CHAPTER 33. Answer to Checkpoint Questions

Transcription

1 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 887 CHAPTE 33 Answer to Checkoint Questions. (a) T; (b) T ; (c) T; (d) T4. (a) 5 V; (b) 50 J 3. (a) ; (b) 4. (a) C, B, A; (b) A, B, 3 S, 4 C; (c) A 5. (a) increases; (b) decreases 6. (a), lags;, leads; 3, in hase; (b) 3 (! d! when X L X C ) 7. (a) increase (circuit is mainly caacitive; increase C to decrease X C to be closer to resonance for maximum P av ); (b) closer 8. ste-u Answer to Questions. (a) T4; (b) T4; (c) T (see Fig. 33-); (d) T (see Eq. 3{40). with n an integer, (a) 0 n; (c) n; (e) n; (g) 3 n 3. (b), (a), (c) 4. (a) less; (b) greater 5. (a) 3,, ; (b), tie of and 3 6. (a) decrease; (b) same (U B equals U E, which has not been changed) 7. slower 8. (a) 3,, ; (b), 3, 9. (a) and 4; (b) and 3 0. (a) leads; (b) caacitive; (c) less. (a) 3, then and tie; (b),, 3. (a) decrease; (b) increase 3. (a) negative; (b) lead

2 888 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 4. (a) less, (b) equal, (c) greater 5. (a) rightward, increase (X L increases, closer to resonance); (b) rightward increase (X C decreases, closer to resonance); (c) rightward, increase (! d! increases, closer to resonance) 6. (a) ositive; (b) decrease L (to decrease X L and get closer to resonance); (c) decrease C (to increase X C and get closer to resonance) Solutions to Exercises & Problems E Solve C from U Q C: C Q U (: C) ( J) 9:4 0 9 F : E Solve I from U LI : I r U L r (0:0 0 6 J) : :5 A : 3E From U LI Q C we nd I Q LC 3: C (:0 0 3 H)(4: F) 4:5 0 A : 4E (a) All the energy in the circuit resides in the caacitor when it has its maximum charge. The current is then zero. If C is the caacitance and Q is the maximum charge on the caacitor then the total energy is U Q C (: C) (3: F) :7 0 6 J :

3 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 889 (b) When the caacitor 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 and I r U L r (: J) H 5: A : 5E (a) T 4(:50 s) 6:00 s. (b) f T (6:00 s) : Hz: (c) Half a eriod, or 3:00 s. 6P efer to Fig (a) t a nt n T n : Hz n(5:00 s) ; where n ; ; 3; 4; : (b) First, it takes T for the charge on the other late to reach its maximum ositive value for the rst time. Later this will reeat once every eriod. So t b T + nt (n + ) (n + )T f (n + ) ( 0 3 Hz) (n + )(:50 s) ; where n 0; ; ; 3; 4; : (c) First, it takes T for the magnetic eld in the inductor to reach its maximum value for 4 the rst time. Later this will reeat every half a eriod. So t c T 4 + nt t b (n + )(:5 s) where n 0; ; ; 3; 4; : 7E (a) (b)! r r k m Fx m T! s 8:0 N (:0 0 3 m)(0:50 kg) 89 rad/s 7:0 0 s : 89 rad/s :

4 890 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS (c) Let! (LC) and solve for C: C! L (89 rad/s) (5:0 H) :5 0 5 F : 8P (a) The mass m corresonds to the inductance, so m :5 kg. (b) The sring constant k corresonds to the recirocal of the caacitance. Since the total energy is given by U Q C, where Q is the maximum charge on the caacitor and C is the caacitance, C Q U C (5: J) : F and k : m/n 37 N/m : (c) The maximum dislacement x m corresonds to the maximum charge, so x m m : (d) The maximum seed v m corresonds to the maximum current. The maximum current is I Q! Q C LC (:5 H)(: F) 3:0 0 3 A : Thus v m 3:0 0 3 m/s. 9E Solve L from f ( LC) : L 4 f C 4 (0 0 3 Hz) (6:7 0 6 F) 3:8 0 5 H : 0E Solve C from f ( LC) : C 4 f L 4 (3: Hz) (: H) : F :

5 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 89 E If T is the eriod of oscillation, then the time required is t T4. The eriod is given by T! LC, where! is the angular frequency of oscillation, L is the inductance, and C is the caacitance. Hence t T 4 LC 4 (0:050 H)(4:0 0 6 F) 4 7:0 0 4 s : E When switch S is closed and the others are oen, the inductor is essentially out of the circuit and what remains is an C circuit. The time constant is C C. When switch S is closed and the others are oen the caacitor is essentially out of the circuit and what remains is an L circuit with time constant L L. When switch S 3 is closed and the others are oen the resistor is essentially out of the circuit and what remains is an LC circuit that oscillates with eriod T LC. Substitute L L and C C to obtain T C L. 3E Aly the loo rule to the LC circuit: Since i dqdt; didt d qdt. Thus E total E L + E C L di dt + q C 0 : L d q dt + q C 0 : 4E Aly the loo rule to the entire circuit: E total E L + E C + E + X j X j L j di dt + q C j + i j L di dt + q C + i 0 ; E Lj + E Cj + E j where L P j L j, C ( P j C j ), and P j j. This is equivalent to the simle LC circuit shown in Fig. 33-3(b).

