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.
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
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