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Unformatted text preview: 750 CHAPTER 28 CIRCUITS CHAPTER 28 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (a) rightward; (b) all tie; (c) b, then a and c tie; (d) b, then a and c tie (a) all tie; (b) R1 , R2 , R3 (a) less; (b) greater; (c) equal (a) V=2, i; (b) V , i=2 (a) 1, 2, 4, 3; (b) 4, tie of 1 and 2, then 3 Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 3, 4, 1, 2 (a) series; (b) parallel; (c) parallel (a) no; (b) yes; (c) all tie (the circuits are the same) (a) 3R; (b) R=3; (c) equal parallel, R2 , R1 , series (a) parallel; (b) series (a) equal; (b) more (a) same; (b) same; (c) less; (d) more (a) less; (b) less; (c) more (a) same; (b) increase; (c) decrease, decrease C1 , 15 V; C2 , 35 V; C3 , 20 V; C4 , 20 V; C5 , 30 V 2:0 A 60 C (a) 16, 14, 12; (b) 16, 14, 12; (c) all tie; (d) 12, 14, 16 c, b, a (a) 1 and 3 tie, then 2; (b) 2, 1, 3 (a) all tie; (b) 1, 3, 2 CHAPTER 28 CIRCUITS 751 18. 19. 20. 2 and 4 tie, then 1 and 3 tie, then 5 1, 3, and 4 tie (8 V on each resistor), then 2 and 5 tie (4 V on each resistor) (a) no; (b) yes; (c) no Solutions to Exercises & Problems (a) The cost is (100 W 8:0 h=2:0 W h)($0:80) = $320: (b) The cost is (100 W 8:0h=103 W h)($0:06) = $0:048 = 4:8 cents: (a) W = eV = e(12 V) = 12 eV = (12 eV)(1:6 10 19 J/eV) = 1:9 10 (b) P = iV = neV = (3:4 1018 =s)(1:6 10 19 C)(12 V) = 6:5 W:
3E 2E 1E 18 J: The chemical energy of the battery is reduced by E = qE , where q is the charge that passes through in time t = 6:0 min and E is the emf of the battery. If i is the current, then q = i t and E = iE t = (5:0 A)(6:0 V)(6:0 min)(60 s/min) = 1:1 104 J. Notice the conversion of time from minutes to seconds.
4P If P is the rate at which the battery delivers energy and t is the time, then E = P t is the energy delivered in time t. If q is the charge that passes through the battery in time t and E is the emf of the battery, then E = qE . Equate the two expressions for E and solve for t: t = qE = (120 A h)(12 V) = 14:4 h = 14 h 24 min : P 100 W (a) Since E1 > E2 the current ows counterclockwise. (b) Battery 1, since the current ows through it from its negative terminal to the positive one. (c) Point B , since the current ows from B to A. 5E CI T AP ER 28 U RC IT S CH . en uit is giv ) = c cir or :0 ive sit the esist )2 (8 the bs po for i: : t nd r A be lve y aabsor E 1, e ou that 0:50 A o 0 rg ar r it t0. So 0:5 wise by = ( ene tery Fo 1 thuit. , es t . y = 6:0 V = lock ated 2, P 2 pli e ba tions atter e circ emf 2 p c p 8:0 nter dissi for R ry su . Th irec In b o th f the 2 V + cou er nd t o e n d att rectio osite 0 W. ergy tion :0 t is pow a b 4 n he e di opp = 3: es en direc e the 1:0 W t r en sam e in V) ppli the cur hen ) = th 0 , t 4:0 E , n themf arA)(6: ry su osite R2 ( f r ) em are i d e 0:50 atte opp ing. A h b s g n 0 : wit emf nt a = ( this y 2 i char J: 7 J at is ry and rre , P 2 so ter is 6 e 80 = ) = th att ent e cu r E 2 emf bat t. It b r h o in in) ergy i a ur f t f he t in cu /m /m en c E i and s t ren cir 0s 0s e i 6 l a r e ( )(6 ma W tion e cu m th in) :0 = :0 c h in er 6 ire T fro :0 m + 5 0 m e th : 2 ) = e d ing. rgy )(2 is th (c) )2 ( 1:0 (a) and (b) e 0V samcharg s en is 2 (5:0 13 J, (2: 10 d dis bsorb = erate V al to PR =iV R (mJ) is 202 t e ed E R i(mA)n 2:0 5:0 equ nce. ya g rr er a sfe t =15 r + ergy + Ric(10is1eV)t V h sis n n :0 ra r P 5 yt = 10 al e 2 t = 1 0 , wh rnal g r m R U r U te ne the E nd its in ee a of 5 Th nt + R en U e to u r mo 0 = 0 etwe ry du200 a 100 300 400 500 200 500 400 e b tte 2 Rt h T i R () R () ce a 0 = (b) ren he b e U di in t e Th rated (c) ene it i if s t to he left i 1 nR . CHAPTER 28 CIRCUITS 753 In the plot shown in the last page the curve in (c) is the power PR consumed by R as a function of R. (a) The potential dierence is V = E + ir = 12 V + (0:040 )(50 A) = 14 V. (b) P = i2 r = (50 A)2 (0:040 ) = 100 W. (c) P 0 = iV = (50 A)(12 V) = 600 W: (d) In this case V = E ir = 12 V (0:040 )(50 A) = 10 V and P = i2 r = 100 W: The current in the circuit is i = (150 V 50 V)=(3:0 + 2:0 ) = 20 A. So from VQ + 150 V (2:0 )i = VP we get VQ = 100 V + (2:0 )(20 A) 150 V = 10 V.
