CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
887
CHAPTER 33
Answer to Checkpoint Questions
1.
2.
3.
4.
5.
6.
7.
8.
(a) T=2; (b) T ; (c) T=2; (d) T=4
(a) 5 V; (b) 150 J
(a) 1; (b) 2
(a) C , B , A; (b) 1 A, 2 B , 3 S , 4 C ; (c) A
(a) increases; (b) decreases
(a) 1, lags; 2, leads; 3, in phase; (b) 3 (!d = ! when XL = XC )
(a) increase (circuit is mainly capacitive; increase C to decrease XC to be closer to
resonance for maximum Pav ); (b) closer
step-up
Answer to Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
(a) T=4; (b) T=4; (c) T=2 (see Fig. 33-2); (d) T=2 (see Eq. 31{40)
with n an integer, (a) 0 2n; (c) =2 2n; (e) 2n; (g) 3=2 2n
(b), (a), (c)
(a) less; (b) greater
(a) 3, 1, 2; (b) 2, tie of 1 and 3
(a) decrease; (b) same (UB equals UE , which has not been changed)
slower
(a) 3, 2, 1; (b) 2, 3, 1
(a) 1 and 4; (b) 2 and 3
(a) leads; (b) capacitive; (c) less
(a) 3, then 1 and 2 tie; (b) 2, 1, 3
(a) decrease; (b) increase
(a) negative; (b) lead
888
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
(a) less, (b) equal, (c) greater
(a) rightward, increase (XL increases, closer to resonance); (b) rightward increase
(XC decreases, closer to resonance); (c) rightward, increase (!d =! increases, closer
to resonance)
16. (a) positive; (b) decrease L (to decrease XL and get closer to resonance); (c) decrease
C (to increase XC and get closer to resonance)
14.
15.
Solutions to Exercises & Problems
1E
Solve C from U = 12 Q2 =C :
2
:60 10 6 C)2 = 9:14 10 9 F :
C = 2QU = (12(140
10 6 J)
2E
Solve I from U = 12 LI 2 :
I=
r
2U =
L
r
2(10:0 10 6 J) = 0:115 A :
1:50 10 3
3E
From U = 12 LI 2 = 12 Q2 =C we nd
3:00 10 6 C
= 4:52 10 2 A :
I = pQ = p
3
6
LC
(1:10 10 H)(4:00 10 F)
4E
(a) All the energy in the circuit resides in the capacitor when it has its maximum charge.
The current is then zero. If C is the capacitance and Q is the maximum charge on the
capacitor then the total energy is
2
:90 10 6 C)2 = 1:17 10 6 J :
U = 2QC = (2
2(3:60 10 6 F)
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
889
(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 and
I=
r
2U =
L
r
2(1:168 10 6 J) = 5:58 10 3 A :
75 10 3 H
5E
(a) T = 4(1:50 s) = 6:00 s.
(b) f = T 1 = (6:00 s) 1 = 1:67 105 Hz:
(c) Half a period, or 3:00 s.
6P
Refer to Fig. 33-1.
(a)
ta = nT = Tn = 2:00 n103 Hz = n(5:00 s) ;
where n = 1; 2; 3; 4; :
(b) First, it takes 12 T for the charge on the other plate to reach its maximum positive value
for the rst time. Later this will repeat once every period. So
1) = (2n + 1) = (2n + 1)(2:50 s) ;
tb = 21 T + nT = 12 (2n + 1)T = (2n2+
f
2(2 103 Hz)
where n = 0; 1; 2; 3; 4; :
(c) First, it takes 14 T for the magnetic eld in the inductor to reach its maximum value for
the rst time. Later this will repeat every half a period. So
1 t = (2n + 1)(1:25 s)
=
tc = T4 + nT
2 2b
where n = 0; 1; 2; 3; 4; :
7E
(a)
(b)
r
r
s
8:0 N
=
! = mk = F=x
m
(2:0 10 3 m)(0:50 kg) = 89 rad/s :
2 = 7:0 10 2 s :
T = 2! = 89 rad/s
890
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
(c) Let ! = (LC ) 1=2 and solve for C :
C = !12 L = (89 rad/s)1 2 (5:0 H) = 2:5 10 5 F :
8P
(a) The mass m corresponds to the inductance, so m = 1:25 kg.
(b) The spring constant k corresponds to the reciprocal of the capacitance. Since the total
energy is given by U = Q2 =2C , where Q is the maximum charge on the capacitor and C is
the capacitance,
2
175 10 6 C 2
Q
C = 2U = 2(5:70 10 6 J) = 2:69 10 3 F
and
k = 2:69 101 3 m/N = 372 N/m :
(c) The maximum displacement xm corresponds to the maximum charge, so
xm = 175 10 6 m :
(d) The maximum speed vm corresponds to the maximum current. The maximum current
is
6
I = Q! = pQ = p 175 10 C 3 = 3:02 10 3 A :
LC
(1:25 H)(2:69 10 F)
Thus vm = 3:02 10 3 m/s.
9E
p
Solve L from f = (2 LC ) 1 :
L = 421f 2 C = 42 (10 103 Hz)1 2 (6:7 10 6 F) = 3:8 10 5 H :
10E
p
Solve C from f = (2 LC ) 1 :
C = 421f 2 L = 42 (3:50 103 Hz)1 2 (1:30 10 3 H) = 1:59 10 6 F :
CHAPTER 33
891
ELECTROMAGNETIC OSCILLATIONS
11E
If T is the period
p of oscillation, then the time required is t = T=4. The period is given by
T = 2=! = 2 LC , where ! is the angular frequency of oscillation, L is the inductance,
and C is the capacitance. Hence
p
p
:0 10 6 F) = 7:0 10 4 s :
t = T4 = 2 4LC = 2 (0:050 H)(4
4
12E
When switch S1 is closed and the others are open, the inductor is essentially out of the
circuit and what remains is an RC circuit. The time constant is C = RC . When switch
S2 is closed and the others are open the capacitor is essentially out of the circuit and what
remains is an LR circuit with time constant L = L=R. When switch S3 is closed and
the others are open the resistor is essentially
p out of the circuit and what remains is an LC
LC . Substitute L = RL and C = C =R to
circuit that oscillates
with
period
T
=
2
p
obtain T = 2 C L .
13E
Apply the loop rule to the LC circuit:
di + q = 0 :
Etotal = EL + EC = L dt
C
Since i = dq=dt; di=dt = d2 q=dt2 . Thus
2
L ddt2q + Cq = 0 :
14E
Apply the loop rule to the entire circuit:
X
Etotal = EL1 + EC1 + ER1 + =
=
X
di + q + iR
Lj dt
j
C
j
ELj + ECj + ERj
j
j
di + q + iR = 0 ;
= L dt
C
P
P
P
where L = j Lj , C 1 = ( j Cj 1 ) 1 , and R = j Rj . This is equivalent to the simple
LRC circuit shown in Fig. 33-23(b).
892
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
15P
(a) Q = CVmax = (1:0 10 9 F)(3:0 V) = 3:0 10 9 C.
