 # Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch31

(Parte 1 de 2)

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

J. c h ch

(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 = LI2 /2 leads to

A. c h 2. According to ULIQC== 1.10 10 H F A. c hc h

F. c h

ch

4. (a) The period is T = 4(1.50 µs) = 6.0 µs. (b) The frequency is the reciprocal of the period:

f T

Hz.

(c) The magnetic energy does not depend on the direction of the current (since UB ∝ i2 ), so this will occur after one-half of a period, or 3.0 µs.

5. (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:

tn T nf

where n = 1, 2, 3, 4, !. The earliest time is (n=1) 5.00s.A tµ=

(b) We note that it takes tT= 12 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,

tT n T n T nf µ−− where n = 1, 2, 3, 4, !. The earliest time is (n=1) 2.50s.tµ=

(c) At tT= 14 , 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 halfperiod (compare steps c and g in Fig. 31-1). Therefore,

Tn T T

where n = 1, 2, 3, 4, !. The earliest time is (n=1) 1.25s.tµ=

6. (a) The angular frequency is

(b) The period is 1/f and f = ω/2π. Therefore,

(c) From ω = (LC) –1/2 , we obtain

ω rad s H F. b g b g.

7. (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,

F c h

c h. .

and

N/m.

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

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

b gc h

Consequently, vmax = 3.02 × 10–3 m/s.

8. We apply the loop rule to the entire circuit:

 with ,,

j jLC R L C R j j j j j j j di q Li R dt C di q Li R L L R R dt C C C where we require εtotal = 0. This is equivalent to the simple LRC circuit shown in Fig. 31- 24(b).

9. The time required is t = T/4, where the period is given by TLC==2ππ/.ω Consequently, t TL C

 4π π HF s.

π π Hz F H. c h c h.

1. (a) Q = CVmax = (1.0 × 10–9 F)(3.0 V) = 3.0 × 10–9 C.

(b) From ULIQC==

HF A. c hc h

(c) When the current is at a maximum, the magnetic field is at maximum:

 UL IB,max == × × = ×−− −

12. The capacitors C1 and C2 can be used in four different ways: (1) C1 only; (2) C2 only; (3) C1 and C2 in parallel; and (4) C1 and C2 in series.

(a) The smallest oscillation frequency is

 −− −π πb g c hc h

(b) The second smallest oscillation frequency is

(c) The second largest oscillation frequency is

(d) The largest oscillation frequency is

13. (a) After the switch is thrown to position b the circuit is an LC circuit. The angular frequency of oscillation is ω=1/LC. Consequently,

 36π π π HF 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 = VC = (34.0 V)(6.20 × 10–6 F) =

2.1 × 10–4 C. The current amplitude is

, and for the second one ω = (L2C2)–1/2 . When the two circuits are connected in series, the new frequency is

15. (a) Since the frequency of oscillation f is related to the inductance L and capacitance C by fLC=12/,π the smaller value of C gives the larger value of f. Consequently, fLCfLCmaxminminmax/,/,==1212ππ and

maxmax min min

10pF

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

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

The solution for C is pF pF pF. b g b g b g

(c) We solve fLC=12/π for L. For the minimum frequency C = 365 pF + 36 pF = 401 pF and f = 0.54 MHz. Thus

π πb g b g c hc hFH z H.. .

16. The linear relationship between θ (the knob angle in degrees) and frequency f is f f

where f0 = 2 × 105 Hz. Since f = ω/2π = 1/2π LC, we are able to solve for C in terms of

2 2 2 π πθ θb g b g with SI units understood. After multiplying by 1012 (to convert to picofarads), this is plotted, below.

17. (a) The total energy U is the sum of the energies in the inductor and capacitor:

qi L U U

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

 QC U== × × = ×−− −2 2 780 10 198 10 556 1066 6 FJ C.c hc h

(c) From U = I2 L/2, we find the maximum current:

A. c h

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

− cos cos qQ C C

For φ = +46.9° the charge on the capacitor is decreasing, for φ = –46.9° it is increasing. To check this, we calculate the derivative of q with respect to time, evaluated for t = 0.

We obtain –ωQ sin φ, 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 φ = –46.9°.

(e) Now we want the derivative to be negative and sin φ to be positive. Thus, we take 46.9.φ=+°

18. (a) Since the percentage of energy stored in the electric field of the capacitor is (175.0%)25.0%−=, then

U qCQC

U LiLI

19. (a) The charge (as a function of time) is given by q = Q sin ωt, where Q is the maximum charge on the capacitor and ω 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 i dq dt Qt==ωωcos,

 QI LC== × × = ×−− −2 0 30 10 2 70 10 180 1036 4 AHF C.b g c hc h

and at t = 0, it is I = ωQ. Since ω=1/,LC (b) The energy stored in the capacitor is given by

qC Qt C sin ω and its rate of change is dUdt Qt tC

2ωω ωsin cos

We use the trigonometric identity cossinsinωωωt= 122bg to write this as dUdt Q C tE = ω ω

The greatest rate of change occurs when sin(2ωt) = 1 or 2ωt = π/2 rad. This means tL C ππ π

(c) Substituting ω = 2π/T and sin(2ωt) = 1 into dUE/dt = (ωQ2 /2C) sin(2ωt), we obtain dUdt QTC QTC EFHG IKJ ==

max

2ππ dUdt EFHG IKJ =

W. c h

chch

We note that this is a positive result, indicating that the energy in the capacitor is indeed increasing at t = T/8.

