Solutions  Manual
 
Chapter  21
 
159
 
 
Solution
 
(a)  Using
 
the  equation
 
A

0.258
Ω

15.0
W
1.00

have

we
,
2




R
P
I
R
I
P
.  So  using  
Ohm’s  L
aw  and
 
Ir
E
V


 
we  have
 
Ω

9.04


Ω

15.0


A

0.258
V

1.54






:
or

,
















R
I
E
r
IR
Ir
E
V
 
(b)  You  can
no
t  
have  negative  resistance.
 
(c)  The  voltage  should  be  less  than  the  emf  of  the  battery;  otherwise  the  internal  
resistance  comes  out  negative.  Therefore,  the  power  delivered  is  too  large  for  the  
given  resistance,  leading  to  a  current  that  is  too  large.
 
21.3  
KIRCHHOFF’S  RULES
 
31.
 
Apply  the  loop  rule  to  loop  abcdefgha  in  
Figure  21.25
.
 
Solution
 
Using  the  loop  rule  for  loop  abcdefgha  in  
Figure  21.25
 
gives:  
 
0

2
2
3
3
3
1
2
1
3
2






- E
r
I

r
I

r
I
E
R
I
 
37.
 
Apply  the  loop  rule  to  loop  akledcba  in  
Figure  21.52
.
 
Solution
 
Using  the  loop  rule  to  loop  akledcba  in  
Figure  21.52
 
gives:  
 
0
1
1
1
1
1
5
1
2
2
2
2
2


R
I

- E
r
I

R
I

R
- I
r
- I
E




 
21.4  
DC  VOLTMETERS  AND  AM
METERS
 
44.
 
Find  the  resistance  that  must  be  placed  in  series  with  a  
Ω
-
0
.
25
 
galvanometer  having  
a  
µA
-
0
.
50
 
sensitivity  (the  same  as  the  one  discussed  in  the  text)  to  allow  it  to  be  used  
as  a  voltmeter  with  a  0.100
-­‐
V  full
-­‐
scale  reading.
 
College  Physics
 
Student  Solutions  Manual
 
Chapter  21
 
160
 
 
Solution
 
We  are  given
 

.
A

0
.
50



and

V,

200
.
0


,


0
.
25





Ω

I
V
r
 
Since  the  resistor
s
 
are  in  series,  the  total  resistance  for  the  voltmeter  is  fou
nd  by  
using  
...
3
2
1
s




R
R
R
R
.  So,  using  Ohm’s  law  we  can  find  the  resistance  
R
:
 
k
Ω
98
.
1

Ω
1975

Ω
0
.
25
A
10
00
.
5
V
100
.
0
that
so

5
tot






r
I
V
R
,
I
V
r
R
R
-



×






 
50.
 
Suppose  you  measure  the  terminal  voltage  of  a  1.585
-­‐
V  alkaline  cell  having  an  
internal  resistance  of  
Ω

100
.
0
 
by  placing  a  
Ω
k
-
00
.
1
 
voltmeter  across  its  terminals.  
(
Figure  21.54
.)  (a)  What  current  flows?  (b)  Find  the  terminal  voltage.  (c)  To  see  how  
close  the  measured  terminal  voltage  is  to  the  emf,  calculate  their  ratio.
 
Solution
 
(a)
   
E
r
V
I
 
Going  counterclockwise  around  the  loop  using  the  loop  rule  gives:
 


A

10


58
.
1

A

10


5848
.
1


100
.
0




10


00
.
1
V

585
.
1



or

0,


3

3
3
×

×

Ω

Ω
×







r
R
E
I
IR
Ir
E
 
(b)  
T
he  terminal  voltage  is  given  by
 
the  equation
 
Ir
E
V


:
 




V

5848
.
1



100
.
0
A

10


5848
.
1
V

585
.
1
3

Ω
×




Ir
E
V
 
Note:  The  answer  is  reported  to  5  
significant  figures  to  see  the  difference.
 
(c)  To  calculate  the  ratio,  divide  the  terminal  voltage  by  the  emf:
 
99990
.
0


V

585
.
1
V

5848
.
1



E
V
 
 
 
College  Physics
 
Student  Solutions  Manual
 
Chapter  21
 
161
 
 
21.5  
NULL  MEASUREMENTS
 
 
58.
 
