12.pdf - W = 2(1_e-HRC =1 e—HRC.’ e-rrac 99 Q Take the...

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Unformatted text preview: W) = 2(1 _e-HRC) % =1_ e—HRC.’ e-rrac 99) Q Take the natural logarithm of both sides. 13% ‘EJ RC Q _ —r C in [ — 9-92] Q —2. 00 R = S 0 990 (5. 00x10"1 F)1r1[l— '—Q] R = 869 O 31.2 Begin with the expression for the potential energy stored in a capacitor. Combine this with the definition of capacitance. '2 0' AV PE = —-—( ) 2 1 Q 2 1 PE = —— — AV = — it? gist > .9 When the capacitor is fully charged, the potential difference across it is equal to the emf of the battery. 1 PE=— 5‘ ZQ 31.3 Start with the equation giving the charge on the capacitor as a function of time for RC circuits. Using the definition of capacitance. derive an expression for the potential difference across the capacitor. q<r>= git—6“”) _ 4(3) me). _C Q — RC AVCt)=-E(1ae ” ) When the capacitor has its maximum charge Q. the potential difference across it equals the emf of the battery. AVG") = 5(1-e_“m ) At this point, use the information given in the problem. The voltage at a certain time is known. The time between flashes. or the period. is the recip‘rocat of the frequency. T=l f 1 . T=—=0.1003 10.0Hz The lamp flashes when the potential across the capacitor is 6.00 voits, and this first happens at a time OH). 100 s. after which the cycle repeats again. Copyright 2007 Kinetic Books Co. Chapter 29 - Pg 34 of 42 500 V = [9-00 V)[1— e—(omu s)fR{3.Dfl><1Et“ 11)] Solve for the resistance. 6.00V [ -(U.1uu s)rn[3.unxin“ 11)] - l—e 9.001: _ e—(umn sjrnlzouxm-‘i F) _ ]_ 5,00 V 9cov {0.100 s)IR[3.00x10‘5 1:): 1:1[1— G'ODV] 9.001; R_ —(o.1co s) [3.00x10'é F)ln[1— 5'00 V] 9.00 v R=3.03x10“' O 31.4 Find the potential difference across the resistor as a function of time by combining Ohm's law and the equation for the current in the RC circuit as a function of time. AV(:)=I(I)R a AV(:)= Ee_HRC R Mn): 35““ AV = (25 v )e‘i‘WKE-D 93(050 F)) 31.5 Use the definition of capacitance, C = q/AV, to substitute forthe charge in the equation for the charge on a capacitor in an RC circuit as a function of time. When the capacitor has its maximum charge, Q. the potential difference across it is equal to the emf of the battery. Copyright 2007 Kinetic Books Co. Chapter 29 ~ Pg 35 of 42 9’3): Q[1_e—HRC) CAM) = Q[1 -e-_”RC) cam) = 05(1— (”RC ) M0 = -m l-e (275 Q)1n 1— 08757”) (1.50V) 0:5.19x10‘9 F 31.6 The RC circuit signats to pop the toast up when the charge on the capacitor (10‘) is 90% of the final charge Q. in other words, q(t) = 0.90Q. Start with the equation giving the charge on the capacitor as a function of time for RC circuits. Solve for the resistance. qcr>= Q(1—e'”R") —HRC'=1I1[ —fl] —£ Cln£ 499] Q (a) Use this equation to calculate the largest resistance needed. This corresponds to the longest amount of time. —[90 s) R: R_ “(90 S) “‘5‘ cm(o.10) Cepyright 2007 Kinetic Books Co. Chapter 29 - Pg 36 of 42 (b) Use the same equation to calculate the smallest resistance needed. which corresponds to the shortest time. _ -(30 s) Rsmeltest - 0909 Q —[30 s) Cin[0.1o) (a) 3.6e+6 9 (b) 1.2e+6 Q 010 1— RsmaflESt = 31.? (a) Look at this circuit as a simple RC circuit with one equivalent resistance and one equivalent capacitance. In equilibrium, the potential difference across the equivalent capacitance will equal the potential difference across the battery. The potential difference across each of the capacitors is always equal, so at equilibrium the potential difference across the 1.00 microfarad capacitor is 72.0 V. Use the definition of capacitance to find the charge. (7:1 AV q = CAT? =[1.00>.:10'ts F)(7z.0 v) 0:120:005 c (0) Use the same equation to find the charge on the 2.00 micrcfarad capacitor. q: 001?: [2.00x10'5 F)[72.0 v) q:1.44x10'4 c (0) Use the same equation to find the charge on the 3.00 microfarad capacitor. 