6 89 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 5P (a) Q CV max (:0 0 9 F)(3:0 V) 3:0 0 9 C. (b) From U LI Q C we get I Q LC 3:0 0 9 C (3:0 0 3 H)(:0 0 9 F) :7 0 3 A : (c) U B;max LI (3:0 0 3 H)(:7 0 3 A) 4:5 0 9 J : 6P (a)! LC (3: H)(0:0 0 6 F) 5:77 03 rad/s : (b) q T! 5: rad/s : s : (c) Use q(t) Q cos(!t). The sketch is shown below. Here Q 0 C and T : s. q max 0 T t -q max 7P (a) Use U LI Q C to solve for L: L Q C I C I (4: F) CVmax Vmax C I 3: H : :50 V 50:0 0 3 A

7 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 893 (b) f (c) From Fig. 33- LC (3: H)(4: F) :33 03 Hz : t 4 T 4f 4(: Hz) : s : 8P (a) After the switch is thrown to osition b the circuit is an LC circuit. frequency of oscillation is! LC and the frequency is The angular f! LC (54:0 0 3 H)(6:0 0 6 F) 75 Hz : (b) When the switch is thrown, the caacitor is charged to V 34:0 V and the current is zero. Thus the maximum charge on the caacitor is Q V C (34:0 V)(6:0 0 6 F) : 0 4 C. The current amlitude is I!Q fq (75 Hz)(: 0 4 C) 0:365 A : 9P The caacitors C and C can be used in four dierent ways: () C only; () C only; (3) C and C in arallel; and (4) C and C in series. The corresonding oscillation frequencies are: f LC (:0 0 H)(5:0 0 6 F) 7: 0 Hz ; and f 3 f 4 f LC (:0 0 H)(:0 0 6 F) : 03 Hz ; L(C + C ) (:0 0 H)(:0 + 5:0)0 6 F 6:0 0 Hz ; s LC C (C + C ) :3 0 3 Hz : (:0 + 5:0)0 6 F (:0 0 H)(:0 0 6 F)(5:0 0 6 F)

8 894 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 0P (a) Solve L from f (LC) : (b) L 4 f C 4 (0:4 0 3 Hz) ( F) 6: H : U LI (6: H)(7:0 0 3 A) :79 0 J : (c) Solve for Q from U Q C: Q CU ( F)(:79 0 J) :0 0 7 C : P (a) At any time the total energy U in the circuit is the sum of the energy U E in the electric eld of the caacitor and the energy U B in the magnetic eld of the inductor. When U E 0:500U B, then U B :00U E and U U E + U B 3:00U E. U E is given by q C, where q is the charge on the caacitor and C is the caacitance. The total energy U is given by Q C, where Q is the maximum charge on the caacitor, so Q C 3:00q C or q Q 3:00 0:577Q. (b) If the caacitor is fully charged at time t 0 then the charge on the caacitor 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! T, where T is the eriod of oscillation, t 0:955T 0:5T. P (a) Since the ercentage of energy stored in the electric eld of the caacitor is ( 75:0%) 5:0%, U E U q C Q C 5:0% ; which gives q 5:0% Q 0:500Q: (b) From U B U Li LI 75:0% we get i 75:0% I 0:866I : 3P The frequency f as a function of, the angle of rotation of the knob, is given by f() f 0 + k, where f Hz and k [f(80 ) f 0 ] (4 0 5 Hz 0 5 Hz) 6:4 0 4 Hz/rad. Thus from f() [LC()] we get C() 4 Lf () 4 (f 0 + k) L :

9 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 895 The lot of C() vs is as follows C (nf) θ (rad) 4P (a) Since the frequency of oscillation f is related to the inductance L and caacitance C by f LC, the smaller value of C gives the larger value of f. Hence, f max LC min, f min LC max, and f max f min Cmax Cmin 365 F 0 F 6:0 : (b) You want to choose the additional caacitance C so the ratio of the frequencies is r :60 MHz 0:54 MHz :96 : Since the additional caacitor is in arallel with the tuning caacitor, its caacitance adds to that of the tuning caacitor. If C is in icofarads, then C F C + 0 F :96 : The solution for C is C (365 F) (:96) (0 F) (:96) 36 F : Solve f LC for L. For the minimum frequency C 365 F + 36 F 40 F and f 0:54 MHz. Thus L () Cf () (40 0 F)(0: Hz) : 0 4 H :

10 896 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 5P (a) The total energy U is the sum of the energies in the inductor and caacitor. If q is the charge on the caacitor, C is the caacitance, i is the current, and L is the inductance, then U U E + U B q C + i L (3: C) (7: F) + (9:0 0 3 A) (5:0 0 3 H) (b) Solve U Q C for the maximum charge Q: : J : Q CU (7: F)(: J) 5: C : (c) Solve U I L for the maximum current I: I r U L r (: J) 5:0 0 3 H :6 0 A : (d) If q 0 is the charge on the caacitor at time t 0, then q 0 Q cos and cos q Q cos 3: C 5: C 46:9 : For +46:9 the charge on the caacitor is decreasing, for 46:9 it is increasing. To check this calculate the derivative of q with resect to time, evaluated for t 0. You should get!q sin. You want this to be ositive. Since sin(+46:9 ) is ositive 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 ositive. Take +46:9. 6P (a) The charge is given by q(t) Q sin!t, where Q is the maximum charge on the caacitor 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!q cos!t dt and at t 0 it is I!Q. Since! LC, Q I LC (:00 A) (3: H)(: F) : C : (b) The energy stored in the caacitor is given by U E q C Q sin!t C

