Halliday-HW-Ch31

Fundamentals of PhysicsHalliday & Resnic

東海大學物理系98學年度第 2學期

系級：□0401物理組 □0409應用物理組

學號：姓名： .

【CH31】Alternating Fields and Current7, 9, 11, 15, 19, 21, 25, 29, 33, 36, 37, 45, 51, 55, 57, 63, 65

【Problem 31-7】The energy in an oscillating LC circuit containing a 1.25 H inductor is . The maximum

charge on the capacitor is . For a mechanical system with the same period, find the (a) mass,

(b) spring constant, (c) maximum displacement, and (d) maximum speed.

<解>：(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,

And

(c) The maximum displacement corresponds to the maximum charge, so

(d) The maximum speed vmax corresponds to the maximum current. The maximum current is

Consequently, vmax = 3.02 10–3 m/s.

【Problem 31-9】In an oscillating LC circuit with and , the current is initially a maximum. How

long will it take before the capacitor is fully charged for the first time?

<解>：The time required is t = T/4, where the period is given by

Consequently,

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【Problem 31-11】In Fig. 31-28, , , and , and the ideal battery has emf . The

switch is kept in position a for a long time and then thrown to position b. What are the (a) frequency

and (b) current amplitude of the resulting oscillations?

（圖 31-28）<解>：(a) After the switch is thrown to position b the circuit is an LC circuit. The angular

frequency of oscillation is . Consequently,

(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

. The current amplitude is

【Problem 31-15】A variable capacitor with a range from 10 to 365 pF is used with a coil to form a variable-frequency

LC circuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum

frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from

0.54 MHz to 1.60 MHz, the ratio computed in (a) is too large. By adding a capacitor in parallel to

the variable capacitor, this range can be adjusted. To obtain the desired frequency range, (b) what

capacitance should be added and (c) what inductance should the coil have?

<解>：(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. Consequently,

and




(b) An additional capacitance C is chosen so the ratio of the frequencies is

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 (pF), then

The solution for C is

(c) We solve for L. For the minimum frequency C = 365 pF + 36 pF = 401

pF and f = 0.54 MHz. Thus

【Problem 31-19】In an oscillating LC circuit, and . At time the current is 9.20 mA, the

charge on the capacitor is , and the capacitor is charging. What are (a) the total energy in the

circuit, (b) the maximum charge on the capacitor, and (c) the maximum current? (d) If the charge on the capacitor is given by , what is the phase angle ? (e) Suppose the data are the

same, except that the capacitor is discharging at . What then is ?

<解>：(a) The total energy U is the sum of the energies in the inductor and capacitor:

(b) We solve U = Q2/2C for the maximum charge:




(c) From U = I2L/2, we find the maximum current:

(d) If q0 is the charge on the capacitor at time t = 0, then q0 = Q cos and

For = +46.9° the charge on the capacitor is decreasing, for = –46.9° it is increasing.

To check this, we calculate the derivative of q with respect to time, evaluated for t = 0.

We obtain –Q sin , which we wish 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 we want the derivative to be negative and sin to be positive. Thus, we take

【Problem 31-21】In an oscillating LC circuit, and . At the charge on the capacitor is zero

and the current is 2.00 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time is the rate at which energy is stored in the capacitor greatest, and (c) what is

that greatest rate?

<解>：(a) The charge (as a function of time) is given by , 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 (as a function of time) is

and at t = 0, it is I = Q. Since

(b) The energy stored in the capacitor is given by

and its rate of change is

We use the trigonometric identity to write this as




The greatest rate of change occurs when sin(2t) = 1 or 2t = /2 rad. This means

(c) Substituting = 2/T and sin(2t) = 1 into dUE/dt = (Q2/2C) sin(2t), we obtain

Now so

We note that this is a positive result, indicating that the energy in the capacitor is indeed

increasing at t = T/8.

【Problem 31-25】What resistance R should be connected in series with an inductance and capacitance

for the maximum charge on the capacitor to decay to 99.0% of its initial value in 50.0

cycles? (Assume .)