9': r:u:«.V=(3.00x10'ES F)[72.0U) q=2.16x10‘4 c 31.8 (a) The rate at which charge is increasing on the capacitor is the current in the RC circuit at that time. 12.0 V —[0500 s)f[{474>€§03 Q)[15.0><1.0" Fl) “—9 474x103 Q I=2.36x10‘5 A I (0.500 s) = (0) Write an expression for the energy stored in the capacitor and differentiate it with respect to time. 2 PE = q— 20 “’(PE) = ea = 261-1 0?: 20 dt C d: The factor dq/dr is the current. Substitute the equation for the charge as a function of time. expressing the maximum charge Q in terms of the capacitance of the capacitor and the emf of the battery. Copyright 2007 Kinetic Books Co. Chapter 29 - Pg 37 of 42 Cg [-1 _ e‘f [RC —= ——I(z)= ——)z(:)=g[1—e-“RC)I(:) d: C C Evaluate the rate of change of stored energy at the requested time. ' — 0.500 I 4M><1U3 9 15mm" F _ d(PE) = (12.0v)[1—e [ s) U M i) [2.35x10 5 A) “'5 r=u.5na 5 «PE? =1.92x10‘5w 5“ t=fl.50El s 31.9 The power dissipated by a resistor is the current squared. multiplied by the resistance. Use the expression for the current in an RC circuit as a function of time. P=flR 2 P[t)=(1[t))2 R: germ“ R ”0:32—2:ch R The energy dissipated by the resistor is the power integrated over time. The system does not reach equilibrium except after an infinite amount of time (though for all practical purposes. the system is in equilibrium after some tens of time constants have elapsed). 133:} ie—zrracdhij e—zuacdr o R R u u: 21ch du=(2fRC)'dr pg = i£§1r53im€ (.113: R 2 0 RC 05:3 m -u PE=TL e du. _ 2 m PE: 05 €_uL 2. _ 2 ‘” PE: CE [em—en) 2 u 2 PE: ‘35 (—1) 2 2 pg=_cg =C_55 2 2 The emf equals the final capacitor charge, divided by the capacitance. Copyright 2007 Kinetic Books Co. Chapter 29 - Pg 38 of 42 E=—— 2 C Ezaa Z 32.1 Use the equation that gives the charge on a capacitor as a function of time in a discharging RC circuit. q (r) = Qe—H-RC The final charge is equal to the magnitude of the charge of one electron qe. Solve for the time, t. —r IRE qe : Qt? Write the initial charge Q as a function of the capacitance C and the charging emf. Q = 05 Substitute for Q In the time equation. r=-R01n[q=] ca. [1.5x1o‘19 c) . __ 4 . . . — t- [2.ox10 QMCQWWWCEW [Capacymceflio V) 32.2 Use the equation for the charge on a discharging capacitor in an RC circuit as a function of time. Solve for t1r2- g: Qe—fl-J'i {RC 2 l = 9-1332ch 2 Take the natural logarithm of both sides (log to the base 6). 1 — . tug = —'RCII1[%) = R0111 (2:1 32.3 (a) Use the equation for the current as a function of time. The current will be at a maximum when the capacitor first begins to discharge. or at time = 0. Copyright 2007 Kinetic Bucks Co. Chapter 29 — Pg-39 of 42 IQ.) = flag—NRC RC 1(o)=£efl=& RC RC -I5 1(0): 17.0x10 on (9959)_(15.0x10 F) I(O)=114OA (b) Again, use the equation for the current as a function of time. IQ) = Q e-rrm RC 17.0x1t3rts c _ x 4 -1: I(25.0x10'9 S): [ J 9 [25.1310 sjrpos o)[15.u=qu F) (995 Q)[15.0x10'12 F) rczscxto'9 s): 213 A (c) Use the equation for the charge on a capacitor as a function of time. 90f) = 92“" _ _ - 2511x104 {995 5: 15mm—12 F q(25.0x10 9 s)=(17.0x10 ‘5 cje ( s“ H J qCZfiflxlO—g s) = 3.18x10"5 0 32.4 Begin with the equation for the charge on the capacitor as a function of time. The question asks for the number of time constants that have elapsed. Write the time variable as m and solve for n. 9(5) 2 QE-Hr qtr) ~—- 96“" N) = Q” n=—h[flflj Q Q n=5.30 32.5 (a) Fifteen minutes is long enough compared to any time constant in the circuit that you may assume that all the transient currents have died down. and there is a steady current in the circuit. Assume there is a clockwise current flowing in the circuit. Apply Kirchhoff‘s loop rule to the outside loop, starting from the lower left corner. 81—I[R1)—82—I(R2)=0 12 V—I[2.0 Q)— 6.0V—I(O.50 Q) = 0 6.0V = 1(25 9] I=Z4A (b) Start at the bottom of the capacitor, and proceed clockwise around the circuit, keeping track of the changes in potential. Copyright 2007 Kinetic Books Co. Chapter 29 — Pg 40 of 42 until the top of the capacitor is reached. th+51—I(RI)= Vtop I'1r’7to;:_'y';'not =51—Ith) AV=51-I(Rx) AV:12V—(2.4 AXZO Q) AV=7.2V 32.6 (a) t "HRC' V(r)=M=—Qe C C -6 100x10 F V(:)=13e‘”RC V(2.00,us)=13e 912.00 1.15): 3.43 v -[2.oax1 0" s)l[[15fl {Woman—9 I?” (b) If there is 3.43 volts across the capacitor, then there is the same amount of voltage drop across the resistor. since there are only two elements in the circuit. Use Ohm's Law. 1:5- 3.431: R _ 15.0 Q I = 0.229 A [1.30x10-6 (3)2 ' = H_— arms PE: 2(100x10'9 F) e FEB = [8. 