(b) From U = 12 LI 2 = 12 Q2 =C we get
3:0 10 9 C
I = pQ = p
= 1:7 10 3 A :
3
9
LC
(3:0 10 H)(1:0 10 F)
(c)
UB;max = 21 LI 2 = 21 (3:0 10 3 H)(1:7 10 3 A)2 = 4:5 10 9 J :
16P
(a)
(b)
1
! = p1 = p
= 5:77 103 rad/s :
LC
(3:00 10 3 H)(10:0 10 6 F)
T = 2! = 5:77 2103 rad/s = 1:09 10 3 s :
(c) Use q(t) = Q cos(!t). The sketch is shown below. Here Q = 220 C and T = 1:09 10 3 s.
q
qmax
0
t
T
-q max
17P
(a) Use U = 12 LI 2 = 12 Q2 =C to solve for L:
2
2
2
Q
CV
1
V
1
max
max
=
=C
L=
C I
C
I
= (4:00 10 6 F) 50:01:5010V 3 A
2
I
= 3:60 10 3 H :
CHAPTER 33
(b)
893
ELECTROMAGNETIC OSCILLATIONS
f=
p1
2 LC
(c) From Fig. 33-1
=
p
2 (3:60 10
1
3 H)(4:00 10 6 F)
= 1:33 103 Hz :
t = 14 T = 41f = 4(1:33 1 103 Hz) = 1:88 10 4 s :
18P
(a) After the switch is thrown to
p position b the circuit is an LC circuit. The angular
frequency of oscillation is ! = 1= LC and the frequency is
f = 2! =
p1
2 LC
=
p
2 (54:0 10
1
3 H)(6:20 10 6 F)
= 275 Hz :
(b) When the switch is thrown, the capacitor is charged to V = 34:0 V and the current is
zero. Thus the maximum charge on the capacitor is Q = V C = (34:0 V)(6:20 10 6 F) =
2:11 10 4 C. The current amplitude is
I = !Q = 2fQ = 2(275 Hz)(2:11 10 4 C) = 0:365 A :
19P
The capacitors C1 and C2 can be used in four dierent ways: (1) C1 only; (2) C2 only;
(3) C1 and C2 in parallel; and (4) C1 and C2 in series. The corresponding oscillation
frequencies are:
f1 =
1
1 =
p
2 LC
p
2 (1:0 10
1
2 H)(5:0 10 6 F)
= 7:1 102 Hz ;
1
= 1:1 103 Hz ;
2
6
2
(1
:
0
10
H)(2
:
0
10
F)
2
1
= p
= 6:0 102 Hz ;
f3 = p 1
2
2 L(C1 + C2 ) 2 (1:0 10 H)(2:0 + 5:0)10 6 F
f2 =
and
1 =
p
2 LC
p
s
(2:0 + 5:0)10 6 F
f4 = p
= 21 (1:0 10 2 H)(2
:0 10 6 F)(5:0 10 6 F)
2 LC1 C2 =(C1 + C2 )
= 1:3 103 Hz :
1
894
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
20P
(a) Solve L from 2f = (LC ) 1=2 :
1
7
L = 421f 2 C = 42 (10:4 103 Hz)
2 (340 10 6 F) = 6:89 10 H :
(b)
U = 21 LI 2 = 12 (6:89 10 7 H)(7:20 10 3 A)2 = 1:79 10 11 J :
(c) Solve for Q from U = 12 Q2 =C :
p
p
Q = 2CU = 2(340 10 6 F)(1:79 10
11 J) = 1:10 10 7 C :
21P
(a) At any time the total energy U in the circuit is the sum of the energy UE in the electric
eld of the capacitor and the energy UB in the magnetic eld of the inductor. When
UE = 0:500UB , then UB = 2:00UE and U = UE + UB = 3:00UE . UE is given by q2 =2C ,
where q is the charge on the capacitor and C is the capacitance. The total energy U is
given by Qp2 =2C , where Q is the maximum charge on the capacitor, so Q2 =2C = 3:00q2 =2C
or q = Q= 3:00 = 0:577Q.
(b) If the capacitor is fully charged at time t = 0 then the charge on the capacitor is given by
q(t) = Q cos !t, where ! is the angular frequency of oscillation. The condition q = 0:577Q
is satised when cos !t = 0:557 or !t = 0:955 rad. Since ! = 2=T , where T is the period
of oscillation, t = 0:955T=2 = 0:152T .
22P
(a) Since the percentage of energy stored in the electric eld of the capacitor is (1 75:0%) =
25:0%,
UE = q2 =2C = 25:0% ;
U Q2 =2C
p
which gives q = 25:0% Q = 0:500Q:
(b) From
p
we get i = 75:0% I = 0:866I :
23P
UB = Li2 =2 = 75:0%
U LI 2 =2
The frequency f as a function of , the angle of rotation of the knob, is given by f () =
f0 + k, where f0 = 2 105 Hz and k = [f (180 ) f0 ]= = (4 105 Hz 2 105 Hz)= =
6:4 104 Hz/rad. Thus from 2f () = [LC ()] 1=2 we get
C () = 42 Lf1 2 () = 42 (f 1+ k)2 L :
0
895
ELECTROMAGNETIC OSCILLATIONS
CHAPTER 33
The plot of C () vs is as follows.
0.8
0.6
C (nF)
0.4
0.2
0
0
24P
0.5
1
1.5
θ (rad)
2
2.5
3
3.5
(a) Since the p
frequency of oscillation f is related to the inductance L and capacitance C
value of C gives the larger value of f . Hence, fmax =
by f p= 1=2 LC , the smaller
p
1=2 LCmin , fmin = 1=2 LCmax , and
p
p
fmax = pCmax = p365 pF = 6:0 :
fmin
10 pF
Cmin
(b) You want to choose the additional capacitance C so the ratio of the frequencies is
60 MHz = 2:96 :
r = 01::54
MHz
Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds
to that of the tuning capacitor. If C is in picofarads, then
pC + 365 pF
pC + 10 pF = 2:96 :
The solution for C is
(2:96)2 (10 pF) = 36 pF :
C = (365 pF)(2:96)
2 1
p
Solve f = 1=2 LC for L. For the minimum frequency C = 365 pF + 36 pF = 401 pF and
f = 0:54 MHz. Thus
L = (2)12 Cf 2 = (2)2 (401 10 121F)(0:54 106 Hz)2 = 2:2 10 4 H :
896
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
25P
(a) The total energy U is the sum of the energies in the inductor and capacitor. If q is the
charge on the capacitor, C is the capacitance, i is the current, and L is the inductance,
then
2
2
U = UE + UB = 2qC + i 2L
:80 10 6 C)2 + (9:20 10 3 A)2 (25:0 10 3 H) = 1:98 10 6 J :
= (3
2(7:80 10 6 F)
2
(b) Solve U = Q2 =2C for the maximum charge Q:
p
p
Q = 2CU = 2(7:80 10 6 F)(1:98 10 6 J) = 5:56 10 6 C :
(c) Solve U = I 2 L=2 for the maximum current I :
I=
r
r
2U =
L
2(1:98 10 6 J) = 1:26 10 2 A :
25:0 10 3 H
(d) If q0 is the charge on the capacitor at time t = 0, then q0 = Q cos and
= cos
1
q = cos
Q
1
3:80 10 6 C = 46:9 :
5:56 10 6 C
For = +46:9 the charge on the capacitor is decreasing, for = 46:9 it is increasing.
To check this calculate the derivative of q with respect to time, evaluated for t = 0. You
should get !Q sin . You want this to be positive. Since sin(+46:9 ) is positive and
sin( 46:9 ) is negative the correct value for increasing charge is = 46:9 .
(e) Now you want the derivative to be negative and sin to be positive. Take = +46:9 .