<解>：Since ', we may write T = 2/ as the period and as the angular

frequency. The time required for 50 cycles (with 3 significant figures understood) is

The maximum charge on the capacitor decays according to (this is called the

exponentially decaying amplitude in §31-5), where Q is the charge at time t = 0 (if we take = 0 in Eq. 31-25). Dividing by Q and taking the natural logarithm of both sides, we obtain

which leads to

【Problem 31-29】




(a) At what frequency would a 6.0 mH inductor and a 10 _F capacitor have the same reactance? (b)

What would the reactance be? (c) Show that this frequency would be the natural frequency of an

oscillating circuit with the same L and C.

<解>：(a) The inductive reactance for angular frequency d is given by , and the

capacitive reactance is given by XC = 1/dC. The two reactances are equal if dL =

1/dC, or . The frequency is

(b) The inductive reactance is

XL = dL = 2fdL = 2(650 Hz)(6.0 10–3 H) = 24 .

The capacitive reactance has the same value at this frequency.

(c) The natural frequency for free LC oscillations is , the same as we

found in part (a).

【Problem 31-33】

An ac generator has emf , where and . The current

produced in a connected circuit is , where . At what time after

does (a) the generator emf first reach a maximum and (b) the current first reach a maximum? (c)

The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a

resistor? Justify your answer. (d) What is the value of the capacitance, inductance, or resistance, as

the case may be?

<解>：(a) The generator emf is a maximum when sin(dt – /4) = 1 or

dt – /4 = (/2) ± 2n [n = integer].

The first time this occurs after t = 0 is when dt – /4 = /2 (that is, n = 0). Therefore,

(b) The current is a maximum when sin(dt – 3/4) = 1, or

dt – 3/4 = (/2) ± 2n [n = integer].

The first time this occurs after t = 0 is when dt – 3/4 = /2 (as in part (a), n = 0).

Therefore,




(c) The current lags the emf by rad, so the circuit element must be an inductor.

(d) The current amplitude I is related to the voltage amplitude VL by VL = IXL, where XL is the

inductive reactance, given by XL = dL. Furthermore, since there is only one element in

the circuit, the amplitude of the potential difference across the element must be the same

as the amplitude of the generator emf: VL = m. Thus, m = IdL and

【Problem 31-36】The current amplitude I versus driving angular frequency for a driven RLC circuit is given in

Fig. 31-29, where the vertical axis scale is set by . The inductance is , and the emf

amplitude is 8.0 V. What are (a) C and (b) R?

（圖 31-29）<解>：(a) The graph shows that the resonance angular frequency is 25000 rad/s, which means

(using Eq. 31-4)

C = (2L)1 = [(25000)2 ×200 × 106]1 = 8.0 F.

(b) The graph also shows that the current amplitude at resonance is 4.0 A, but at resonance

the impedance Z becomes purely resistive (Z = R) so that we can divide the emf

amplitude by the current amplitude at resonance to find R: 8.0/4.0 = 2.0 .

【Problem 31-37】Remove the capacitor from the circuit in Fig. 31-7 and set , , , and

. What are (a) , (b) , and (c) ? (d) Draw a phasor diagram.

（圖 31-7）<解>：(a) Now XC = 0, while R = 200 and XL = L = 2fdL = 86.7 remain unchanged.

Therefore, the impedance is




(b) The phase angle is, from Eq. 31-65,

(c) The current amplitude is now found to be

(d) We first find the voltage amplitudes across the circuit elements:

This is an inductive circuit, so m leads I. The phasor diagram is drawn to scale below.

【Problem 31-45】An RLC circuit such as that of Fig. 31-7 has , , , and . (a) At

what angular frequency will the current amplitude have its maximum value, as in the resonance

curves of Fig. 31-16? (b) What is this maximum value? At what (c) lower angular frequency

and (d) higher angular frequency will the current amplitude be half this maximum value? (e)

What is the fractional half-width of the resonance curve for this circuit?