5x19"5 J)e-2HRC T“ =%[[8.45x10_5 J)e‘2”RC) if“ = %(8.45x10'6 germ“ dpg _ —2[8.45x1o"S I) a [2.00 #5) — W d? (2.00 315) = —o.78 w Copyright 2007 Kinetic Books Co. 6 —2[2mxw‘“ s)t[(15.u {apnoea—9 F)] Chapter 29 - Pg 41 of 42 The derivative is negative. meaning that the energy stored in the capacitor is decreasing. This is the same rate at which energy is appearing in the resistor. Copyright 2007 Kinetic Books Co. Chapter 29 - Pg 42 of 42 8.1 6.2 6.3 on 0.5 6.6 6.7 0-8 0.9 The magnetic pole in the Northern hemisphere is a south magnetic pole, so it will attract the north end of the compass needle, which will point down. (a) Yes, a nonzero magnetic field can exist in the region. (b) The magnitude of-the force on a moving charged particle in a magnetic field is given by the following formula. F: (1123 sin 9 The force. if it exists. is perpendicular to the velocity and will change the particle's direction. The charge and velocity are nonzero. If the magnitude of the field- is also nonzero. the force will be nonzero unless the sine of 6 iszero. This happens when the magnetic field is parallel or antiparallel with the direction of motion (0 = 0° or 6-: 180°). (a) No. The direction your left thumb points will be opposite the direction your thumb would have pointed had you used the right hand. (b) The "left-hand rule" Works for electrons, because they have the opposite charge of protons. Positrons 'feel a force in the. same direction as protons (they are also positively charged), and neutrons and neutrinos feel no force when moving in a uniform magnetic field. because they have no electric charge. Because the magnetic force is always perpendicular to the velocity vector. the acoeleration is always perpendicular to the Velocity. A perpendicular aCCeleratlon will cause the velocity vector to change direction, not magnitude. The force on a charged particle from a magnetic field is proportional to the velocity of the particle. If the particle is not moving, there is no force. With no force, the particle wiil stay at rest. Use the right—hand rule. Point the fingers of your right hand in the direction of the velocity vector, and orient your hand so that the palm faces in the direction of the magnetic field. Now you can curl your fingers from the velocity “into" the magnetic field direction. The right thumb points in the direction of the force on a positively charg'Ed moving particle. For each part: (a) toward you. (b) away from you. (c) down. and (d) down. For a negatively charged particle, if the velocity v and magnetic field Bare given, and you use the right-hand rule, curling the fingers of your right hand from v into B. your right thumb will point directly opposite to the force on that negatively charged particle. Take care to always remember to reverse the direction that the right-hand rule indicates. (Alternatively, you could use a "left-hand rule" for negative particles.) in this problem. v and F are given. and you work "backwards” to figure out B. This can be done by experimentation: try each of the proposed magnetic fields in turn. and select the one such that v X B points in the direction of the force (remember the negative charge). The correct answer in .e'ach case is: (a) toward you. (in) left. ((2) down. and (d) down. (a) Assume there is a (nonzero) electric field. It would exert an electric force on the charged particle, accelerating it. Since there is no magnetic field. there can be no opposing magnetic force on the particle, so its velocity could not be constant. Therefore. the original assumption must be'wrong. and there is no electric field. (b). There could be no magnetic field, or there could be a magnetic field. parallel to the v'elooity. These would not change the particle's velocity. Assume there is a magnetic fietd with a component perpendicular to the charge. The magnetic force would accelerate the charge. Since there is no electric field, there would be no countervailing electric force on the particle, so its velocity could not be constant. Therefore the assumption must have been wrong and there cannot be a magnetic field with a component perpendicular to the particle's velocity. (c) The particle feels a force, perpendicular to the velocity from the magnetic field, so there must be an electric force to balance the magnetic force. There is a nonzero electric field that is perpendicular to the particle's veldcity. (a) Path A is that of a negatively charged particle. (b) Path B is that of a positively charged particle. Copyright 2007 Kinetic Books Co. Chapter 30 - Pg 1 of 24 ...
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