26P
(a) The charge is given by q(t) = Q sin !t, where Q is the maximum charge on the capacitor
and ! is the angular frequency of oscillation. A sine function was chosen so that q = 0 at
time t = 0. The current is
i(t) = dq
dt = !Q cos !t
p
and at t = 0 it is I = !Q. Since ! = 1= LC ,
p
p
Q = I LC = (2:00 A) (3:00 10 3 H)(2:70 10 6 F) = 1:80 10 4 C :
(b) The energy stored in the capacitor is given by
2
2 sin2 !t
q
Q
UE = 2C = 2C
CHAPTER 33
897
ELECTROMAGNETIC OSCILLATIONS
and its rate of change is
dUE = Q2 ! sin !t cos !t :
dt
C
Use the trigonometric identity cos !t sin !t = 12 sin(2!t) to write this
dUE = !Q2 sin(2!t) :
dt
2C
The greatest rate of change occurs when sin(2!t) = 1 or 2!t = =2 rad. This means
T = T ;
t = 4! = 4(2
) 8
where T is the period of oscillation. The relationship ! = 2=T was used.
(c) Substitute ! = 2=T and sin(2!t) = 1 into dUE =dt = (!Q2 =2C ) sin(2!t) to obtain
2Q2 = Q2 :
dUE
=
dt max 2TC TC
p
p
Now T = 2 LC = 2 (3:00 10 3 H)(2:70 10 6 F) = 5:655 10 4 s, so
dUE
dt
(1:80 10 4 C)2
= (5:655 10 4 s)(2:70 10 6 F) = 66:7 W :
max
Notice that this is a positive result, indicating that the energy in the capacitor is indeed
increasing at t = T=8.
27P
(a) Consider the joint point P as shown. The
current i0 owing out of point P into the middle inductor is i0 = i i = 0. Since no current
goes through the middle inductor the circuit is
equivalent to two L's and C 's in series. The loop
equation reads
or
i
i
P
i'
di + q + L di + q = 0 ;
L dt
C
dt C
di + q = 0 :
2L dt
C=2
This is equivalent with a single LC circuit with Leq = 2L and Ceq = C=2. Thus
= p1 :
!=p 1 =p 1
Leq Ceq
LC
(2L)(C=2)
898
CHAPTER 33
(b) Now i0 = i + i = 2i (see the gure to the
right). Apply the loop equation to the left half
of the circuit containing the capacitor on the left
and the inductors on the left and in the middle
to obtain
or
ELECTROMAGNETIC OSCILLATIONS
i
P
i
i'
di + q + L di0 = L di + q + 2L di = 0 ;
L dt
C
dt
dt C
dt
di + q = 0 ;
Leq dt
C
eq
where Leq = 3L and Ceq = C . The corresponding oscillation angular frequency is then
!=p 1 =p1 :
Leq Ceq
3LC
p
(Note that in a single-loop LC circuit there is only one frequency, ! = (1)= Leq Ceq . This
contradicts with our case here so it is impossible to reduce our circuit to a single-loop one.)
28P
For the rst circuit ! = (L1 C1 ) 1=2 , and for the second one ! = (L2 C2 ) 1=2 . When the
two circuits are connected in series the new frequency is
1
1
!0 = p 1 = p
=p
Leq Ceq
(L1 + L2 )C1 C2 =(C1 + C2 )
(L1 C1 C2 + L2 C2 C1 )=(C1 + C2 )
1
p
=p
= !;
L1 C1 (C1 + C2 )=(C1 + C2 )
p
p
where we used ! 1 = L1 C1 = L2 C2 :
29P
Compare the expression for i here with i = I sin(!t+0 ). Let (!t+) = 2500t+0:680 = =2
to obtain t = 3:56 10 4 s.
(b) Since ! = 2500 rad/s = (LC ) 1=2 ,
1
3
L = !21C = (2500 rad/s)2 (64
:0 10 6 F) = 2:50 10 H :
(c)
U = 12 LI 2 = 21 (2:50 10 3 H)(1:60 A)2 = 3:20 10 3 J :
CHAPTER 33
899
ELECTROMAGNETIC OSCILLATIONS
30P
The energy needed to charge the 100 F capacitor to 300 V is
1 C V 2 = 1 (100 10 6 F)(300 V)2 = 4:50 J :
2 100
2
The energy originally in the 900 F capacitor is
1 C V 2 = 1 (900 10 6 F)(100 V)2 = 4:5 J :
2 900
2
All the energy originally in the 900 F capacitor must be transferred to the 100 F capacitor.
The plan is to store it temporarily in the inductor. To do this leave switch S1 open and
close switch S2 . Wait until the 900 F capacitor is completely discharged and the current
in the right-hand loop is a maximum. This is one quarter of the period of oscillation. Since
p
p
T900 = 2 LC900 = 2 (10:0 H)(900 10 6 F) = 0:596 s ;
you should wait (0:596 s)=4 = 0:149 s. At that instant close switch S1 and open switch S2
so the current is in the left-hand loop. Now wait one quarter of the period of oscillation
of the left-hand LC circuit and open switch S1 . The 100 F capacitor then has maximum
charge and all the energy resides in it. The period of oscillation is
p
p
T100 = 2 LC100 = 2 (10:0 H)(100 10 6 F) = 0:199 s
and you must keep S1 closed for (0:199 s)=4 = 0:0497 s before opening it again.
31E
In Eq. 33-21, let
p
q = e RnT=2L = 99:0% ;
Q
where n = 50:0 and T 2 LC , we get
r
r
:0%) L = ln(99:0%) 220 10 3 H = 8:66 10 3 :
R = ln(99
n
C
50:0
12:0 10 6 F
p
32E
(a) Since T = 2=! = 2 LC , we may rewrite the power on the exponential factor as
r
Thus e
p
r
R CL Tt = R CL pt = 2RtL :
2 LC
Rt=2L = e R C=L(t=T ) :
900
CHAPTER 33
p
ELECTROMAGNETIC OSCILLATIONS
p
(b) Since R C=Lp(t=T ) must be unitlss and so is t=T , R pC=L must also be unitless.
must be . p
Thus the SI unit of C=L must be 1 , i.e., the SI unit for L=C
pC=L
R
(T=T ) = e R C=L
(c) Since the amplitude of oscillation reduces by apfactor of e
p
after each cycle, the condition is equivalent to R C=L 1, or R L=C .
33P
Since the maximum amount of energy stored in the capacitor during each cycle is given by
1 Q2 =C , where Q is the maximum charge and C is the capacitance, you want the time for
2
which
q2 = 1 Q2 :
2C 2 2C
p
This means q = Q= 2. Now q is given by
q = Qe Rt=2L ;
where R is the resistance and L is the inductance in the circuit. Divide by Q and take the
natural logarithm of both sides to obtain
ln Qq = 2RtL :
Solve for t:
t = 2RL ln Qq = 2RL ln p1 = RL ln 2 :
2
p
p
The identities ln(1= 2) = ln 2 = 12 ln 2 were used to obtain the last form of the result.
34P
The charge q after N cycles is obtained by stbstituting t = NT = 2N=!0 into Eq. 33-21:
p
q = Qe Rt=2L cos(!0 t) = Qe RNT=2L = Qe RN (2 L=C )=2L
p
= Qe NR C=L :
So
p
3:20 F=12:0 H = 5:85 C;
qjN =5 = (6:20 C)e 5(7:20 ) p
qjN =10 = (6:20 C)e 10(7:20 ) 3:20 Fp=12:0 H = 5:52 C;
and qjN =100 = (6:20 C)e 100(7:20 ) 3:20 F=12:0 H = 1:93 C:
35P
(a) In Eq. 33-21, let q = 0 and t = 0 to obtain 0 = Q cos . This gives = =2.
CHAPTER 33
901
ELECTROMAGNETIC OSCILLATIONS
(b) First, calculate the current i(t) from Eq. 33-21:
i
h
dq
d
Rt=
2
L
0
i(t) = dt = dt Qe
cos(! t + )
Rt=2L cos(!0 t + ) !0 Qe Rt=2L sin(!0 t + )
= QR
2L e
s
= Q !02 +
R 2 e Rt=2L sin(!0 t + + ) ;
2L
where tan = R=2L!0 . The current amplitude is then
s
2
R
0
2
I (t) = Q ! + 2L e Rt=2L = Q!e Rt=2L = Ie Rt=2L :
Thus
I e Rt=2L cos !0 t + !