11 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 897 and its rate of change is du E dt Q! sin!t cos!t C Use the trigonometric identity cos!t sin!t sin(!t) to write this du E dt!q C sin(!t) : The greatest rate of change occurs when sin(!t) or!t rad. This means t 4! T 4() T 8 ; where T is the eriod of oscillation. The relationshi! T was used. (c) Substitute! T and sin(!t) into du E dt (!Q C) sin(!t) to obtain due Q dt T C Q T C : Now T LC (3: H)(: F) 5: s, so due (: C) dt (5: s)(: F) 66:7 W : max max Notice that this is a ositive result, indicating that the energy in the caacitor is indeed increasing at t T8. : 7P (a) Consider the joint oint P as shown. The current i 0 owing out of oint P into the middle inductor is i 0 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 loo equation reads i P i' i or L di dt + q C + L di dt + q C 0 ; L di dt + q C 0 : This is equivalent with a single LC circuit with L eq L and C eq C. Thus! Leq C eq (L)(C) LC :

12 898 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS (b) Now i 0 i + i i (see the gure to the right). Aly the loo equation to the left half of the circuit containing the caacitor on the left and the inductors on the left and in the middle to obtain i P i' i or L di dt + q C + L di0 dt L di dt + q C L eq di dt + q C eq 0 ; + L di dt 0 ; where L eq 3L and C eq C. The corresonding oscillation angular frequency is then! : Leq C eq 3LC (Note that in a single-loo LC circuit there is only one frequency,! () L eq C eq. This contradicts with our case here so it is imossible to reduce our circuit to a single-loo one.) 8P For the rst circuit! (L C ), and for the second one! (L C ). When the two circuits are connected in series the new frequency is! 0 Leq C eq (L + L )C C (C + C ) (L C C + L C C )(C + C ) L C (C + C )(C + C )! ; where we used! L C L C : 9P Comare the exression for i here with i I sin(!t+ 0 ). Let (!t+) 500t+0:680 to obtain t 3: s. (b) Since! 500 rad/s (LC), L! C (500 rad/s) (64:0 0 6 F) : H : (c) U LI (: H)(:60 A) 3:0 0 3 J :

13 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS P The energy needed to charge the 00 F caacitor to 300 V is C 00V ( F)(300 V) 4:50 J : The energy originally in the 900 F caacitor is C 900V ( F)(00 V) 4:5 J : All the energy originally in the 900 F caacitor must be transferred to the 00 F caacitor. The lan is to store it temorarily in the inductor. To do this leave switch S oen and close switch S. Wait until the 900 F caacitor is comletely discharged and the current in the right-hand loo is a maximum. This is one quarter of the eriod of oscillation. Since T 900 LC 900 (0:0 H)( F) 0:596 s ; you should wait (0:596 s)4 0:49 s. At that instant close switch S and oen switch S so the current is in the left-hand loo. Now wait one quarter of the eriod of oscillation of the left-hand LC circuit and oen switch S. The 00 F caacitor then has maximum charge and all the energy resides in it. The eriod of oscillation is T 00 LC 00 (0:0 H)( F) 0:99 s and you must kee S closed for (0:99 s)4 0:0497 s before oening it again. 3E In Eq. 33-, let where n 50:0 and T LC, we get q Q e ntl 99:0% ; ln(99:0%) n r L C ln(99:0%) 50:0 r H :0 0 6 F 8: : 3E (a) Since T! LC, we may rewrite the ower on the exonential factor as r C Thus e tl e CL(tT ) : L t T r C L t LC t L :

14 900 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS (b) Since CL(tT ) must be unitlss and so is tt, CL must also be unitless. Thus the SI unit of CL must be, i.e., the SI unit for LC must be. (c) Since the amlitude of oscillation reduces by a factor of e CL(TT ) e CL after each cycle, the condition is equivalent to CL, or LC. 33P Since the maximum amount of energy stored in the caacitor during each cycle is given by Q C, where Q is the maximum charge and C is the caacitance, you want the time for which q C Q C : This means q Q. Now q is given by q Qe tl ; where is the resistance and L is the inductance in the circuit. Divide by Q and take the natural logarithm of both sides to obtain q ln t Q L : Solve for t: t L ln q Q L ln L ln : The identities ln( ) ln ln were used to obtain the last form of the result. 34P The charge q after N cycles is obtained by stbstituting t NT N! 0 into Eq. 33-: q Qe tl cos(! 0 t) Qe NTL Qe N( LC)L Qe N CL : So qj N5 (6:0 C)e 5(7:0 ) 3:0 F:0 H 5:85 C; qj N0 (6:0 C)e 0(7:0 ) 3:0 F:0 H 5:5 C; and qj N00 (6:0 C)e 00(7:0 ) 3:0 F:0 H :93 C: 35P (a) In Eq. 33-, let q 0 and t 0 to obtain 0 Q cos. This gives.