11E 10E 9E Since the energy of the charge decreases, point A is at a higher potential than point B; that is VA VB = 50 V. (b) The endtoend potential dierence is given by VA VB = +iR + E , where E is the emf of element C and is taken to be positive if it is to the left in the diagram. Thus E = VA VB iR = 50 V (1:0 A)(2:0 ) = 48 V. (c) A positive value was obtained for E , so it is toward the left. The negative terminal is at B . The potential dierence across R2 is
12E (a) If i is the current and V is the potential dierence then the power absorbed is given by P = i V . Thus 50 V = P = 1:0W = 50 V : i A R (12 ) V2 = iR2 = R +ER 2+ R = 3:0 +V)(4:0+ 5:0 = 4:0 V : 4:0 1 2 3
From Va E1 ir1 = Vc and i = (E2 E1 )=(R + r1 + r2 ), we get (2 1 Va Vc = E1 + ir1 = E1 + RE+ r E+)r1 1 r2 (4:4 V 2:1 V)(1:8 ) = 2:5 V : = 2:1 V + 5:5 + 1:8 + 2:3 13E 754 CHAPTER 28 CIRCUITS (a) Since Rtank = 140 , i = 12 V=(10 + 140 ) = 8:0 10 2 A. (b) Now Rtank = (140 + 20 )=2 = 80 , so i = 12 V=(10 + 80 ) = 0:13 A: (c) Now Rtank = 20 so i = 12 V=(10 + 20 ) = 0:40 A. (a) and (b) Denote L = 10 km and = 13 =km. Measured from the east end we have R1 = 100 = 2(L x) + R, and measured from the west end R2 = 200 = 2x + R. Solve for x and R: km x = R2 4R1 + L = 200 =100 + 102 = 6:9 km ; 2 4(13 km)
15P 14E R = R1 + R2 L = 100 + 200 (13 =km)(10 km) = 20 : 2 2
(a) Solve i = (E2 E1 )=(r1 + r2 + R) for R: 0V R = E2 i E1 r1 r2 = 3::0 V 102:3 A 3:0 3:0 = 9:9 102 : 10 (b) P = i2 R = (1:0 10 3 A)2 (9:9 102 ) = 9:9 10 4 W. Let the emf be V . Then V = iR = (5:0 A)R = i0 (R + R0 ) = (4:0 A)(R + 2:0 ). Solve for R: R = 8:0 . (a) From P = V 2 =R we nd V = PR = (10 W)(0:10 ) = 1:0 V. (b) From i = V=R = (E V )=r we nd r = R(E V )=V = (0:10 )(1:5 V 1:0 V)=(1:0 V) = 0:050 : Let the power supplied be Ps and that dissipated be Pd . Since Pd = i2 R and i = Ps =E , we have Pd = Ps2 =E 2 R / E 2 . The ratio is then
19P 18P 17P 16P p p Pd (E = 110; 000 V) = 110 V Pd (E = 110 V) 110; 000 V 2 = 1:0 10 6 : CHAPTER 28 CIRCUITS 755 (a) 20P i V V JA = JB = A = (R + R )A = (R +4R )D2 1 2 1 2 :0 V) = (0:127 + 0:4(60
)(2:60 10 3 m)2 = 1:32 107 A/m2 : 729
(b) VA = V R1 =(R1 + R2 ) = (60:0 V)(0:127 )=(0:127 + 0:729 ) = 8:90 V, and VB = V VA = 60:0 V 8:9 V = 51:1 V. (c) Calculate the resistivity from R = L=A for both materials: A = RA A=LA = RA D2 =4LA = (0:127 )(2:60 10 3 m)2 =[4(40:0 m)] = 1:69 10 8 m. So A is made of copper. Similarly we nd B = 9:68 10 8 m, so B is made of iron. The internal resistance of the battery is r = (12 V 11:4 V)=50 A = 0:012 < 0:020 , so the battery is OK. The resistance of the cable is R = 3:0 V=50 A = 0:060 > 0:040 , so the cable is defective. (a) The circuit is shown in the diagram to the right. i The current is taken to be positive if it is clockwise. The potential dierence across battery 1 is given by r1 V1 = E ir1 and for this to be 0 the current must be i = r2 E =r1 . Kirchho's loop rule gives 2E ir1 ir2 iR = 0. Substitute i = E =r1 and solve for R. You should get R = r1 r2 . (b) Since R must be positive, r1 must be greater than r2 . The potential dierence across the battery with the larger internal resistance can be made to vanish with the proper choice of R, the potential dierence across the battery with the smaller potential dierence cannot be made to vanish. (a) and (b) Let the emf of the solar cell be E and the output voltage be V , then V = E ir = E (V=R)r for both cases. Numerically we get 0:10 V = E (0:10 V=500 )r and 0:15 V = E (0:15 V=1000 )r: Solve for E and r: E = 0:30 V, r = 1000 . (c) The eciency is
