CHAPTER 31
SOLUTION FOR PROBLEM 15
(a) Since √the frequency of oscillation f is related to the inductance L and capacitance
√ C by
the
smaller
value
of
C
gives
the
larger
value
of
f
.
Hence,
f
=
1/2π
LCmin ,
f = 1/2π LC,
max
√
fmin = 1/2π LCmax, and
√
√
Cmax
365 pF
fmax
= 6.0 .
= √
= √
fmin
10 pF
Cmin
(b) You want to choose the additional capacitance C so the ratio of the frequencies is
r=
1.60 MHz
= 2.96 .
0.54 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
√
C + 365 pF
√
= 2.96 .
C + 10 pF
The solution for C is
C=
(365 pF) − (2.96)2 (10 pF)
= 36 pF .
(2.96)2 − 1
√
(c) Solve f = 1/2π LC for L. For the minimum frequency, C = 365 pF + 36 pF = 401 pF and
f = 0.54 MHz. Thus,
L=
1
1
=
= 2.2 × 10−4 H .
2
2
2
−12
(2π) Cf
(2π) (401 × 10
F)(0.54 × 106 Hz)2
CHAPTER 31
SOLUTION FOR PROBLEM 35
(a) For a given amplitude Em of the generator emf, the current amplitude is given by
I=
Em
Em
=0
,
Z
R2 + (ωd L − 1/ωd C)2
where R is the resistance, L is the inductance, C is the capacitance, and ωd is the angular
frequency. To find the maximum, set the derivative with respect to ωd equal to zero and solve
for ωd . The derivative is
]
}
]}
J 2
o
1
dI
1
2 −3/2
= −Em R + (ωd L − 1/ωd C)
.
ωd L −
L+ 2
dωd
ωd C
ωd C
√
The only factor that can equal zero is ωd L − (1/ωd C) and it does for ωd = 1/ LC. For the
given circuit,
1
1
= 224 rad/s .
ωd = √
=0
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
I=
Em 30.0 V
=
= 6.00 A .
R
5.00 Ω
(c) and (d) You want to find the values of ωd for which I = Em /2R. This means
0
Em
R2 + (ωd L − 1/ωd C)2
=
Em
.
2R
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
w
W2
1
2
R + ωd L −
= 4R2 .
ωd C
Thus
w
ωd L −
1
ωd C
W2
= 3R2 .
Now take the square root of both sides and multiply by ωd C to obtain
p√
Q
2
ωd (LC) ± ωd
3CR − 1 = 0 ,
where the symbol ± indicates the two possible signs for the square root. The last equation is a
quadratic equation for ωd . Its solutions are
√
√
± 3CR ± 3C 2 R2 + 4LC
ωd =
.
2LC
You want the two positive solutions. The smaller of these is
√
√
− 3CR + 3C 2 R2 + 4LC
ω2 =
2LC
√
− 3(20.0 × 10−6 F)(5.00 Ω)
=
2(1.00 H)(20.0 × 10−6 F)
0
3(20.0 × 10−6 F)2 (5.00 Ω)2 + 4(1.00 H)(20.0 × 10−6 F)
+
2(1.00 H)(20.0 × 10−6 F)
= 219 rad/s
and the larger is
√
√
+ 3CR + 3C 2 R2 + 4LC
ω1 =
2LC
√
+ 3(20.0 × 10−6 F)(5.00 Ω)
=
2(1.00 H)(20.0 × 10−6 F)
0
3(20.0 × 10−6 F)2 (5.00 Ω)2 + 4(1.00 H)(20.0 × 10−6 F)
+
2(1.00 H)(20.0 × 10−6 F)
= 228 rad/s .
(e) The fractional width is
ω1 − ω2 228 rad/s − 219 rad/s
= 0.04 .
=
ω0
224 rad/s
CHAPTER 31
SOLUTION FOR PROBLEM 43
(a) The power factor is cos φ, where φ is the phase angle when the current is written i =
I sin(ωd t − φ). Thus φ = −42.0◦ and cos φ = cos(−42.0◦ ) = 0.743.
