14.A: Inductance (Majibu)

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Angalia Uelewa Wako

14.1. $$\displaystyle 4.77×10^{−2}V$$

14.2. a. kupungua;

b. kuongezeka; Kwa kuwa sasa inapita katika mwelekeo kinyume cha mchoro, ili kupata emf nzuri upande wa kushoto wa mchoro (a), tunahitaji kupungua sasa kwa upande wa kushoto, ambayo inajenga emf iliyoimarishwa ambapo mwisho mzuri ni upande wa kushoto. Ili kupata emf nzuri upande wa kulia wa mchoro (b), tunahitaji kuongeza sasa kwa upande wa kushoto, ambayo inajenga emf iliyoimarishwa ambapo mwisho mzuri ni upande wa kulia.

14.3. 40 A/s

14.4. a$$\displaystyle 4.5×10^{−5}H$$;.

b.$$\displaystyle 4.5×10^{−3}V$$

14.5. a$$\displaystyle 2.4×10^{−7}Wb$$;.

b.$$\displaystyle 6.4×10^{−5}m^2$$

14.6. 0.50 J

14.8. a. 2.2 s;

b. 43 H;

c. 1.0 s

14.10. a$$\displaystyle 2.5μF$$;.

b.$$\displaystyle π/2rad$$ au$$\displaystyle 3π/2rad$$;

c.$$\displaystyle 1.4×10^3$$ rad/s

14.11. a. overdamped;

b. 0.75 J

Maswali ya dhana

1. $$\displaystyle \frac{Wb}{A}=\frac{T⋅m^2}{A}=\frac{V⋅s}{A}=\frac{V}{A/s}$$

3. Sasa inayotokana na betri 12-V inapita kupitia inductor, kuzalisha voltage kubwa.

5. Self-inductance ni sawia na flux magnetic na inversely sawia na sasa. Hata hivyo, tangu flux magnetic inategemea sasa mimi, madhara haya kufuta nje. Hii ina maana kwamba inductance binafsi haitegemei sasa. Kama emf ni ikiwa katika kipengele, haina hutegemea jinsi mabadiliko ya sasa na wakati.

7. Fikiria mwisho wa waya sehemu ya mzunguko wa RL na ueleze kujitegemea kutoka kwa mzunguko huu.

9. Shamba la magnetic litatoka mwishoni mwa solenoid kwa hiyo kuna mtiririko mdogo kupitia upande wa mwisho kuliko katikati ya solenoid.

11. Kama sasa inapita kwa njia ya inductor, kuna sasa nyuma na sheria ya Lenz ambayo imeundwa ili kuweka sasa ya sasa kwenye amps zero, sasa ya awali.

13. hapana

15. Katika$$\displaystyle t=0$$, au wakati kubadili ni kwanza kutupwa.

17. 1/4

19. Awali,$$\displaystyle I_{R1}=\frac{ε}{R_1}$$ na$$\displaystyle I_{R2}=0$$, na baada ya muda mrefu umepita,$$\displaystyle I_{R1}=\frac{ε}{R_1}$$ na$$\displaystyle I_{R2}=\frac{ε}{R_2}$$.

21. ndiyo

23. Ukubwa wa oscillations ya nishati hutegemea nishati ya awali ya mfumo. Mzunguko katika mzunguko wa LC inategemea maadili ya inductance na capacitance.

25. Hii inajenga mzunguko wa RLC ambao hupunguza nishati, na kusababisha oscillations kupungua kwa amplitude polepole au haraka kulingana na thamani ya upinzani.

27. Ungependa kuchukua upinzani kwamba ni ndogo ya kutosha ili kituo kimoja tu kwa wakati ni ilichukua, lakini kubwa ya kutosha ili tuner haina kuweka katika hasa frequency sahihi. Inductance au capacitance ingekuwa tofauti ya tune ndani ya kituo, hata hivyo kwa kawaida kuzungumza, capacitors variable ni rahisi sana kujenga katika mzunguko.

Matatizo

29. $$\displaystyle M=3.6×10^{−3}H$$

31. a$$\displaystyle 3.8×10^{−4}H$$;.

b.$$\displaystyle 3.8×10^{−4}H$$

33. $$\displaystyle M_{21}=2.3×10^{−5}H$$

35. 0.24 H

37. 0.4 A/s

39. $$\displaystyle ε=480πsin(120πt−π/2)V$$

41. 0.15 V. hii ni polarity sawa na EMF kuendesha gari sasa.

43. a. 0.089 H/m;

b. 0.44 V/m

45. $$\displaystyle \frac{L}{l}=4.16×10^{−7}H/m$$

47. 0.01 A

49. 6.0 g

51. $$\displaystyle U_m=7.0×10^{−7}J$$

53. a. 4.0 A;

b. 2.4 A;

c. juu ya R:$$\displaystyle V=12V$$; juu ya L:$$\displaystyle V=7.9V$$

55. $$\displaystyle 0.69τ$$

57. a. 2.52 ms;

b.$$\displaystyle 99.2Ω$$

59. a$$\displaystyle I_1=I_2=1.7A$$;.

b$$\displaystyle I_1=2.73A,I_2=1.36A$$;

c$$\displaystyle I_1=0,I_2=0.54$$;

d.$$\displaystyle I_1=I_2=0$$

61. ushahidi

63. $$\displaystyle ω=3.2×10^{7}rad/s$$

65. a$$\displaystyle 7.9×10^{−4}s$$;.

b.$$\displaystyle 4.0×10^{−4}s$$

67. $$\displaystyle q=\frac{qm}{\sqrt{2}},I=\frac{q_m}{\sqrt{2LC}}$$

69. $$\displaystyle C=\frac{1}{4π^2f^2L}$$

$$\displaystyle f_1=540Hz;$$$$\displaystyle C_1=3.5×10^{−11}F$$

$$\displaystyle f_2=;1600Hz$$;$$\displaystyle C_2=4.0×10^{−12}F$$

71. 6.9 ms

73. ushahidi

Nje,$$\displaystyle B=\frac{μ_0I}{2πr}$$ Ndani,$$\displaystyle B=\frac{μ_0Ir}{2πa^2}$$

$$\displaystyle U=\frac{μ_0I^2l}{4π}(\frac{1}{4}+ln\frac{R}{a})$$

Hivyo,$$\displaystyle \frac{2U}{I^2}=\frac{μ_0l}{P2π}(\frac{1}{4}+ln\frac{R}{a})$$ na$$\displaystyle L=∞$$

75. $$\displaystyle M=\frac{μ_0l}{π}ln\frac{d+a}{d}$$

77. a. 100 T;

b. 2 A;

c. 0.50 H

79. a. 0 A;

b. 2.4 A

81. $$\displaystyle a. 2.50×10^6V$$;

(b) Voltage ni ya juu sana kwamba arcing itatokea na sasa haiwezi kupunguzwa kwa kasi sana.

(c) Sio busara kuzima sasa kubwa sana katika inductor kubwa kama hiyo kwa muda mfupi sana.

Changamoto Matatizo

83. ushahidi

85. a$$\displaystyle \frac{dB}{dt}=6×10^{−6}T/s$$;.

b$$\displaystyle Φ=\frac{μ_0aI}{2π}ln(\frac{a+b}{b})$$;

c. 4.0 nA

Wachangiaji na Majina

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