2
Rt=2L cos(!0 t + ) =
q(t) = Qe
= !I e Rt=2L sin(!0 t) :
36P
(a) From Eq. 33-21
and
dq = d hQe Rt=2L cos(!0 t + )i
dt dt
Rt=2L cos(!0 t + ) !0 Qe Rt=2L sin(!0 t + )
= RQ
2L e
d2 q = R e Rt=2L RQ cos(!0 t + ) !0 Q sin(!0 t + )
dt2
2L
2L
0
RQ!
Rt=
2
L
0
0
2
0
+e
2L sin(! t + ) ! Q cos(! t + ) :
Substitute these expressions and Eq. 33-21 into Eq. 33-20:
Qe
Rt=2L
!02 L
R
2L
2
+ 1c cos(!0 t + ) = 0 :
Since this equation is valid at any time t, we must have
!02 L
R
2L
2
+ C1 = 0
902
ELECTROMAGNETIC OSCILLATIONS
CHAPTER 33
or
!0 =
s
1
LC
R
2L
2
s
=
!2
R
2L
2
:
(b) The fractional shift in frequency is
r
p
! = ! !0 = 1
!
!
2
(1=LC
) (R=2L)2 = 1
p
1 R4LC
1=LC
s
2 (7:30 10 6 F)
= 0:210% :
1 (100 )4(4
=1
:40 H)
37P
Let t be a time at which the capacitor is fully charged in some cycle and let qmax 1 be the
charge on the capacitor then. The energy in the capacitor at that time is
2
2
1 = Q e Rt=L ;
U (t) = qmax
2C
2C
where
qmax 1 = Q e Rt=2L
was used. Here Q is the charge at t = 0. One cycle later the maximum charge is
qmax 2 = Q e R(t+T )=2L
and the energy is
2
2
2 = Q e R(t+T )=L ;
U (t + T ) = qmax
2C
2C
where T is the period of oscillation. The fractional loss in energy is
U = U (t) U (t + T ) = e Rt=L e R(t+T )=L = 1 e RT=L :
U
U (t)
e Rt=L
Assume that RT=L is small compared to 1 (the resistance is small) and use the binomial
theorem to expand the exponential. The rst two terms are:
e RT=L 1 RT
L :
Replace T with 2=!, where ! is the angular frequency of oscillation. Thus
U 1
U
RT = 2R :
1 RT
=
L
L
!L
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
903
38E
Use I = E =Xc = !C E .
(a) I = !C Em = 2fC Em = (2)(1:00 103 Hz)(1:50 10 6 F)(30:0 V) = 0:283 A:
(b) I = (2)(8:00 103 Hz)(1:50 10 6 F)(30:0 V) = 2:26 A.
39E
(a) The current amplitude I is given by I = VL =XL , where VL is the voltage amplitude
across the inductor and XL is the inductive reactance. The reactance is given by XL =
!L = 2fL, where ! is the angular frequency, f is the frequency, and L is the inductance.
Since the circuit contains only the inductor and a sinusoidal generator, VL = Em , where Em
is the generator emf amplitude. Thus
Em =
30:0 V
I = XVL = 2fL
3
2(1:00 10 Hz)(50:0 10 3 H) = 0:0955 A :
L
(b) The frequency is now eight times larger than in part (a), so the inductive reactance is
eight times larger and the current is one-eighth as much, or = (0:0955 A)=8 = 0:0119 A.
40E
(a) and (b) Regardless of the frequency of the emf,
I = ERm = 3050:0
V = 0:60 A :
41E
(a) Solve L from XL = !L = 2fL:
1:30 103 L =
3
f = 2XL
(2)(45:0 10 3 H) = 4:60 10 Hz :
(b) Solve C from XC = (!C ) 1 = (2fC ) 1 :
1
1 =
C = 2fX
= 2:66 10 8 F :
3
C 2(4:60 10 Hz)(1:30 103 )
(c) Since XL / f and XC / f 1 , when f is doubled XL = 2(1:30 103 ) = 2:60 103 and XC = 1:30 103 =2 = 6:50 102 .
42E
(a)
1 =
1
f = 2CX
= 8:84 103 Hz :
C 2(1:50 10 6 F)(12:0 )
904
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
(b) Since XC / f 1 , if the frequency is doubled then XC = 12:0 =2 = 6:00 .
43E
(a) The inductive reactance for angular frequency ! is given by XL = !L, where L is
the inductance, and the capacitive reactance is given by XC = 1=!C
p , where C is the
capacitance. The two reactances are equal if !L = 1=!C , or ! = 1= LC . The frequency
is
1
f = 2! = p1 = p
= 650 Hz :
3
2 LC 2 (6:0 10 H)(10 10 6 F)
(b) The inductive reactance is XL = !L = 2fL = 2(650 Hz)(6:0 10 3 H) = 24 . The
capacitive reactance has the same value for this frequency.
p
(c) The natural frequency for free LC oscillations is f = !=2 = 1=2 LC , the same as
that for which the reactances are equal.
44P
(a)
Em = Em =
25:0 V
3 A:
I=X
=
5
:
22
10
L !d L (377 rad/s)(12:7 H)
(b) Since E (t) and i(t) has a 90 -phase dierence, E (t) is zero when i(t) = I .
(c) In this case !t must be (2n =p6) (where n is an integer), so i = I sin(2n =6 +
=2) = I sin(=3) = (5:22 10 3 A)( 3=2) = 4:51 10 3 A.
45P
(a)
I = XEm = !C Em = (377 rad/s)(4:15 10 6 F)(25:0 V) = 3:91 10 2 A :
C
(b) Since E (t) and i(t) again has a 90 -phase dierence, E (t) is zero when i(t) = I .
(c) In this case !t must
p again be 2n =2 6, so i = I sin(2n =6 =2) = I sin(2=3) =
3
(3:91 10 A)( 3=2) = 3:38 10 A.
46P
(a) The generator emf is a maximum when sin(!t =4) = 1 or !t =4 = =2 2n, where
n is an integer, including zero. The rst time this occurs after t = 0 is when !t =4 = =2
or
3
3
t = 43! = 4(350
s 1 ) = 6:73 10 s :
(b) The current is a maximum when sin(!t 3=4) = 1, or !t 3=4 = =2 2n. The
rst time this occurs after t = 0 is when
5
2 s:
=
1
:
12
10
t = 45! = 4(350
1
s )
(c) The current lags the inductor by =2 rad, so the circuit element must be an inductor.
CHAPTER 33
905
ELECTROMAGNETIC OSCILLATIONS
(d) The current amplitude I is related to the voltage amplitude VL by VL = IXL , where
XL is the inductive reactance, given by XL = !L. Furthermore, since there is only one
element in the circuit the amplitude of the potential dierence across the element must be
the same as the amplitude of the generator emf: VL = Em . Thus Em = I!L and
0V
L = EI!m = (620 10 303 :A)(350
rad/s) = 0:138 H :
47P
(a) Let !t =4 = =2 to obtain t = 3=4! = 3=[4(350 rad/s)] = 6:73 10 3 s.
(b) Let !t + =4 = =2 to obtain t = =4! = =[4(350 rad/s)] = 2:24 10 3 s.
(c) Since i leads E in phase by =2, the element must be a capacitor.