(b) Since φ < 0, ωd t − φ > ωd t and the current leads the emf.
(c) The phase angle is given by tan φ = (XL − XC )/R, where XL is the inductive reactance,
XC is the capacitive reactance, and R is the resistance. Now tan φ = 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 is in resonance, XL is the same as XC , tan φ is zero, and φ would be zero.
Since φ is not zero, we conclude the circuit is not in resonance.
(e), (f), and (g) Since tan φ is negative and finite, neither the capacitive reactance nor the resistance
is 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.
(h) The average power is
1
1
Pav = Em I cos φ = (75.0 V)(1.20 A)(0.743) = 33.4 W .
2
2
(i) The answers above depend on the frequency only through the phase angle φ, which is given.
If values are given for R, L, and C, then the value of the frequency would also be needed to
compute the power factor.
CHAPTER 31
HINT FOR PROBLEM 11
When the capacitor has maximum charge Q the potential difference across it is a maximum. Use
Q = CV to compute the maximum charge. Here V is the maximum potential difference.
The
√
maximum current is related to the maximum charge by I = ωQ, where ω (= 1/ LC) is the
angular frequency of oscillation. The maximum energy stored in the inductor is 12 LI 2 .
J
ans: (a) 3.0 nC; (b) 1.7 mA; (c) 4.5 nJ
o
CHAPTER 31
HINT FOR PROBLEM 17
The total energy is the sum of the energy stored in the capacitor and the energy stored in the
inductor. The total energy is also the energy in the capacitor when it has maximum charge and
is the energy in the inductor when the current is maximum. If q = Q cos(ωt + φ) the charge on
the capacitor at time t = 0 is q0 = Q cos φ. There are two solutions for φ. To answer part (a)
you want the one for which the current is positive and to answer part (b) you want the one for
which the current is negative. The current at t = 0 is −ωQ sin φ.
J
ans: (a) 1.98 µJ; (b) 5.56 µC; (c) 12.6 mA; (d) −46.9◦ ; (e) +46.9◦
o
CHAPTER 31
HINT FOR PROBLEM 25
The current amplitude is given by Em /XL , where XL (= ωL) is the inductive reactance. The
angular frequency ω is related to the frequency f by ω = 2πf .
J
ans: (a) 95.5 mA; (b) 11.9 mA
o
CHAPTER 31
HINT FOR PROBLEM 31
Removing the capacitor is equivalent to setting the capacitance C equal to infinity (think about a
parallel-plate capacitor with a plate separation of zero). Set the capacitive reactance equal to zero
in the equations for the impedance and phase angle. Use I = Em /Z, where Em is the maximum
emf and Z is the impedance, to compute the maximum current I. Use VR = IR and VL = IXL
to compute the voltage amplitudes for the resistor and inductor, then draw the phasor diagram.
Here XL (ωL) is the inductive reactance, L is the inductance, ω is the angular frequency, and R
is the resistance.
J
ans: (a) 218 Ω; (b) 23.4◦ ; (c) 165 mA
o
CHAPTER 31
HINT FOR PROBLEM 34
The phase constant φ obeys tan φ = (XL − XC )/R, where XL (= ωL) is the inductive reactance,
XC (= 1/ωC) is the capacitive reactance, R is the resistance, L is the inductance, C is the
capacitance, and ω is the angular frequency. Recall that the angular frequency and frequency F
are related by ω = 2πf .
J
ans: 89 Ω
o
CHAPTER 31
HINT FOR PROBLEM 38
Replace the three capacitors, which are in parallel, with their equivalent capacitor and the two
inductors, which are in series, with their equivalent inductor. The circuit is now a series LC
circuit and its resonant frequency is determined by the equivalent capacitance and equivalent
inductance.
J
ans: (a) 796 Hz; (b) no change; (c) decreased; (d) increased
o
CHAPTER 31
HINT FOR PROBLEM 47
The power supplied by the generator is the product of the rms voltage Vt and the rms current
irms . The rate of energy dissipation is i2rms , where R is the total resistance of the two cables.
J
ans: (a) 2.4 V; (b) 3.2 mA; (c) 0.16 A
o