2 0:15 V e = PV =R = (1000 )(5:0 cm2 )(2:0 10 3 W/cm2 ) = 2:3 10 3 : received
23P 22P 21P R 756 CHAPTER 28 CIRCUITS 24P (a) The current in the circuit is i = E =(r + R), so the rate of energy dissipation in R is You want to nd the value of R that maximizes P . To do this nd an expression for the derivative with respect to R, set it equal to 0, and solve for R. The derivative is dP = E 2 2E 2 R = E 2 (r R) : dR (r + R)2 (r + R)3 (r + R)3 Thus r R = 0 and R = r. For R very small P increases with R. For R large P decreases with increasing R. So R = r is a maximum, not a minimum. (b) Substitute R = r into P = E 2 R=(r + R)2 to obtain P = E 2 =4r. (a) The power delivered by the motor is P = (2:00 V)(0:500 m/s) = 1:00 W. From P = i2 Rmotor and E = i(r + Rmotor ) we then nd i2 r iE + P = 0 (which also follows directly from the conservation of energy principle). Solve for i: p p E E 2 4rP = 2:00 V (2:00 V)2 4(0:500 )(1:00 W) : i= 2r 2(0:500 ) The answer is either 3:41 A or 0:586 A. (b) Use V = E ir = 2:00 V i(0:500 ). Substitute the two values of i obtained in (a) into the above formula to get V = 0:293 V or 1:71 V. (c) The power P delivered by the motor is the same for either solution. Since P = iV we may have a lower i and higher V or, reversely, a lower V and higher i, while keeping P = iV a constant. In fact you should check for yourself that the two sets of solutions for i and V above do yield the same power P = iV . Denote silicon with subscript s and iron with i. Let T0 = 20 . If R(T ) = Rs (T ) + Ri (T ) = Rs (T0 )[1 + (T T0 )] + Ri (T0 )[1 + i (T T0 )] = [Rs (T )0 s + Ri (T0 )i ]+(temperatureindependent terms) is to be temperatureindependent, we must require that Rs (T0 )s + Ri (T0 )i = 0: Also note that Rs (T0 ) + Ri (T0 ) = R = 1000 : Solve for Rs (T0 ) and Ri (T0 ) to obtain 10 3 Rs (T0 ) = Ri = 6:(1000 )(6:570 10 ) 3 = 85:0 ; 5 10 3 + and Ri (T0 ) = 1000 85:0 = 915 :
i s
27E 26P 25P 2 P = i2 R = (r E R )2 : +R The potential dierence across each resistor is V = 25:0 V. Since the resistors are identical, the current in each one is i = V=R = (25:0 V)=(18:0 ) = 1:39 A. The total current through the battery is then itotal = 4(1:39 A) = 5:56 A. CHAPTER 28 CIRCUITS 757 When a potential dierence of 25:0 V is applied to the equivalent resistor the current through it is the same as the total current through the 4 resistors in parallel. Thus itotal = V=Req = 4V=R = 4(25:0 V)=(18:0 ) = 5:56 A. Since Req < R the two resistors (R = 12:0 and Rx ) must be connceted in parallel: Req = 3:00 = Rx R=(R +Rx ) = Rx (12:0 )=(12:0 +Rx ). Solve for Rx : Rx = Req R=(R Req ) = (3:00 )(12:0 )=(12:0 3:00 ) = 4:00 . Let the resistances of the two resistors be R1 and R2 , respectively. Note that the smallest value of the possible Req must be the result of connecting R1 and R2 in parallel, while the largest one must be that of connecting them in series. Thus R1 R2 =(R1 + R2 ) = 3:0 and R1 + R2 = 16 . So R1 ; R2 must be 4:0 and 12 . (a) Req (AB ) = 20:0 =3 = 6:67 (three 20:0 resistors in parallel). (b) Req (AC ) = 20:0
=3 = 6:67 (Three 20:0 resistors in parallel). (c) Req (BC ) = 0 (as B and C are connected by a conducting wire).