(d) Solve C from XC = (!C ) 1 = Em =I : C = I=Em ! = (6:20 10 3 A)=[(30:0 V)
(350 rad/s)] = 5:90 10 5 F:
48P
(a) and (b) Consider the following combinations: V12 = V1 V2 , V13 = V1 V3 , and
V23 = V2 V3 . For V12
2!t 120 120
V12 = A sin(!t) A sin(!t 120 ) = 2A sin 2 cos
2
p
= 3 A cos(!t 60 ) ;
p
where we used sin sin = 2 sin[( )=2] cos[( + )=2] and sin 60 = 3=2. This
expression indicates
p that V12 oscillates sinusoidally with angular frequency !, and has an
amplitude of 3A. Similarly,
2!t 240 240
V13 = A sin(!t) A sin(!t 240 ) = 2A sin 2 cos
2
p
= 3 A cos(!t 120 )
and
2!t 360 120
V23 = A sin(!t 120 ) A sin(!t 240 ) = 2A sin 2 cos
2
p
= 3 A cos(!t 180 ) ;
p
both sinusoidal functions of t with an amplitude of 3A.
49E
(a) Now XC = 0 and R and XL remain unchanged, so
q
p
Z = R2 + XL2 = (160 )2 + (86:7 )2 = 182 ;
906
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
:0 V = 0:198 A ;
I = EZm = 36
182 X
X
86
:
7
0
L
C
1
1
= tan
= 28:5 :
= tan
R
160 and
(b)
εm
VL
I
VR
φ
ωt
50E
(a) Now XL = 0 and R and XC remain unchanged, so
q
and
(b)
p
Z = R2 + XC2 = (160 )2 + (177 )2 = 239 ;
:0 V = 0:151 A ;
I = EZm = 36
239 X
X
0
177
L
C
1
1
:
=
tan
=
47
:
9
= tan
R
160 I
VR
εm
φ
ωt
VC
51E
(a) The capacitive reactance is
1 = 1 =
1
XC = !C
2fC 2(60:0 Hz)(70:0 10 6 F) = 37:9 :
CHAPTER 33
907
ELECTROMAGNETIC OSCILLATIONS
The inductive reactance is unchanged, 86:7 . The new impedance is
p
p
Z = R2 + (XL XC )2 = (160 )2 + (37:9 86:7 )2 = 167 :
The current amplitude is
:0 V = 0:216 A :
I = EZm = 36
167 The phase angle is
= tan
1
XL XC = tan
R
1
86:7 37:9 = 17:0 :
160 (b) The voltage amplitudes are
VR = IR = (0:216 A)(160 ) = 34:6 V,
VL = IXL = (0:216 A)(86:7 ) = 18:7 V,
and
εm
VL
VR
VL - V L
VC = IXC = (0:216 A)(37:9 ) = 8:19 V.
φ
VC
Note that XL > XC , so that Em leads I . The
phasor diagram is shown on the right.
52E
f = f0 =
p1
2 LC
=
1
= 1:84 103 Hz :
3
6
2 (2:50 10 H)(3:00 10 F)
p
53P
(a) The resonance frequency f0 of the circuit is about (1:50 KHz 1:30 KHz)=2 = 1:40 KHz:
Thus from 2f0 = (LC ) 1=2 we get
1
3
L = 421f 2 C = 42 (1:40 103 Hz)
2 (5:50 10 6 F) = 2:35 10 H :
0
(b) From Fig. 33-13 we see that as R increases the resonance curve gets more spread out,
so the two frequencies at which the amplitude is at half-maximum level will move away
from each other.
908
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
54P
From Fig. 33-29 we see that the x and y components of the Z -vector are Zx = R and
Zy = XC XL . Thus
q
p
jZ j = Zx2 + Zy2 = R2 + (XC XL )2
and
tan = ZZy = XL R XC ;
x
which indeed are the same as Eqs. 33-52 and 33-56.
55P
The amplitude of the voltage across the inductor in an RLC series circuitpis given by
VL = IXL , where XL (= !L) is the inductive reactance. At resonance ! = 1= LC , where
L is the inductance and C is the capacitance. For the given circuit
1:0 H
XL = pL = p
= 1000 :
LC
(1:0 H)(1:0 10 6 F)
At resonance the capacitive reactance has the same value as the inductive reactance, so
XC = 1000 . For XL = XC , Eq. 33-52 gives Z = R. Hence
V = 1:0 A :
I = ERm = 10
10 Thus
VL = IXL = (1:0 A)(1000 ) = 1000 V :
This is much larger than the amplitude of the generator emf (10 V).
56P
The resistanc R of the coil satises
XL XC = !L 1=!C = tan ;
R
R
which we solve for R:
1
1
R = tan !L !C
1
1
2
= tan 75 (2)(930 Hz)(8:8 10 H) (2)(930 Hz)(0:94 10 6 F)
= 89 :
CHAPTER 33
909
ELECTROMAGNETIC OSCILLATIONS
57P
(a) 36:0 V, by denition.
(b) VR = IR cos = (0:196 A)(160 ) cos(29:4 ) = 27:3 V:
(c) Refer to the gure to the right.
VC = IXC sin = (0:196 A)(177 ) sin(29:4 )
= 17:0 V :
(d)
VL = IXL sin = (0:196 A)(86:7 ) sin(29:4 )
= 8:34 V :
I
VR
VL
| φ|
|φ|
εm
|φ|
VC
(e) Since VR + VC + VL = 27:3 V + 17:0 V 8:34 V = 36:0 V = Em , the loop theorem is
satised.
58P
(a)
(b)
1 =
1
XC = 2fC
2(400 Hz)(24:0 10 6 F) = 16:6 :
p
p
Z = R2 + (XL XC )2 = R2 + (2fL XC )2
p
= (220 )2 + [2(400 Hz)(150 10 3 H) 16:6 ]2 = 422 :
(c)
V = 0:521 A :
I = EZm = 220
422 1
(d) Now XC / Ceq will increase since Ceq decreases.
(e) Now Ceq = C=2 and the new impedance is
p
Z = (220 )2 + [2(400 Hz)(150 10 3 H) 2(16:6 )]2 = 408 < 422 ;
i.e., the impedance decreases.
(f) Since I / Z 1 , it increases.
59P
(a) For a given amplitude Em of the generator emf, the current amplitude is given by
;
I = EZm = p 2 Em
R + (!L 1=!C )2
910
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
where R is the resistance, L is the inductance, C is the capacitance, and ! is the angular
frequency. To nd the maximum set the derivative with respect to ! equal to zero and
solve for !. The derivative is
1
dI = E R2 + (!L 1=!C )2 3=2 !L 1
L+
:
d!
m
!C
!2 C
p
The only factor that can equal zero is !L (1=!C ) and it does for ! = 1= LC . For the
given circuit
1
! = p1 = p
= 224 rad/s :
LC
(1:00 H)(20:0 10 6 F)
(b) For this value of the angular frequency the impedance is Z = R and the current
amplitude is
:0 V = 6:00 A :
I = ERm = 30
5:00 (c) You want to nd the values of ! for which I = Em =2R. This means
Em
2
R + (!L
p
1=!C )2
m:
= 2ER
Cancel the factors Em that appear on both sides, square both sides, and set the reciprocals
of the two sides equal to each other to obtain
2
1
2
= 4R2 :
R + !L
!C
Thus
1
!L !C
2
= 3R2 :
Now take the square root of both sides and multiply by !C to obtain
!2 (LC ) !
