ch31 - Chapter 31 1(a All the energy in the circuit resides...

1202
Chapter 31
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum
charge. The current is then zero. If Q is the maximum charge on the capacitor, then the
total energy is
U 
Q
2
2 C

2.90  10  6 C
 
2
2 3.60  10  6 F
 
 1.17  10  6 J.
(b) When the capacitor is fully discharged, the current is a maximum and all the energy
resides in the inductor. If I is the maximum current, then U = LI
2
/2 leads to
I 
2 U
L

2 1.168  10  6 J
 
75  10  3 H
 5.58  10
 3
A.
2. (a) We recall the fact that the period is the reciprocal of the frequency. It is helpful to
refer also to Fig. 31-1. The values of t when plate A will again have maximum positive
charge are multiples of the period:
t
A
 nT 
n
f

n
2.20  10 3 Hz
 n 4.55  s
 
,
where n = 1, 2, 3, 4,  . The earliest time is ( n = 1) t
A
 4.54  s.
(b) We note that it takes t T 
1
2
for the charge on the other plate to reach its maximum
positive value for the first time (compare steps a and e in Fig. 31-1). This is when plate A
acquires its most negative charge. From that time onward, this situation will repeat once
every period. Consequently,
t 
1
2
T  ( n  1) T 
1
2
2 n  1
 
T 
2 n  1
 
2 f

2 n  1
 
2 2.2  10 3 Hz
 
 2 n  1
 
2.27  s
 
,
where n = 1, 2, 3, 4,  . The earliest time is ( n = 1) t  2.27  s.
(c) At t T 
1
4
, the current and the magnetic field in the inductor reach maximum values
for the first time (compare steps a and c in Fig. 31-1). Later this will repeat every half-
period (compare steps c and g in Fig. 31-1). Therefore,

1203
t
L

T
4

( n  1) T
2
 2 n  1
 
T
4
 2 n  1
 
1.14  s
 
,
where n = 1, 2, 3, 4,  . The earliest time is ( n = 1)
t  1.14  s.
3. (a) The period is T = 4(1.50  s) = 6.00  s.
(b) The frequency is the reciprocal of the period: f
T
   
1 1
6 00
167 10 5
.
.
 s
Hz.
(c) The magnetic energy does not depend on the direction of the current (since U
B
 i
2
),
so this will occur after one-half of a period, or 3.00  s.
4. We find the capacitance from U Q C 
1
2
2 :
C 
Q
2
2 U

2.40  10  6 C
 
2
2 140  10
 6
J
 
 2.06  10  8 F.
5. According to
U LI Q C   1
2
2
1
2
2
,
the current amplitude is
I 
Q
LC

3.00  10
 6
C
1.10  10  3 H
 
4.00  10  6 F
 
 4.52  10  2 A.
6. (a) The angular frequency is
 
k
m

F x
m

8.0N
2.0  10
 3
m
 
0.25kg
 
 126rad s .
(b) The period is 1/ f and f =  /2  . Therefore,
T 
2 


2 
 rad s
 5.0  10
 2
s.
(c) From  = ( LC )
–1/2
, we obtain
C 
1
 2 L

1
126rad s
 
2
5.0H
 
 1.3  10
 5
F.

CHAPTER 31 1204
7. Table 31-1 provides a comparison of energies in the two systems. From the table, we
see the following correspondences:
2
2 2 2
1
, , , ,
1 1 1
, .