30E 29E 28E You might use the idea of equivalent resistance. For 4 identical resistors in parallel the equivalent resistance is given by 1 =4: R R
eq Req = 2:50 + (4:00 )(4:00 )=(4:00 + 4:00 ) = 4:50 :
Let i1 be the current in R1 and take it to be positive if it is to the right. Let i2 be the current in R2 and take it to be positive if it is upward. When the loop rule is applied to the lower loop, the result is E2 i1 R1 = 0 : and when it is applied to the upper loop, the result is
32E 31E E1 E2 E3 i2 R2 = 0 :
The rst equation yields E 5:0 i1 = R2 = 100V = 0:050 A : 1 758 CHAPTER 28 CIRCUITS The second yields E V i2 = E1 R2 E3 = 6:0 V 5:0 4:0 V = 0:060 A : 50 2
The negative sign indicates that the current in R2 is actually downward. If Vb is the potential at point b, then the potential at point a is Va = Vb + E3 + E2 , so Va Vb = E3 + E2 = 4:0 V + 5:0 V = 9:0 V. S1 ; S2 and S3 all open: ia = 0:00 A. S1 closed, S2 and S3 open: ia = E =2R1 = 120 V=40:0 = 3:00 A: S2 closed, S1 and S3 open: ia = E =(2R1 + R2 ) = 120 V=50:0 = 2:40 A: S3 closed, S1 and S2 open: ia = E =(2R1 + R2 ) = 120 V=60:0 = 2:00 A: S1 open, S2 and S3 closed: Req = R1 + R2 + R1 (R1 + R2 )=(2R1 + R2 ) = 20:0 + 10:0 + (20:0 )(30:0 )=(50:0 ) = 42:0 , so ia = E =Req = 120 V=42:0 = 2:86 A: S2 open, S1 and S3 closed: Req = R1 + R1 (R1 + 2R2 )=(2R1 + 2R2 ) = 20:0 + (20:0 ) (40:0 )=(60:0 ) = 33:3 , so ia = E =Req = 120 V=33:3 = 3:60 A: S3 open, S1 and S2 closed: Req = R1 + R1 (R1 + R2 )=(2R1 + R2 ) = 20:0 + (20:0 ) (30:0 )=(50:0 ) = 32:0 , so ia = E =Req = 120 V=32:0 = 3:75 A: S1 , S2 and S3 all closed: Req = R1 + R1 R0 =(R1 + R0 ) where R0 = R2 + R1 (R1 + R2 )=(2R1 + R2 ) = 22:0 , i.e., Req = 20:0 + (20:0 )(22:0 )=(20:0 + 22:0 ) = 30:5 ; so ia = E =Req = 120 V=30:5 = 3:94 A: 33E (a) Let E be the emf of the battery. When the bulbs are connected in parallel the potential dierence across them is the same and is the same as the emf of the battery. The power dissipated by bulb 1 is P1 = E 2 =R1 and the power dissipated by bulb 2 is P2 = E 2 =R2 . Since R1 is greater than R2 , bulb 2 dissipates energy at a greater rate than bulb 1 and is the brighter of the two. (b) When the bulbs are connected in series the current in them is the same. The power dissipated by bulb 1 is now P1 = i2 R1 and the power dissipated by bulb 2 is P2 = i2 R2 . Since R1 is greater than R2 greater power is dissipated by bulb 1 than by bulb 2 and bulb 1 is the brighter of the two. 34E CHAPTER 28 CIRCUITS 759 The currents i1 , i2 and i3 are obatined from Eqs. 2815 through 2817:
8 > > > > > > > > > > < 35E E( + :0 ) (1:0 V)(5:0 ) 3 i1 = R 1RR2 + RR) +E2 R3 = (10(4:0 V)(10+ (105
)(5:0 ) + (10 )(5:0 ) )(10 ) 1 2 + R2 3 R1 R3 = 0:275 A; E1 R3 E2 (R1 + R2 ) = (4:0 V)(5:0 ) (1:0 V)(10 + 5:0 ) > i2 = > > R1 R2 + R2 R3 + R1 R3 (10 )(10 ) + (10 )(5:0 ) + (10 )(5:0 ) > > > > = 0:025 A; > > > : i3 = i2 i1 = 0:025 A 0:275 A = 0:250 A:
Vd Vc can now be calculated by taking various paths. Two examples: from Vd i2 R2 = Vc we get Vd Vc = i2 R2 = (0:0250 A)(10 ) = +0:25 V; or from Vd + i3 R3 + E2 = Vc we get Vd Vc = i3 R3 E2 = ( 0:250 A)(5:0 ) 1:0 V = +0:25 V :