p
3CR
1 = 0;
where the symbol indicates the two possible signs for the square root. The last equation
is a quadratic equation for !. Its solutions are
p 2 2
p
3
CR
3C R + 4LC :
!=
2LC
You want the two positive solutions. The smaller of these is
p 2 2
p
3
CR
+
3C R + 4LC
!2 =
2LC
p
:0 10 6 F)(5:00 )
= 2(13(20
:00 H)(20:0 10 6 F)
p
6 F)2 (5:00 )2 + 4(1:00 H)(20:0 10 6 F)
+ 3(20:0 10 2(1
:00 H)(20:0 10 6 F)
= 219 rad/s
CHAPTER 33
911
ELECTROMAGNETIC OSCILLATIONS
and the larger is
p
p
3C 2 R2 + 4LC
!1 = + 3CR + 2LC
p
+
:0 10 6 F)(5:00 )
= 2(13(20
:00 H)(20:0 10 6 F)
p
6 F)2 (5:00 )2 + 4(1:00 H)(20:0 10 6 F)
+ 3(20:0 10 2(1
:00 H)(20:0 10 6 F)
= 228 rad/s :
(d) The fractional width is
!1 !2 = 228 rad/s 219 rad/s = 0:039 :
!0
224 rad/s
60P
(a)
V = 39:1 :
Z = EIm = 3125
:20 A
(b) from VR = IR = Em cos we get
= (125 V) cos(0:982 rad) = 21:7 :
R = Em cos
I
3:20 A
(c) Since XL XC / sin = sin( 0:982 rad) < 0, i.e., XL < XC , the circuit is predominantly capacitive.
61P
p
From the problem statement 2f0 = (LC ) 1=2 = 6:00 kHz, Z = R2 + (2f1 L 1=2f1 C )2
= 1:00 k
where f1 = 8:00 kHz, and cos = R=Z = cos 45 . Solve these equations for R,
L and C :
(a) R = Z cos 45 = (1:00 k
)(cos 45 ) = 707 ;
(b)
p
p
2
2
(1:00 k
)2 (707 )2
L = 2(fZ f 2R=f ) = 2[8:00 kHz
(6:00 kHz)2 =8:00 kHz]
1
0 1
= 3:22 10 2 H ;
(c)
C = 421f 2 L = 42 (6:00 kHz)21(3:22 10 2 H) = 2:19 10 8 F :
0
912
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
62P
(a) The phase angle is given by
= tan VL R VC = tan 1 VL V =V2L:=002:00 = tan 1 (1:00) = 45:0 :
L
(b) Solve R from Em cos = IR:
= (30:0 V)(cos 45 ) = 70:7 :
R = Em cos
I
300 10 3 A
1
p
63P
(a) The resonance frequency is f0 = 1=2 LC; at which XL = XC , which is not satised
here.
(b) Let the additional capacitance be C 0 . Then the new impedance for the combined
capacitors should satisfy
1
1
=
1
0
!(C + C ) XC + !C 0 = XL ;
which gives
1
1
1
1
1
1
0
C = ! X X = 2(60:0 Hz) 86:7 177 = 1:56 10 5 F :
L
C
(c) At resonance XL XC = 0 so Z = R. Thus
:0 V = 0:255 A :
I = EZm = ERm = 36
160 64P
Four possibilties exist: (1) C1 = 4:00 F is used alone; (2) C2 = 6:00 F is used alone;
(3) C1 and C2 are connected in series; and (4) C1 and C2 are connected in parallel. The
corresponding resonant frequencies are
1
= 1:78 103 Hz ;
f1 = p1 = p
2 LC1 2 (2:00 10 3 H)(4:00 10 6 F)
1
= 1:45 103 Hz ;
f2 = p1 = p
3
6
2 LC2 2 (2:00 10 H)(6:00 10 F)
f3 =
p
1
2 LC1 C2 =(C1 + C2 )
= 2:30 103 Hz ;
=
1
2 (2:00 10 3 H)(4:00)(6:00)10 6 F=(4:00 + 6:00)
p
CHAPTER 33
and
f4 =
913
ELECTROMAGNETIC OSCILLATIONS
p
1
2 L(C1 + C2 )
=
1
= 1:13 103 Hz :
3
6
2 (2:00 10 H)(4:00 + 6:00)10 F
p
65P
(a) Since Leq = L1 + L2 and Ceq = C1 + C2 + C3 for the circuit, the resonant frequency is
1
= p
! = p1
2 Leq Ceq 2 (L1 + L2 )(C1 + C2 + C3 )
1
= 796 Hz :
= p
2 (1:70 + 2:30)(10 3 H)(4:00 + 2:50 + 3:50)(10 6 F)
(b) The resonant frequency does not depend on R so it will not change as R increases.
(c) Since ! / (L1 + L2 ) 1=2 , it will decrease as L1 is increased.
(d) Since ! / Ceq1=2 and Ceq decreases as C3 is removed, ! will increase.
66P
When the two circuits are separated, !0 = (L1 C1 ) 1=2 = (L2 C2 ) 1=2 . When combined,
the resonance frequency is
1
1
=p
!=p 1 =p
Leq Ceq
(L1 + L2 )C1 C2 =(C1 + C2 )
(L1 C1 C2 + C1 L2 C2 )=(C1 + C2 )
1
p
=p
=! ;
L1 C1 (C2 + C1 )=(C1 + C2 ) 0
where we used !0 = (L1 C1 ) 1=2 = (L2 C2 ) 1=2 :
67P
Use the result of 59P:
and
Also use
Thus
p
p
p
p
3C 2 R2 + 4LC
!1 = + 3CR + 2LC
!2 =
3CR + 3C 2 R2 + 4LC :
2LC
!0 = p 1 :
LC
p
p
r
! = !1 !2 = 2 3CR LC = R 3C :
!0
!0
2LC
L
914
CHAPTER 33
For the data of 59P
ELECTROMAGNETIC OSCILLATIONS
r
! = (5:00 ) 3(20:0 10 6 F) = 3:87 10 2 :
!0
1:00 H
This is in agreement with the result of 59P. The method of 59P, however, gives only one
signicant gure since two numbers that are close in value (!1 and !2 ) are subtracted.
Here the subtraction is done algebraically and 3 signicant gures are obtained.
68P*
When the switch is open we have an LRC circuit with L; R; C in series, with
VL Vc = !L 1=!C = tan = tan( 20 ) :
R
1
R
When the switch is in position 1 the equivalent capacitance of the LRC circuit becomes
2C . So now
!L 1=2!C = tan = tan 10:0 :
2
R
Finally, with the switch in position 2 the circuit is an LC circuit so
I = EXm = p
Em
(!L)2 + (1=!C )2
Solve for L, R and C from these three equations:
R=
:
Em
q
2I (tan 2 tan 1 )2 + (tan 2 12 tan 1 )2
120 V
q
=
2(2:00 A) ( tan 20:0 + tan 10:0 )2 + (tan 10:0 + 12 tan 20:0 )2
= 1:65 102 ;
L = 2!R tan 2
1 tan 1
2
2 ) 1
2(1
:
65
10
= 2(60:0 Hz) tan 10:0 + 2 tan 20:0 = 0:313 H ;
and
C = 2!R(tan 1 tan )
2
1
= 2(2)(60:0 Hz)(1:65 1021
)(tan 10:0 + tan 20:0 )
= 1:49 10 5 F :
CHAPTER 33
69E
915
ELECTROMAGNETIC OSCILLATIONS
p
p
From Eq. 33-63 Vmax = 2Vrms = 2(100 V) = 141 V:
70E
Since the impedance of the voltmeter is large, it will not aect the impedance of the circuit
when connected in parallel with the circuit. So the reading will be 100 V in all three cases.
71E
(a) Refer to 44P, part (c). The power delivered by the generator at this instant is P =
E (t)i(t) = Em sin(2n =6)I sin(=3) = Em I sin(=6) sin(=3) < 0, so it is taking energy
from the rest of the circuit.