15 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 90 (b) First, calculate the current i(t) from Eq. 33-: h i i(t) dq dt d Qe tl cos(! 0 t + ) dt Q L e tl cos(! 0 t + )! 0 Qe tl sin(! 0 t + ) s Q! 0 + e tl sin(! 0 t + + ) ; L where tan L! 0. The current amlitude is then s I(t) Q! 0 + e tl Q!e tl Ie tl : L Thus q(t) Qe tl cos(! 0 t + ) I! e tl cos! 0 t + I! e tl sin(! 0 t) : 36P (a) From Eq. 33- and dq d q dt dt d dt h i Qe tl cos(! 0 t + ) Q L e tl cos(! 0 t + )! 0 Qe tl sin(! 0 t + ) L Q e tl cos(! 0 t + )! 0 Q sin(! 0 t + ) L + e tl Q! 0 L sin(!0 t + )! 0 Q cos(! 0 t + ) Substitute these exressions and Eq. 33- into Eq. 33-0: Qe tl! 0 L + cos(! 0 t + ) 0 : L c Since this equation is valid at any time t, we must have :! 0 L L + C 0

16 90 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS or! 0 s LC (b) The fractional shift in frequency is!!!!0! s L s! L : (LC) (L) LC (00 ) (7: F) 4(4:40 H) r 0:0% : C 4L 37P Let t be a time at which the caacitor is fully charged in some cycle and let q max be the charge on the caacitor then. The energy in the caacitor at that time is U(t) q max C Q C e tl ; where q max Q e tl was used. Here Q is the charge at t 0. One cycle later the maximum charge is q max Q e (t+t )L and the energy is U(t + T ) q max C Q C e (t+t )L ; where T is the eriod of oscillation. The fractional loss in energy is U U U(t) U(t + T ) U(t) (t+t )L e tl e e tl e TL : Assume that TL is small comared to (the resistance is small) and use the binomial theorem to exand the exonential. The rst two terms are: e TL T L : elace T with!, where! is the angular frequency of oscillation. Thus U U T T L L!L :

17 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS E Use I EX c!ce. (a) I!CE m fce m ()(: Hz)(: F)(30:0 V) 0:83 A: (b) I ()(8: Hz)(: F)(30:0 V) :6 A. 39E (a) The current amlitude I is given by I V L X L, where V L is the voltage amlitude across the inductor and X L is the inductive reactance. The reactance is given by X L!L fl, 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, V L E m, where E m is the generator emf amlitude. Thus I V L X L E m fl 30:0 V (: Hz)(50:0 0 3 H) 0:0955 A : (b) The frequency is now eight times larger than in art (a), so the inductive reactance is eight times larger and the current is one-eighth as much, or (0:0955 A)8 0:09 A. 40E (a) and (b) egardless of the frequency of the emf, I E m 30:0 V 50 0:60 A : 4E (a) Solve L from X L!L fl: f X L L (b) Solve C from X C (!C) (fc) : C fx C : ()(45:0 0 3 H) 4:60 03 Hz : (4: Hz)(: ) : F : (c) Since X L / f and X C / f, when f is doubled X L (: ) : and X C : : E (a) f CX C (: F)(:0 ) 8:84 03 Hz :

18 904 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS (b) Since X C / f, if the frequency is doubled then X C :0 6:00. 43E (a) The inductive reactance for angular frequency! is given by X L!L, where L is the inductance, and the caacitive reactance is given by X C!C, where C is the caacitance. The two reactances are equal if!l!c, or! LC. The frequency is f! LC 650 Hz : (6:0 0 3 H)(0 0 6 F) (b) The inductive reactance is X L!L fl (650 Hz)(6:0 0 3 H) 4. The caacitive reactance has the same value for this frequency. (c) The natural frequency for free LC oscillations is f! LC, the same as that for which the reactances are equal. 44P (a) I E m X L E m! d L 5:0 V (377 rad/s)(:7 H) 5: 0 3 A: (b) Since E(t) and i(t) has a 90 -hase dierence, E(t) is zero when i(t) I. (c) In this case!t must be (n 6) (where n is an integer), so i I sin(n 6 + ) I sin(3) (5: 0 3 A)( 3) 4:5 0 3 A. 45P (a) I E m X C!CE m (377 rad/s)(4:5 0 6 F)(5:0 V) 3:9 0 A : (b) Since E(t) and i(t) again has a 90 -hase dierence, E(t) is zero when i(t) I. (c) In this case!t must again be n 6, so i I sin(n 6 ) I sin(3) (3:9 0 3 A)( 3) 3:38 0 A. 46P (a) The generator emf is a maximum when sin(!t 4) or!t 4 n, where n is an integer, including zero. The rst time this occurs after t 0 is when!t 4 or t 3 4! 3 4(350 s ) 6: s : (b) The current is a maximum when sin(!t 34), or!t 34 n. The rst time this occurs after t 0 is when t 5 4! 5 4(350 s ) : 0 s : (c) The current lags the inductor by rad, so the circuit element must be an inductor.

19 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 905 (d) The current amlitude I is related to the voltage amlitude V L by V L IX L, where X L is the inductive reactance, given by X L!L. Furthermore, since there is only one element in the circuit the amlitude of the otential dierence across the element must be the same as the amlitude of the generator emf: V L E m. Thus E m I!L and L E m I! 30:0 V ( A)(350 rad/s) 0:38 H : 47P (a) Let!t 4 to obtain t 34! 3[4(350 rad/s)] 6: s. (b) Let!t + 4 to obtain t 4! [4(350 rad/s)] :4 0 3 s. (c) Since i leads E in hase by, the element must be a caacitor. (d) Solve C from X C (!C) E m I: C IE m! (6:0 0 3 A)[(30:0 V) (350 rad/s)] 5: F: 48P (a) and (b) Consider the following combinations: V V V, V 3 V V 3, and V 3 V V 3. For V 0 V A sin(!t) A sin(!t 0 ) A sin 3 A cos(!t 60 ) ;!t cos 0 where we used sin sin sin[( )] cos[( + )] and sin This exression indicates that V oscillates sinusoidally with angular frequency!, and has an amlitude of 3A. Similarly, 40 V 3 A sin(!t) A sin(!t 40 ) A sin 3 A cos(!t 0 )!t cos 40 and 0 V 3 A sin(!t 0 ) A sin(!t 40 ) A sin 3 A cos(!t 80 ) ;!t cos 360 both sinusoidal functions of t with an amlitude of 3A. 49E (a) Now X C 0 and and X L remain unchanged, so Z q + X L (60 ) + (86:7 ) 8 ;