20. (a) We use ULIQC==

63max max 3

(b) Since f = ω/2π, the frequency is

 π π HF Hz.

(c) Referring to Fig. 31-1, we see that the required time is one-fourth of a period (where the period is the reciprocal of the frequency). Consequently,

e j

21. (a) We compare this expression for the current with i = I sin(ωt+φ0). Setting (ωt+φ) =

2500t + 0.680 = π/2, we obtain t = 3.56 × 10–4 s.

(b) Since ω = 2500 rad/s = (LC)–1/2 ,

ω rad /s F H. b g c h.

(c) The energy is

2. (a) From V = IXC we find ω = I/CV. The period is then T = 2π/ω = 2πCV/I = 46.1 µs.

(b) 1

CV2 = 6.8 nJ.

 (d) We apply Eq. 30-35 as V = L(di/dt)max We can substitute L = CV2/I2

(c) The answer is again 6.8 nJ (see Fig. 31-4). (combining what we found in part (a) with Eq. 31-4) into Eq. 30-35 (as written above) and solve for

 (di/dt)max Our result is 1.02 × 103

A/s.

(e) The derivative of U = 1

Li2 leads to dU/dt = LI2ω sin(ωt)cos(ωt) =

2 LI2ω sin(2ωt).

Therefore, (dU/dt)max = 1

2 LI2ω =

2 IV = 0.938 mW.

23. The loop rule, for just two devices in the loop, reduces to the statement that the magnitude of the voltage across one of them must equal the magnitude of the voltage across the other. Consider that the capacitor has charge q and a voltage (which we’l consider positive in this discussion) V = q/C. Consider at this moment that the current in the inductor at this moment is directed in such a way that the capacitor charge is increasing (so i = +dq/dt). Eq. 30-35 then produces a positive result equal to the V across the capacitor: V = −L(di/dt), and we interpret the fact that −di/dt > 0 in this discussion to mean that d(dq/dt)/dt = d2q/dt2 < 0 represents a “deceleration” of the charge-buildup process on the capacitor (since it is approaching its maximum value of charge). In this way we can “check” the signs in Eq. 31-1 (which states q/C = − L d2q/dt2 ) to make sure we have implemented the loop rule correctly.

24. The charge q after N cycles is obtained by substituting t = NT = 2πN/ω' into Eq. 31-25:

cos cos 2 /

Rt L RNT L RN L C L

qQ e t Qe N

Qe N Qe

We note that the initial charge (setting N = 0 in the above expression) is q0 = Q cos φ, where q0 = 6.2 µC is given (with 3 significant figures understood). Consequently, we write the above result as qqeN NR C L= −0

(a) For N = 5,

(b) For N = 10,

(c) For N = 100,

25. Since ω ≈ ω', we may write T = 2π/ω as the period and ω=1/LC as the angular frequency. The time required for 50 cycles (with 3 significant figures understood) is

0.5104s.

tT LC ω

The maximum charge on the capacitor decays according toqQe Rt L the exponentially decaying amplitude in §31-5), where Q is the charge at time t = 0 (if we take φ = 0 in Eq. 31-25). Dividing by Q and taking the natural logarithm of both sides, we obtain ln maxqQ RtLFHG IKJ =− 2 which leads to qL R

26. The assumption stated at the end of the problem is equivalent to setting φ = 0 in Eq.

31-25. Since the maximum energy in the capacitor (each cycle) is given byqCmax/2, where qmax is the maximum charge (during a given cycle), then we seek the time for which

qC Q C q Qmax max .

Now qmax (referred to as the exponentially decaying amplitude in §31-5) is related to Q (and the other parameters of the circuit) by

qQ e qQ RtL

Rt L max

Setting qQmax=/2, we solve for t:

LR qQ LR L R =− FHG IKJ =− FHG IKJ =

result.

27. 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

q C Q C where qQ e Rt L

(see the discussion of the exponentially decaying amplitude in §31-5). One period later the charge on the fully charged capacitor is

Rt T L qQ e T

and the energy is

Rt T Lq Q Ut T e

The fractional loss in energy is

Ut Ut TUt e e

Rt L R t T L

Rt L

Assuming that RT/L is very small compared to 1 (which would be the case if the resistance is small), we expand the exponential (see Appendix E). The first few terms are:

If we approximate ω ≈ ω', then we can write T as 2π/ω. As a result, we obtain

28. (a) We use I = ε/Xc = ωdCε:

(b) I = 2π(8.0 × 103 Hz)(1.50 × 10–6 F)(30.0 V) = 2.26 A.