(b) Refer to 45P, part (c). The power delivered by the generator at this instant is P =
E (t)i(t) = Em sin(2n =6)I sin( 2=3) = Em I sin(=6) sin(2=3) > 0, so it is supplying
energy to the rest of the system.
72E
The average power dissipated in resistance R when the current is alternating
p is given by
2
Pav = Irms R, where Irms is the root-mean-square current. Since Irms = I= 2, where I is
the current amplitude, this can be written Pav = I 2 R=2. The power dissipated in the same
resistor when the current is direct is given by P = i2 R, where i is the current. Set the two
powers equal to each other and solve for i. You should get
60 A = 1:84 A :
i = pI = 2:p
2
2
73E
2 R = I 2 R=2:
Use Pav = Irms
(a) Pav = 0, since R = 0.
(b) Pav = I 2 R=2 = (0:600 A)2 (50 )=2 = 9:0 W:
(c) Pav = I 2 R=2 = (0:198 A)2 (160 )=2 = 3:14 W:
(d) Pav = I 2 R=2 = (0:151 A)2 (160 )=2 = 1:82 W:
74E
75E
2 R
Erms R = Erms
:
Pav = Erms Irms cos = Erms Z
Z
Z2
2 Re = Pmechanical , or
The eective resistance Re satises Irms
(0:100 hp)(746 W/hp) = 177 :
Re = Pmechanical
=
2
Irms
(0:650 A)2
916
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
This is not the same as the resistance R of its coils but just the eective resistance for power
2 R would not give you Pmechanical
transfer from electrical to mechanical form. In fact Irms
but rather the rate of energy loss due to thermal dissipation.
76E
(a) The impedance is given by
p
Z = R2 + (XL XC )2 ;
where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.
Thus
p
Z = (12:0 )2 + (1:30 0)2 = 12:1 :
(b) The average rate at which energy is supplied to the air conditioner is given by
Pav = Erms
Z cos ;
2
where cos is the power factor. Now
12 = 0:994 ;
=
cos = R
Z 12:07 so
77E
V)2 (0:994) = 1:19 103 W :
Pav = (120
12:07 420 V
Erms = p
p
Irms = Erms
= 7:61 A :
=
2
2
Z
(45:0 )2 + (32:0 )2
R + XL
78P
sin2 (!t
Z
2
1
) av = nT=2
sin2 (!t ) dt
0
Z nT
2
1 cos(2!t 2) dt
2
= nT
2
0
nT
2
2
t
1
= nT 2 4! sin(2!t 2) 0 = 21 41! sin(n!T 2) + sin 2 :
nT
CHAPTER 33
917
ELECTROMAGNETIC OSCILLATIONS
Since n!T = n!(2=!) = 2n, we have sin(n!T
[sin(n!T 2) + sin 2] = 0. Thus
2) = sin(2n 2) = sin 2 so
sin2 (!t ) av = 21 :
79P
(a) The energy stored in the capacitor is given by UE = q2 =2C . Since q is a periodic
function of t with period T , so must be UE . So UE will not be changed over one complete
cycle.
(b) Similarly, the energy stored in the inductor is UB = 21 i2 L. Since i is a periodic function
of t with period T , so must be UB .
(c) The energy supplied by the generator is
1
Pav T = (Irms Erms cos )T = 2 T Em I cos ;
p
p
where we used Irms = I= 2 and Erms = Em = 2.
(d) The energy dissipated by the resistor is
Pav, resistor T = (Irms VR )T = Irms (Irms R)T = 21 T I 2 R ;
p
where we used Irms = I= 2.
(e) Since Em I cos = Em I (VR =Em ) = Em I (IR=Em ) = I 2 R, the two quantities are indeed
the same.
80P
The current in the circuit satises i(t) = I sin(!t ), where
I = EZm = p 2 Em
R + (!L 1=!C )2
45:0 V
=q
(16:0 )2 + f(3000 rad/s)(9:20 mH) 1=[(3000 rad/s)(31:2 F)]g2
= 1:93 A
and
= tan XL R XC = tan 1 !L R1=!C
(3000
rad/s)(9
:
20
mH)
1
1
= tan
16:0 (3000 rad/s)(16:0 )(31:2 F)
= 46:5 :
1
918
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
(a) The power supplied by the generator is
Pg = i(t)E (t) = I sin(!t )Em sin !t
= (1:93 A)(45:0 V) sin[(3000 rad/s)(0:442 ms)] sin[(3000 rad/s)(0:442 ms) 46:5 ]
= 41:4 W :
(b) The rate at which the energy in the capacitor changes is
2
Pc = dtd 2qC = i Cq = iVc
2
I
= I sin(!t ) !C cos(!t ) = 2I!C sin[2(!t )]
(1:93 A)2
= 2(3000 rad/s)(31
:2 10 6 F) sin[2(3000 rad/s)(0:442 ms) 2(46:5 )]
= 17:0 W :
(c) The rate at which the energy in the inductor changes is
1
d
di = LI sin(!t ) d [I sin(!t )]
2
Pi = dt 2 Li = Li dt
dt
= 21 !LI 2 sin[2(!t )]
= 12 (3000 rad/s)(1:93 A)2 (9:20 mH) sin[2(3000 rad/s)(0:442 ms) 2(46:5 )]
= 44:1 W :
(d) The rate at which energy is being dissipated by the resistor is
Pr = i2 R = I 2 R sin2 (!t )
= (1:93 A)2 (16:0 ) sin2 [(3000 rad/s)(0:442 ms) 46:5 ]
= 14:4 W :
(e) The negative result for Pi means that energy is being taken away from the inductor at
this particular time.
(f) Pi + Pr + Pc = 44:1W 17:0 W + 14:4 W = 41:5 W = Pg .
81P
Since
PR
to maximize PR we set
= i2 R =
Em
2
r +R R;
dPR = Em2 [(r + R)2 2(r + R)R] = Em2 (r R) = 0 ;
dR
(r + R)4
(r + R)3
CHAPTER 33
919
ELECTROMAGNETIC OSCILLATIONS
which gives R = r.
82P
(a) The power factor is cos , where is the phase angle when the current is written
i = I sin(!t ). Thus = 42:0 and cos = cos( 42:0 ) = 0:743.
(b) Since < 0, !t > !t and the current leads the emf.
(c) The phase angle is given by tan = (XL XC )=R, where XL is the inductive reactance,
XC is the capacitive reactance, and R is the resistance. Now tan = tan( 42:0 ) = 0:900,
a negative number. This means XL XC is negative, or XC > XL . The circuit in the box
is predominantly capacitive.
(d) If the circuit were in resonance XL would be the same as XC , tan would be zero, and
would be zero. Since is not zero, we conclude the circuit is not in resonance.
(e) Since tan is negative and nite, neither the capacitive reactance nor the resistance are
zero. This means the box must contain a capacitor and a resistor. The inductive reactance
may be zero, so there need not be an inductor. If there is an inductor its reactance must
be less than that of the capacitor at the operating frequency.
(f) The average power is
1
1
Pav = Em I cos = (75:0 V)(1:20 A)(0:743) = 33:4 W :
2
2
(g) The answers above depend on the frequency only through the phase angle , which is
given. If values were given for R, L and C then the value of the frequency would also be
needed to compute the power factor.
