Download Homework Set 4

Homework Set #4 Solutions Chapter 18
18.11
The equivalent resistance is Req  R  Rp , where R p is the total resistance of the three
parallel branches;
 30   R  5.0  
1
1
1
 1

 1

Rp  







R  35 
 120  40  R  5.0  
 30  R  5.0  
1
Thus, 75   R 
1
 30   R  5.0    R 2   65   R  150 2
R  35 
R  35 
which reduces to R2  10   R  2 475 2  0 or  R  55   R  45    0 .
Only the positive solution is physically acceptable, so R  55 
18.12
The resistance of the parallel combination of the
3.00  and 1.00  resistors is
1
1 
 1
Rp  

  0.750 
 3.00  1.00  
The equivalent resistance of the circuit connected to
the battery is
Req  2.00   Rp  4.00   6.75 
and the current supplied by the battery is
I
V 18.0 V

 2.67 A
Req 6.75 
The power dissipated in the 2.00- resistor is
P2  I 2 R2   2.67 A   2.00    14.2 W
2
and that dissipated in the 4.00- resistor is
P4  I 2 R4   2.67 A   4.00    28.4 W
2
The potential difference across the parallel combination of the 3.00  and 1.00 
resistors is
 V  p  I Rp   2.67 A0.750   2.00 V
Thus, the power dissipation in these resistors is given by
 V  p  2.00 V 2
P

 1.33 W
 V  p  2.00 V 2
P

 4.00 W
2
3
3.00 
R3
2
and
18.27
1
1.00 
R1
Qmax  C    5.0  106 F   30 V   1.5 104 C , and
  RC  1.0  106   5.0  106 F   5.0 s
Thus, at t  10 s  2

 


Q  Qmax 1  et   1.5 104 C 1  e2  1.3 104 C
18.30
(a)
I max 

, so the resistance is
R
R

I max

48.0 V
 9.60  104 
0.500 10-3 A
The time constant is   RC , so the capacitance is found to be
C

R

0.960 s
 1.00  105 F  10.0  F
4
9.60  10 
(b) Qmax  C   10.0  F  48.0 V   480 C , so the charge stored in the capacitor at
t  1.92 s is
1.92 s



Q  Qmax 1  e t    480 C  1  e 0.960 s    480 C  1  e 2  415 C






18.34
The maximum power available from this line is
Pmax   V  I max  120 V 15 A   1800 W
Thus, the combined power requirements (2 400 W) exceeds the available power, and you
cannot operate the two appliances together .
18.35
(a) The area of each surface of this axon membrane is
A
 2 r   0.10 m 2 10 106 m  2 106 m2
and the capacitance is
C   0
 2  106 m2 
A
8
 3.0 8.85  1012 C2 N  m2  
  1.67  10 F
-8
d
1.0

10
m


In the resting state, the charge on the outer surface of the membrane is



Qi  C  V i  1.67 108 F 70 103 V  1.17 109 C  1.2 109 C
The number of potassium ions required to produce this charge is
N K 
Qi 1.17  109 C

 7.3  109 K + ions
e
1.6  10-19 C
and the charge per unit area on this surface is

20
2
Qi 1.17  109 C 
1e
1e
  10 m 





-6
2 
-19
2
A 2  10 m  1.6  10 C   1 Å  8.6  104 Å 2
1e
 290 Å 
2
This corresponds to a low charge density of one electronic charge per square of side
290 Å, compared to a normal atomic spacing of one atom per several Å2 .
(b) In the resting state, the net charge on the inner surface of the membrane is
 Qi  1.17 109 C , and the net positive charge on this surface in the excited state
is
Q f  C  V  f  1.67  108 F  30  103 V    5.0  1010 C
The total positive charge which must pass through the membrane to produce the
excited state is therefore
Q  Q f  Qi
  5.0  1010 C    1.17  109 C   1.67  109 C  1.7  109 C
corresponding to
N Na + 
Q
1.67  109 C

 1.0  1010 Na + ions
-19
+
e
1.6  10 C Na ion
(c) If the sodium ions enter the axon in a time of t  2.0 ms , the average current is
I
Q 1.67  109 C

 8.3  107 A  0.83  A
t
2.0  103 s
(d) When the membrane becomes permeable to sodium ions, the initial influx of
sodium ions neutralizes the capacitor with no required energy input. The energy
input required to charge the now neutral capacitor to the potential difference of the
excited state is
2
1
1
2
W  C  V  f  1.67 108 F  30 103 V   7.5 1012 J
2
2
18.36
The capacitance of the 10 cm length of axon was found to be C  1.67  108 F in the
solution of Problem 18.35.
(a) When the membrane becomes permeable to potassium ions, these ions flow out of
the axon with no energy input required until the capacitor is neutralized. To
maintain this outflow of potassium ions and charge the now neutral capacitor to the
resting action potential requires an energy input of
2
1
1
2
W  C  V   1.67 108 F  70 103 V   4.1 1011 J .
2
2
(b) As found in the solution of Problem 18.35, the charge on the inner surface of the
membrane in the resting state is 1.17 109 C and the charge on this surface in the
excited state is  5.0 1010 C . Thus, the positive charge which must flow out of the
axon as it goes from the excited state to the resting state is
Q  5.0 1010 C  1.17 109 C  1.67 109 C ,
and the average current during the 3.0 ms required to return to the resting state is
I
18.37
From Figure 18.24, the duration of an action potential pulse is 4.5 ms. From the solution
Problem 18.35, the energy input required to reach the excited state is W1  7.5 1012 J .
The energy input required during the return to the resting state is found in Problem
18.36 to be W2  4.1 1011 J . Therefore, the average power input required during an
action potential pulse is
P
18.40
Q 1.67  109 C

 5.6  107 A  0.56  A
t
3.0  103 s
Wtotal W1  W2 7.5  1012 J+4.1  1011 J


 1.1  108 W  11 nW
3
t
t
4.5  10 s
(a) The circuit reduces as shown below to an equivalent resistance of Req  14  .
(b) The power dissipated in the circuit is
 V 
P
Req
2
 28 V 

14 
2
 56 W
(c) The current in the original 5.0- resistor (in Figure 1) is the total current supplied
by the battery. From Figure 6, this is
I
18.48
V 28 V

 2.0 A
Req 14 
The total resistance in the circuit is
1
1
 1
 1
1 
1 
R   

  1.2 k
 2.0 k 3.0 k 
 R1 R2 
and the total capacitance is C  C1  C2  2.0  F+3.0  F=5.0  F
Thus, Qmax  C   5.0  F 120 V   600 C
and
  RC  1.2  103  5.0  106 F   6.0 103 s 
6.0 s
1 000
The total stored charge at any time t is then
Q  Q1  Q2  Qmax 1  et   or Q1  Q2   600 C  1  e1 000 t 6.0 s 
(1)
Since the capacitors are in parallel with each other, the same potential difference exists
across both at any time.
Therefore,
 V C 
C 
Q1 Q2
, or Q2   2  Q1  1.5Q1

C1 C2
 C1 
Solving equations (1) and (2) simultaneously gives
Q1   240 C  1  e1 000 t 6.0 s 
and
Q2   360 C 1  e1 000 t 6.0 s 
(2)