29. (a) The current amplitude I is given by I = VL/XL, where XL = ωdL = 2πfdL. Since the circuit contains only the inductor and a sinusoidal generator, VL = εm. Therefore,

Xf L

(b) The frequency is now eight times larger than in part (a), so the inductive reactance XL is eight times larger and the current is one-eighth as much. The current is now

I = (0.0955 A)/8 = 0.019 A = 1.9 mA.

30. (a) The current through the resistor is

31. (a) The inductive reactance for angular frequency ωd is given by XL = ωdL, and the capacitive reactance is given by XC = 1/ωdC. The two reactances are equal if ωdL = 1/ωdC, or 1/dLCω=. The frequency is d f

(b) The inductive reactance is

XL = ωdL = 2πfdL = 2π(650 Hz)(6.0 × 10–3 H) = 24 Ω. The capacitive reactance has the same value at this frequency.

(c) The natural frequency for free LC oscillations isfLC=ω/2π=1/2π, the same as we found in part (a).

32. (a) The circuit consists of one generator across one inductor; therefore, εm = VL. The current amplitude is

(b) When the current is at a maximum, its derivative is zero. Thus, Eq. 30-35 gives εL = 0 at that instant. Stated another way, since ε(t) and i(t) have a 90° phase difference, then ε(t) must be zero when i(t) = I. The fact that φ = 90° = π/2 rad is used in part (c).

(c) Consider Eq. 32-28 with εε=− 1

2m. In order to satisfy this equation, we require sin(ωdt) = –1/2. Now we note that the problem states that ε is increasing in magnitude, which (since it is already negative) means that it is becoming more negative. Thus, differentiating Eq. 32-28 with respect to time (and demanding the result be negative) we must also require cos(ωdt) < 0. These conditions imply that ωt must equal (2nπ – 5π/6) [n = integer]. Consequently, Eq. 31-29 yields (for all values of n) iI n=− FHG IKJ =× F

 KJ =×−−sin ( 2 522 10 451 1033π−

A) 3

3. (a) The generator emf is a maximum when sin(ωdt – π/4) = 1 or ωdt – π/4 = (π/2) ±

2nπ [n = integer]. The first time this occurs after t = 0 is when ωdt – π/4 = π/2 (that is, n = 0). Therefore,

t d

(b) The current is a maximum when sin(ωdt – 3π/4) = 1, or ωdt – 3π/4 = (π/2) ± 2nπ [n = integer]. The first time this occurs after t = 0 is when ωdt – 3π/4 = π/2 (as in part (a), n = 0). Therefore,

t d

(d) The current amplitude I is related to the voltage amplitude VL by VL = IXL, where XL is the inductive reactance, given by XL = ωdL. 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 = εm. Thus, εm = IωdL and

34. (a) The circuit consists of one generator across one capacitor; therefore, εm = VC. Consequently, the current amplitude is

I X Cm

(b) When the current is at a maximum, the charge on the capacitor is changing at its largest rate. This happens not when it is fully charged (±qmax), but rather as it passes through the (momentary) states of being uncharged (q = 0). Since q = CV, then the voltage across the capacitor (and at the generator, by the loop rule) is zero when the current is at a maximum. Stated more precisely, the time-dependent emf ε(t) and current i(t) have a φ = –90° phase relation, implying ε(t) = 0 when i(t) = I. The fact that φ = –90° = –π/2 rad is used in part (c).

(c) Consider Eq. 32-28 with εε=− 1

2m. In order to satisfy this equation, we require sin(ωdt) = –1/2. Now we note that the problem states that ε is increasing in magnitude, which (since it is already negative) means that it is becoming more negative. Thus, differentiating Eq. 32-28 with respect to time (and demanding the result be negative) we must also require cos(ωdt) < 0. These conditions imply that ωt must equal (2nπ – 5π/6) [n = integer]. Consequently, Eq. 31-29 yields (for all values of n) iI n

35. (a) Now XC = 0, while R = 200 Ω and XL = ωL = 2πfdL = 86.7 Ω 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 εm leads I. The phasor diagram is drawn to scale below.

36. (a) The graph shows that the resonance angular frequency is 25000 rad/s, which means (using Eq. 31-4)

(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 Ω.

37. (a) Now XL = 0, while R = 200 Ω and XC = 1/2πfdC = 177 Ω. Therefore, the impedance is

(b) The phase angle is

=tan tan 41.5200 LC

(c) The current amplitude is

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

The circuit is capacitive, so I leads εm. The phasor diagram is drawn to scale next.

38. (a) The circuit has a resistor and a capacitor (but no inductor). Since the capacitive reactance decreases with frequency, then the asymptotic value of Z must be the resistance:

(Parte 1 de 2)