83P
(a) The average power is given by
Pav = Erms Irms cos ;
where Erms is the root-mean-square emf of the generator, Irms is the root-mean-square
current, and cos is the power factor. Now
Irms =
pI = pEm
;
2
2Z
and Z is the
where I is the current amplitude, Em is the maximum emf of the generator,
p
impedance of the circuit. I = Em =Z was used. In addition, Erms = Em = 2 and cos = R=Z ,
where R is the resistance. Thus
E2 R
Em2 R
Pav = m 2 =
2Z
2 [R2 + (!L 1=!C )2 ] :
p
Here the expression Z = R2 + (!L 1=!C )2 for the impedance in terms of the angular
frequency was substituted.
920
CHAPTER 33
ELECTROMAGNETIC OSCILLATIONS
Considered as a function of C , Pav has its largest value when the factor R2 +(!L 1=!C )2
has the smallest possible value. This occurs for !L = 1=!C , or
1
1
=
1
:17 10 4 F :
C= 2 =
2
2
3
! L (2) (60:0 Hz) (60:0 10 H)
The circuit is then at resonance.
(b) Now you want Z 2 to be as large as possible. Notice that it becomes large without
bound as C becomes small. Thus the smallest average power occurs for C = 0.
(c) When !L = 1=!C the expression for the average power becomes
E
Pav = m ;
2R
2
so the maximum average power is
Pav =
(30:0 V)2 = 90:0 W :
2(5:00 )
The minimum average power is Pav = 0.
(d) At maximum power XL = XC , where XL is the inductive reactance and XC is the
capacitive reactance. The phase angle is
tan = XL R XC = 0 ;
so = 0. At minimum power XC is innite, so tan = 1 and = 90 .
(e) At maximum power the power factor is cos = cos 0 = 1 and at minimum power it is
cos = cos( 90 ) = 0.
84P
The power consumed by the light bulb is P = I 2 R=2. So we must let Pmax =Pmin =
(I=Imin )2 = 5, or
I
Imin
2
=Zmin 2
E
Zmax 2
m
= E =Z
= Z
=
m max
min
p
R2 + (!Lmax )2 2
= 5:
R
Solve for Lmax :
Lmax =
(b) Now we must let
2R = 2(120 V)2 =1000 W = 7:64 10 2 H :
!
2(60:0 Hz)
Rmax + Rbulb 2
= 5;
Rbulb
CHAPTER 33
921
ELECTROMAGNETIC OSCILLATIONS
or
p
p
V)2 = 17:8 :
1)Rbulb = ( 5 1) (120
1000 W
This is not done because the resistors would consume, rather than temporarily store, electromegnetic energy.
Rmax = ( 5
85P
(a)
E
Irms = rms =
Z
Erms
p
R2 + (2fL
1=2fC )2
75:0 V
=q
(15:0 )2 + f2(550 Hz)(25:0 mH) 1=[2(550 Hz)(4:70 F)]g2
= 2:59 A :
(b) The various rms voltages are
Vab = Irms R = (2:59 A)(15:0 ) = 38:8 V ;
2:59 A
I
Vbc = Irms XC = rms =
2fC 2(550 Hz)(4:70 F) = 159 V ;
Vcd = Irms XL = 2Irms fL = 2(2:59 A)(550 Hz)(25:0 mH) = 224 V ;
Vbd = jVbc Vcd j = j159:5 V 223:7 Vj = 64:2 V ;
Vad =
q
Vab2 + Vbd2 =
p
(38:8 V)2 + (64:2 V)2 = 75:0 V :
(c) For L and C the rate is zero since they do not dissipate energy. For R
V 2 (38:8 V)2
PR = ab =
R
15:0 = 100 W :
86E
Use Vs Np = Vp Ns to nd
N
500 = 1:00 103 V :
Vs = Vp s = (100 V)
Np
50
87E
(a)
N
10 = 2:4 V :
Vs = Vp s = (120 V)
Np
500
922
CHAPTER 33
(b)
and
ELECTROMAGNETIC OSCILLATIONS
V
2:4 V = 0:16 A
Is = s =
Rs 15 N
10 = 3:2 10 3 A :
Ip = Is s = (0:16 A)
Np
500
88E
Step up: (i) Use T1 T2 as primary and T1 T3 as secondary coil: V13 =V12 = (800+200)=200 =
5:00; (ii) use T1 T2 as primary and T2 T3 as secondary coil: V23 =V13 = 800=200 = 4:00;
(iii) use T2 T3 as primary and T1 T3 as secondary coil: V13 =V23 = (800 + 200)=800 = 1:25.
Step down: By exchanging the primary and secondary coils in each of the three cases
above we get the following possible ratios: (i) 1=5:00 = 0:200; (ii) 1=4:00 = 0:250; and (iii)
1=1:25 = 0:800.
89P
(a) The rms current in the cable is Irms = P=Vt = 250 103 W=(80 103 V) = 3:125 A.
The rms voltage drop is then V = Irms R = (3:125 A)(2)(0:30 ) = 1:9 V and the rate of
2 R = (3:125 A)(2)(0:60 ) = 5:9 W:
energy dissipation is Pd = Irms
(b) Now Irms = 250 103 W=(8:0 103 V) = 31:25 A, so V = (31:25 A)(0:60 ) = 19 V
and Pd = (3:125 A)2 (0:60 ) = 5:9 102 W.
(c) Now Irms = 250 103 W=(0:80 103 V) = 312:5 A, so V = (312:5 A)(0:60 ) =
1:9 102 V and Pd = (312:5 A)2 (0:60 ) = 5:9 104 W.
Obviously, both the rate of energy dissipation and the voltage drop increase as Vt decreases.
So to minimize these eects the best choice among the three Vt 's above is Vt = 80 kV.
90P
The amplier is connected across the primary windings of a transformer and the resistor R
is connected across the secondary windings. If Is is the rms current in the secondary coil
then the average power delivered to R is Pav = Is2 R. Now Is = (Np =Ns )Ip , where Np is
the number of turns in the primary coil, Ns is the number of turns in the secondary coil,
and Ip is the rms current in the primary coil. Thus
Ip Np 2
Pav =
R:
Ns
Now nd the current in the primary circuit. It acts like a circuit consisting of a generator
and two resistors in series. One resistance is the resistance r of the amplier and the other
is the equivalent resistance Req of the secondary circuit. Thus Ip = E =(r + Req ), where E
is the rms emf of the amplier. According to Eq. 33-73 Req = (Np =Ns )2 R, so
E
Ip =
r + (Np =Ns )2 R
CHAPTER 33
923
ELECTROMAGNETIC OSCILLATIONS
and
Pav =
E 2 (Np =Ns )2 R
:
[r + (Np =Ns )2 R]2
You wish to nd the value of Np =Ns so that Pav is a maximum. To do so, let x = (Np =Ns )2 .
Then
E 2 Rx
Pav =
(r + xR)2
and the derivative with respect to x is
dPav E 2 R(r xR)
= (r + xR)3 :
dx
This is zero for x = r=R = (1000 )=(10 ) = 100. Notice that for small x, Pav increases
linearly with x and for large x it decreases in proportion to 1=x. Thus x = r=R is indeed
a maximum, not a minimum.
Since x = (Np =Ns )2 , maximum power is achieved for (Np =Ns )2 = 100, or Np =Ns = 10.
91
(a) XC = (2)(45 10 6 F)f
(c) 17:7 Hz
92
(b) 17:7 Hz;
(c) 71 V;
(d) 10 Hz;
(e) 87 V;
(f) 31 Hz;
(g) 87 V
93
(a) XL = (2)(40 10 3 H)f ;
(c) 796 Hz
94
(b) 796 Hz;
(c) 71 V;
(d) 2250 Hz;
(e) 94 V;
(f) 281 Hz;
(g) 94 V
95
(b) 61 Hz;
(c) 90 and 61 Hz
1
;