（圖 31-7）<解>：(a) For a given amplitude m of the generator emf, the current amplitude is given by




We find the maximum by setting the derivative with respect to equal to zero:

The only factor that can equal zero is ; it does so for .

For this

(b) When , the impedance is Z = R, and the current amplitude is

(c) We want to find the (positive) values of for which

This may be rearranged to yield

Taking the square root of both sides (acknowledging the two ± roots) and multiplying by , we obtain

Using the quadratic formula, we find the smallest positive solution

(d) The largest positive solution




(e) The fractional width is

【Problem 31-51】In Fig. 31-33, a generator with an adjustable frequency of oscillation is connected to resistance

, inductances and , and capacitances , ,

and . (a) What is the resonant frequency of the circuit? (Hint: See Problem 47 in

Chapter 30.) What happens to the resonant frequency if (b) R is increased, (c) is increased, and

(d) is removed from the circuit?

（圖 31-33）<解>：(a) Since Leq = L1 + L2 and Ceq = C1 + C2 + C3 for the circuit, the resonant frequency is

(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 increases.

(d) Since and Ceq decreases as C3 is removed, will increase.

【Problem 31-55】An air conditioner connected to a 120 V rms ac line is equivalent to a resistance and a

inductive reactance in series. Calculate (a) the impedance of the air conditioner and (b) the average

rate at which energy is supplied to the appliance.

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<解>：(a) Using Eq. 31-61, the impedance is

(b) The average rate at which energy has been supplied is

【Problem 31-57】Figure 31-35 shows an ac generator connected to a “black box” through a pair of terminals. The box

contains an RLC circuit, possibly even a multiloop circuit, whose elements and connections we do

not know. Measurements outside the box reveal that

and

(a) What is the power factor? (b) Does the current lead or lag the emf? (c) Is the circuit in the box

largely inductive or largely capacitive? (d) Is the circuit in the box in resonance? (e) Must there be a

capacitor in the box? (f) An inductor? (g) A resistor? (h) At what average rate is energy delivered to

the box by the generator? (i) Why don’t you need to know to answer all these questions?

（圖 31-35）<解>：(a) The power factor is cos , where is the phase constant defined by the expression i = I

sin(t – ). Thus, = – 42.0° and cos = cos(– 42.0°) = 0.743.

(b) Since < 0, t – > t. The current leads the emf.

(c) The phase constant is related to the reactance difference by tan = (XL – XC)/R. We have

tan = tan(– 42.0°) = –0.900,

a negative number. Therefore, XL – XC is negative, which leads to 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 finite, neither the capacitive reactance nor the resistance are

zero. This means the box must contain a capacitor and a resistor.

(f) The inductive reactance may be zero, so there need not be an inductor.

(g) Yes, there is a resistor.

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(h) The average power is

(i) The answers above depend on the frequency only through the phase constant , 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.

【Problem 31-63】A transformer has 500 primary turns and 10 secondary turns. (a) If is 120 V (rms), what is

with an open circuit? If the secondary now has a resistive load of , what is the current in the (b)

primary and (c) secondary?

<解>：(a) The stepped-down voltage is

(b) By Ohm’s law, the current in the secondary is

We find the primary current from Eq. 31-80:

(c) As shown above, the current in the secondary is

【Problem 31-65】An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission

line. At the factory a step-down transformer reduces the voltage from its (rms) transmission value

to a much lower value that is safe and convenient for use in the factory. The transmission line

resistance is , and the power of the generator is 250 kW. If , what are (a) the voltage decrease along the transmission line and (b) the rate Pd at which energy is dissipated in

the line as thermal energy? If , what are (c) and (d) ? If , what are (e)

and (f) ?

<解>：(a) The rms current in the cable is

Therefore, the rms voltage drop is .

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(b) The rate of energy dissipation is

(c) Now , so

(d)

(e) , so .

(f)

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Halliday-HW-Ch31