20 906 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS and (b) tan XL I E m Z X C 36:0 V 8 0:98 A ; tan 86: :5 : ε m I V L φ ωt V 50E (a) Now X L 0 and and X C remain unchanged, so Z q + X C (60 ) + (77 ) 39 ; and (b) tan XL I E m Z X C 36:0 V 39 0:5 A ; tan :9 : I V φ ωt ε m V C 5E (a) The caacitive reactance is X C!C fc (60:0 Hz)(70:0 0 6 F) 37:9 :

21 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 907 The inductive reactance is unchanged, 86:7. The new imedance is Z + (X L X C ) (60 ) + (37:9 86:7 ) 67 : The current amlitude is The hase angle is tan XL (b) The voltage amlitudes are I E m Z X C 36:0 V 67 0:6 A : 86:7 37:9 tan 7:0 : 60 V I (0:6 A)(60 ) 34:6 V, V L ε m V L IX L (0:6 A)(86:7 ) 8:7 V, and V L - V L φ V V C IX C (0:6 A)(37:9 ) 8:9 V. Note that X L > X C, so that E m leads I. The hasor diagram is shown on the right. V C 5E f f 0 LC (: H)(3: F) :84 03 Hz : 53P (a) The resonance frequency f 0 of the circuit is about (:50 KHz Thus from f 0 (LC) we get :30 KHz) :40 KHz: L 4 f 0 C 4 (: Hz) (5: F) : H : (b) From Fig we see that as increases the resonance curve gets more sread out, so the two frequencies at which the amlitude is at half-maximum level will move away from each other.

22 908 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 54P From Fig we see that the x and y comonents of the Z-vector are Z x and Z y X C X L. Thus jzj q Z x + Z y + (X C X L ) and tan Z y Z x X L which indeed are the same as Eqs and X C ; 55P The amlitude of the voltage across the inductor in an LC series circuit is given by V L IX L, where X L (!L) is the inductive reactance. At resonance! LC, where L is the inductance and C is the caacitance. For the given circuit X L L LC :0 H (:0 H)(:0 0 6 F) 000 : At resonance the caacitive reactance has the same value as the inductive reactance, so X C 000. For X L X C, Eq gives Z. Hence Thus I E m 0 V 0 :0 A : V L IX L (:0 A)(000 ) 000 V : This is much larger than the amlitude of the generator emf (0 V). 56P The resistanc of the coil satises X L X C!L!C which we solve for :!L tan!c ()(930 Hz)(8:8 0 H) tan : tan ; ()(930 Hz)(0: F)

23 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS P (a) 36:0 V, by denition. (b) V I cos (0:96 A)(60 ) cos(9:4 ) 7:3 V: (c) efer to the gure to the right. (d) V C IX C sin (0:96 A)(77 ) sin(9:4 ) 7:0 V : V L φ φ φ V ε m I V L IX L sin (0:96 A)(86:7 ) sin(9:4 ) V C 8:34 V : (e) Since V + V C + V L 7:3 V + 7:0 V satised. 8:34 V 36:0 V E m, the loo theorem is 58P (a) (b) X C fc (400 Hz)(4:0 0 6 F) 6:6 : (c) Z + (X L X C ) + (fl X C ) (0 ) + [(400 Hz)( H) 6:6 ] 4 : I E m Z 0 V 4 0:5 A : (d) Now X C / Ceq will increase since C eq decreases. (e) Now C eq C and the new imedance is Z (0 ) + [(400 Hz)( H) (6:6 )] 408 < 4 ; i.e., the imedance decreases. (f) Since I / Z, it increases. 59P (a) For a given amlitude E m of the generator emf, the current amlitude is given by I E m Z E m + (!L!C) ;

24 90 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS where is the resistance, L is the inductance, C is the caacitance, and! is the angular frequency. To nd the maximum set the derivative with resect to! equal to zero and solve for!. The derivative is di d! E m + (!L!C) 3!L!C L + :! C The only factor that can equal zero is!l (!C) and it does for! LC. For the given circuit! 4 rad/s : LC (:00 H)(0:0 0 6 F) (b) For this value of the angular frequency the imedance is Z and the current amlitude is I E m 30:0 V 5:00 6:00 A : (c) You want to nd the values of! for which I E m. This means E m + (!L!C) E m : Cancel the factors E m that aear on both sides, square both sides, and set the recirocals of the two sides equal to each other to obtain +!L 4 :!C Thus!L!C 3 : Now take the square root of both sides and multily by!c to obtain! (LC)! 3C 0 ; where the symbol indicates the two ossible signs for the square root. The last equation is a quadratic equation for!. Its solutions are! 3C 3C + 4LC LC You want the two ositive solutions. The smaller of these is : 3C + 3C! + 4LC LC 3(0:0 0 6 F)(5:00 ) (:00 H)(0:0 0 6 F) 3(0:0 0 6 F) + (5:00 ) + 4(:00 H)(0:0 0 6 F) (:00 H)(0:0 0 6 F) 9 rad/s