Calculate  the  
x
emf
of
 
a  dry  cell  for  which  a  potentiometer  is  balanced  when  
Ω


200
.
1
x
R
,  while  an  alkaline  standard  cell  with  an  emf  of  1.600  V  requires  
Ω


247
.
1
s
R
 
to  balance  the  potentiometer.
 
Solution
 
We  know
 

,

and

s
s
x
x
IR
E
IR
E


so  that  


V

540
.
1


247
.
1

200
.
1

V

600
.
1

or

,
s
x
s
x
s
x
s
x
s
x

⎟
⎠
⎞
⎜
⎝
⎛
Ω
Ω

⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛



R
R
E
E
R
R
I
I
E
E
 
21.6  
DC  CIRCUITS  CONTAINI
NG  RESISTORS  AND  CAP
ACITORS
 
63.
 
The  timing  device  in  an  automobile’s  intermittent  wiper  system  is  based  on  an  
RC
 
time  constant  and  utilizes  a  
µF
-
.500
0
 
capacitor  and  a  variable  resistor.  Over  what  
range  must  
R
 
be  made  to  vary  to  achieve  time  constants  from  2.00  to  15.0  s?
 
Solution
 
From  
the  equation  

,
RC
τ

 
we  know  that:
 
Ω
×

×


Ω
×

×



10


00
.
3


F

10


5.00
s

0
.
15


and


10


00
.
4


F

10


5.00
s

00
.
2
7
7
-
6
7
-
C
τ
R
C
τ
R
 
Therefore,  the  range  for  
R
 
is:
 
Ω

Ω
×

Ω
×
M

30.0

to
00
.
4
10


00
.
3


10

00
.
4

7
6
 
69.
 
A  heart  defibrillator  being  used  on  a  patient  has  an  
RC
 
time  constant  of  10.0  ms  due  
to  the  resistance  of  the  patient  and  the  capacitance  of  the  defibrillator.  (a)  If  the  
defibrillator  has  an  
µF
-
00
.
8
 
capacitance,  what  is  the  resistance  of  the  path  through  
the  patient?  (You  may  neglect  the  capaci
tance  of  the  patient  and  the  resistance  of  the  
defibrillator.)  (b)  If  the  initial  voltage  is  12.0  kV,  how  long  does  it  take  to  decline  to  
V

10
0
0
.
6
2
×
?
 
Solution
 
(a)  Using
 
the  equation
 
RC

τ
we  can  calculate  the  resistance
:
 
College  Physics
 
Student  Solutions  Manual
 
Chapter  21
 
162
 
 
Ω

Ω
×

×
×



k

25
.
1


10
25
.
1

F

10
00
.
8
s

10
00
.
1
3
6
-
2
C
τ
R
 
(b)  
U
sing  
the  equation  
,

0
RC
-
e
V
V
τ

 
we  can  calculate  the  time  it  takes  for  the  
voltage  to  drop  from  
:
V

600

to
kV

12.0
 




ms

0
.
30


s

10

996
.
2

V

10
1.20
V

600
ln

F

10
00
.
8


10

25
.
1

ln
2
-
4
6
-
3
0

×

⎟
⎠
⎞
⎜
⎝
⎛
×
×
Ω
×


⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛


V
V
-RC
τ
 
74.
 
Integrated  Concepts
 
If  you  wish  to  take  a  picture  of  a  bullet  traveling  at  500  m/s,  then  
a  very  brief  flash  of  light  produced  by  an  
RC
 
discharge  through  a  flash  tube  can  limit  
blurring.  Assuming  1.00  mm  of  motion  during  one  
RC
 
constant  
is  acceptable,  and  
given  that  the  flash  is  driven  by  a  
µF
-
600
 
capacitor,  what  is  the  resistance  in  the  flash  
tube?
 
Solution
 
Using  
v
x
t
t
x
c


or

   
and  
the  equation  for  the  time  constant,
 
we  can  write  the  time  
constant  as  
RC

τ
,  so  getting  these  two  times  equal  gives  an  expression  from  which  
we  can  solve  for  the  required  resistance:
Ω

10


3.33


F)

10


m/s)(6.00

(500
m

10


1.00



or
,


3
4
3



×

×
×



vC
x
R
RC
v
x