25 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 9 and the larger is! + 3C + 3C + 4LC LC + 3(0:0 0 6 F)(5:00 ) (:00 H)(0:0 0 6 F) 3(0:0 0 6 F) + (5:00 ) + 4(:00 H)(0:0 0 6 F) (:00 H)(0:0 0 6 F) 8 rad/s : (d) The fractional width is!!! 0 8 rad/s 9 rad/s 4 rad/s 0:039 : 60P (a) Z E m I (b) from V I E m cos we get 5 V 3:0 A 39: : E m cos I (5 V) cos(0:98 rad) 3:0 A :7 : (c) Since X L X C / sin sin( 0:98 rad) < 0, i.e., X L < X C, the circuit is redominantly caacitive. 6P From the roblem statement f 0 (LC) 6:00 khz, Z + (f L f C) :00 k where f 8:00 khz, and cos Z cos 45. Solve these equations for, L and C: (a) Z cos 45 (:00 k)(cos 45 ) 707 ; (b) L Z (:00 k) (f f0 f ) (707 ) [8:00 khz (6:00 khz) 8:00 khz] 3: 0 H ; (c) C 4 f 0 L 4 (6:00 khz) (3: 0 H) :9 0 8 F :

26 9 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 6P (a) The hase angle is given by tan VL V C (b) Solve from E m cos I: tan VL V L :00 tan (:00) 45:0 : V L :00 E m cos I (30:0 V)(cos 45 ) A 70:7 : 63P (a) The resonance frequency is f 0 LC; at which X L X C, which is not satised here. (b) Let the additional caacitance be C 0. Then the new imedance for the combined caacitors should satisfy which gives C 0! X L X C!(C + C 0 ) X C + X L ;!C0 (60:0 Hz) 86:7 : F : 77 (c) At resonance X L X C 0 so Z. Thus I E m Z E m 36:0 V 60 0:55 A : 64P Four ossibilties exist: () C 4:00 F is used alone; () C 6:00 F is used alone; (3) C and C are connected in series; and (4) C and C are connected in arallel. The corresonding resonant frequencies are f 3 f f LC LC (: H)(4: F) :78 03 Hz ; (: H)(6: F) :45 03 Hz ; LC C (C + C ) (: H)(4:00)(6:00)0 6 F(4:00 + 6:00) : Hz ;

27 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 93 and f 4 L(C + C ) (: H)(4:00 + 6:00)0 6 F :3 03 Hz : 65P (a) Since L eq L + L and C eq C + C + C 3 for the circuit, the resonant frequency is! L eq C eq (L + L )(C + C + C 3 ) (:70 + :30)(0 3 H)(4:00 + :50 + 3:50)(0 6 F) 796 Hz : (b) The resonant frequency does not deend on so it will not change as increases. (c) Since! / (L + L ), it will decrease as L is increased. (d) Since! / Ceq and C eq decreases as C 3 is removed,! will increase. 66P When the two circuits are searated,! 0 (L C ) (L C ). When combined, the resonance frequency is! Leq C eq (L + L )C C (C + C ) (L C C + C L C )(C + C ) L C (C + C )(C + C )! 0 ; where we used! 0 (L C ) (L C ) : 67P Use the result of 59P: and Also use Thus! + 3C + 3C + 4LC LC! 3C + 3C + 4LC : LC! 0 LC :!!! 3C LC! 0! 0 LC r 3C L :

28 94 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS For the data of 59P r! 3(0:0 0 6 F) (5:00 )! 0 :00 H 3:87 0 : 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 (! and! ) are subtracted. Here the subtraction is done algebraically and 3 signicant gures are obtained. 68P* When the switch is oen we have an LC circuit with L; ; C in series, with V L V c!l!c tan tan( 0 ) : When the switch is in osition the equivalent caacitance of the LC circuit becomes C. So now!l!c tan tan 0:0 : Finally, with the switch in osition the circuit is an LC circuit so I E m X Solve for L, and C from these three equations: E m (!L) + (!C) : E m q I (tan tan ) + (tan tan ) 0 V q (:00 A) ( tan 0:0 + tan 0:0 ) + (tan 0:0 + tan 0:0 ) :65 0 ; L! tan tan (:65 0 ) (60:0 Hz) tan 0:0 + tan 0:0 0:33 H ; and C!(tan tan ) ()(60:0 Hz)(:65 0 )(tan 0:0 + tan 0:0 ) : F :

29 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 95 69E From Eq V max V rms (00 V) 4 V: 70E Since the imedance of the voltmeter is large, it will not aect the imedance of the circuit when connected in arallel with the circuit. So the reading will be 00 V in all three cases. 7E (a) efer to 44P, art (c). The ower delivered by the generator at this instant is P E(t)i(t) E m sin(n 6)I sin(3) E m I sin(6) sin(3) < 0, so it is taking energy from the rest of the circuit. (b) efer to 45P, art (c). The ower delivered by the generator at this instant is P E(t)i(t) E m sin(n 6)I sin( 3) E m I sin(6) sin(3) > 0, so it is sulying energy to the rest of the system. 7E The average ower dissiated in resistance when the current is alternating is given by P av I rms, where I rms is the root-mean-square current. Since I rms I, where I is the current amlitude, this can be written P av I. The ower dissiated in the same resistor when the current is direct is given by P i, where i is the current. Set the two owers equal to each other and solve for i. You should get i I :60 A :84 A : 73E Use P av I rms I : (a) P av 0, since 0. (b) P av I (0:600 A) (50 ) 9:0 W: (c) P av I (0:98 A) (60 ) 3:4 W: (d) P av I (0:5 A) (60 ) :8 W: 74E P av E rms I rms cos E rms Erms Z Z E rms Z : 75E The eective resistance e satises I rms e P mechanical, or e P mechanical I rms (0:00 h)(746 W/h) (0:650 A) 77 :

30 96 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS This is not the same as the resistance of its coils but just the eective resistance for ower transfer from electrical to mechanical form. In fact I rms would not give you P mechanical but rather the rate of energy loss due to thermal dissiation. 76E (a) The imedance is given by Z + (X L X C ) ; where is the resistance, X L is the inductive reactance, and X C is the caacitive reactance. Thus Z (:0 ) + (:30 0) : : (b) The average rate at which energy is sulied to the air conditioner is given by where cos is the ower factor. Now P av E rms Z cos ; so cos Z :07 0:994 ; (0 V) P av (0:994) :9 0 3 W : :07 77E I rms E rms Z E rms + XL 40 V (45:0 ) + (3:0 ) 7:6 A : 78P sin (!t ) av nt nt nt Z nt 0 t 4! Z nt 0 sin (!t ) dt cos(!t ) dt nt 4! sin(!t ) sin(n!t 0 ) + sin :

31 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 97 Since n!t n!(!) n, we have sin(n!t ) sin(n ) sin so [sin(n!t ) + sin ] 0. Thus sin (!t ) av : 79P (a) The energy stored in the caacitor is given by U E q C. Since q is a eriodic function of t with eriod T, so must be U E. So U E will not be changed over one comlete cycle. (b) Similarly, the energy stored in the inductor is U B i L. Since i is a eriodic function of t with eriod T, so must be U B. (c) The energy sulied by the generator is P av T (I rms E rms cos )T where we used I rms I and E rms E m. (d) The energy dissiated by the resistor is T E m I cos ; P av, resistor T (I rms V )T I rms (I rms )T T I ; where we used I rms I. (e) Since E m I cos E m I(V E m ) E m I(IE m ) I, the two quantities are indeed the same. 80P The current in the circuit satises i(t) I sin(!t ), where I E m Z E m + (!L!C) 45:0 V q (6:0 ) + f(3000 rad/s)(9:0 mh) [(3000 rad/s)(3: F)]g :93 A and tan XL X C (3000 rad/s)(9:0 mh) tan 6:0 46:5 :!L!C tan (3000 rad/s)(6:0 )(3: F)

32 98 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS (a) The ower sulied by the generator is P g i(t)e(t) I sin(!t )E m sin!t (:93 A)(45:0 V) sin[(3000 rad/s)(0:44 ms)] sin[(3000 rad/s)(0:44 ms) 46:5 ] 4:4 W : (b) The rate at which the energy in the caacitor changes is P c d q i q dt C C iv c I I I sin(!t ) cos(!t ) sin[(!t )]!C!C (:93 A) (3000 rad/s)(3: 0 6 F) sin[(3000 rad/s)(0:44 ms) (46:5 )] 7:0 W : (c) The rate at which the energy in the inductor changes is P i d dt Li Li di dt LI sin(!t ) d [I sin(!t )] dt!li sin[(!t )] (3000 rad/s)(:93 A) (9:0 mh) sin[(3000 rad/s)(0:44 ms) (46:5 )] 44: W : (d) The rate at which energy is being dissiated by the resistor is P r i I sin (!t ) (:93 A) (6:0 ) sin [(3000 rad/s)(0:44 ms) 46:5 ] 4:4 W : (e) The negative result for P i means that energy is being taken away from the inductor at this articular time. (f) P i + P r + P c 44:W 7:0 W + 4:4 W 4:5 W P g. 8P Since to maximize P we set P i Em ; r + dp d E m[(r + ) (r + )] E m(r ) 0 ; (r + ) 4 (r + ) 3

33 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 99 which gives r. 8P (a) The ower factor is cos, where is the hase angle when the current is written i I sin(!t ). Thus 4:0 and cos cos( 4:0 ) 0:743. (b) Since < 0,!t >!t and the current leads the emf. (c) The hase angle is given by tan (X L X C ), where X L is the inductive reactance, X C is the caacitive reactance, and is the resistance. Now tan tan( 4:0 ) 0:900, a negative number. This means X L X C is negative, or X C > X L. The circuit in the box is redominantly caacitive. (d) If the circuit were in resonance X L would be the same as X C, 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 caacitive reactance nor the resistance are zero. This means the box must contain a caacitor 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 caacitor at the oerating frequency. (f) The average ower is P av E mi cos (75:0 V)(:0 A)(0:743) 33:4 W : (g) The answers above deend on the frequency only through the hase angle, which is given. If values were given for, L and C then the value of the frequency would also be needed to comute the ower factor. 83P (a) The average ower is given by P av E rms I rms cos ; where E rms is the root-mean-square emf of the generator, I rms is the root-mean-square current, and cos is the ower factor. Now I rms I E m ; Z where I is the current amlitude, E m is the maximum emf of the generator, and Z is the imedance of the circuit. I E m Z was used. In addition, E rms E m and cos Z, where is the resistance. Thus P av E m Z E m [ + (!L!C) ] : Here the exression Z + (!L frequency was substituted.!c) for the imedance in terms of the angular

34 90 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS Considered as a function of C, P av has its largest value when the factor + (!L!C) has the smallest ossible value. This occurs for!l!c, or C! L () (60:0 Hz) (60:0 0 3 H) :7 0 4 F : The circuit is then at resonance. (b) Now you want Z to be as large as ossible. Notice that it becomes large without bound as C becomes small. Thus the smallest average ower occurs for C 0. (c) When!L!C the exression for the average ower becomes so the maximum average ower is P av The minimum average ower is P av 0. P av E m ; (30:0 V) (5:00 ) 90:0 W : (d) At maximum ower X L X C, where X L is the inductive reactance and X C is the caacitive reactance. The hase angle is tan X L X C 0 ; so 0. At minimum ower X C is innite, so tan and 90. (e) At maximum ower the ower factor is cos cos 0 and at minimum ower it is cos cos( 90 ) 0. 84P The ower consumed by the light bulb is P I. So we must let P max P min (II min ) 5, or Solve for L max : I I min Em Z min Zmax E m Z max Z min + (!L max ) 5 : L max! (0 V) 000 W (60:0 Hz) 7:64 0 H : (b) Now we must let max + bulb bulb 5 ;

35 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 9 or max ( 5 ) bulb ( (0 V) 5 ) 000 W 7:8 : This is not done because the resistors would consume, rather than temorarily store, electromegnetic energy. 85P (a) I rms E rms Z E rms + (f L f C) q :59 A : (b) The various rms voltages are 75:0 V (5:0 ) + f(550 Hz)(5:0 mh) [(550 Hz)(4:70 F)]g V ab I rms (:59 A)(5:0 ) 38:8 V ; V bc I rms X C I rms f C :59 A (550 Hz)(4:70 F) 59 V ; V cd I rms X L I rms f L (:59 A)(550 Hz)(5:0 mh) 4 V ; V bd jv bc V cd j j59:5 V 3:7 Vj 64: V ; V ad q V ab + V bd (38:8 V) + (64: V) 75:0 V : (c) For L and C the rate is zero since they do not dissiate energy. For P V ab (38:8 V) 5:0 00 W : 86E Use V s N V N s to nd V s V Ns N 500 (00 V) 50 : V : 87E (a) V s V Ns N 0 (0 V) :4 V : 500

36 9 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS (b) and I s V s :4 V s 5 0:6 A Ns 0 I I s (0:6 A) 3: 0 3 A : N E Ste u: (i) Use T T as rimary and T T 3 as secondary coil: V 3 V (800+00)00 5:00; (ii) use T T as rimary and T T 3 as secondary coil: V 3 V :00; (iii) use T T 3 as rimary and T T 3 as secondary coil: V 3 V 3 ( )800 :5. Ste down: By exchanging the rimary and secondary coils in each of the three cases above we get the following ossible ratios: (i) 5:00 0:00; (ii) 4:00 0:50; and (iii) :5 0: P (a) The rms current in the cable is I rms PV t W( V) 3:5 A. The rms voltage dro is then V I rms (3:5 A)()(0:30 ) :9 V and the rate of energy dissiation is P d Irms (3:5 A)()(0:60 ) 5:9 W: (b) Now I rms W(8:0 0 3 V) 3:5 A, so V (3:5 A)(0:60 ) 9 V and P d (3:5 A) (0:60 ) 5:9 0 W. (c) Now I rms W(0: V) 3:5 A, so V (3:5 A)(0:60 ) :9 0 V and P d (3:5 A) (0:60 ) 5:9 0 4 W. Obviously, both the rate of energy dissiation and the voltage dro increase as V t decreases. So to minimize these eects the best choice among the three V t 's above is V t 80 kv. 90P The amlier is connected across the rimary windings of a transformer and the resistor is connected across the secondary windings. If I s is the rms current in the secondary coil then the average ower delivered to is P av I s. Now I s (N N s )I, where N is the number of turns in the rimary coil, N s is the number of turns in the secondary coil, and I is the rms current in the rimary coil. Thus P av I N N s : Now nd the current in the rimary circuit. It acts like a circuit consisting of a generator and two resistors in series. One resistance is the resistance r of the amlier and the other is the equivalent resistance eq of the secondary circuit. Thus I E(r + eq ), where E is the rms emf of the amlier. According to Eq eq (N N s ), so I E r + (N N s )

37 CHAPTE 33 ELECTOMAGNETIC OSCILLATIONS 93 and P av E (N N s ) [r + (N N s ) ] : You wish to nd the value of N N s so that P av is a maximum. To do so, let x (N N s ). Then and the derivative with resect to x is P av E x (r + x) dp av dx E (r x) (r + x) 3 : This is zero for x r (000 )(0 ) 00. Notice that for small x, P av increases linearly with x and for large x it decreases in roortion to x. Thus x r is indeed a maximum, not a minimum. Since x (N N s ), maximum ower is achieved for (N N s ) 00, or N N s 0. 9 (a) X C ()( F)f ; (c) 7:7 Hz 9 (b) 7:7 Hz; (c) 7 V; (d) 0 Hz; (e) 87 V; (f) 3 Hz; (g) 87 V 93 (a) X L ()( H)f; (c) 796 Hz 94 (b) 796 Hz; (c) 7 V; (d) 50 Hz; (e) 94 V; (f) 8 Hz; (g) 94 V 95 (b) 6 Hz; (c) 90 and 6 Hz