36E or R = r=9. Now r = 4`=d2 and R = 4`=D2 , where is the resistivity of copper. A = d2 =4 was used for the crosssectional area of a single wire and a similar expression was used for the crosssectional area of the thick wire. Since the single thick wire is to have the same resistance as the composite, 4` = 4` : D2 9d2 Solve for D to obtain D = 3d. The maximum power output is (120 V)(15 A) = 1800 W. Since 1800 W=500 W = 3:6, the maximum number of 500W lamps allowed is 3. Consider the lowest branch with the two resistors R1 (= 3:0 ) and the R2 (= 5:0 ). The voltage dierence across the 5:0 resistor is
38E 37E Let r be the resistance of each of the narrow wires. Since they are in parallel the resistance R of the composite is given by 1 = 9; R r R V)(5 V = i2 R2 = R E+ 2R = (120 + 5::0 ) = 7:5 V : 3:0 0 1 2 760 CHAPTER 28 CIRCUITS (a) Req (FH ) = (10:0 )(10:0 )(5:00 )=[(10:0 )(10:0 ) + 2(10:0 )(5:00 )] = 2:50 . (b) Req (FG) = (5:00 )R=(R + 5:00 ), where R = 5:00 + (5:00 )(10:0 )=(5:00 + 10:0 ) = 8:33 . So Req (FG) = (5:00 )(8:33 )=(5:00 + 8:33 ) = 3:13 . When connected in series, the rate at which electric energy dissipates is Ps = E 2 =(R1 + R2 ). When connected in parallel the corresponding rate is Pp = E 2 (R1 +R2 )=R1 R2 . Let Pp =Ps = 5, we get (R1 + R2 )2 =R1 R2 = 5, where R1 = 100 . Solve for R2 : R2 = 38 or 260 . Divide the resistors into groups of n resistors each, with all the resistors in the same group connected in series. Suppose there are m such groups that are connected in parallel with each other. The scheme is shown in the diagram on the right for n = 3 and m = 2. Let R be the resistance of any one of the resistors. Then the equivalent resistance of any group is nR, and Req , the equivalent resistance of the whole array, satises 1 = m: Since we want Rtotal = R, we must select n = m. The current is the same in every resistor and there are n2 resistors, so the maximum total power that can be dissipated is Ptotal = n2 P , where P is the maximum power that can be dissipated by any one of the resistors. You want Ptotal > 5:0P , so n2 must be larger than 5:0. Since n must be an integer, the smallest it can be is 3. The least number of resistors is n2 = 9. (a) The batteries are identical and, because they are connected in parallel, the potential dierences across them are the same. This means the currents in them are the same. Let i be the current in either battery and take it to be positive to the left. According to the junction rule the current in R is 2i and it is positive to the right. The loop rule applied to either loop containing a battery and R yields E ir 2iR = 0, so The power dissipated in R is
42P 41P 40P 39P Rtotal nR E i = r + 2R : 2 P = (2i)2 R = (r 4E 2R )2 : + R CHAPTER 28 CIRCUITS 761 It vanishes and P is a maximum if R = r=2. (b) Substitute R = r=2 into P = 4E 2 R=(r + 2R)2 to obtain
2( 2 Pmax = [r 4E 2(r=2) 2 = Er : + r=2)] 2 Find the maximum by setting the derivative with respect to R equal to zero. The derivative is 16E 2 R = 4E 2 (r 2R) : dP = 4E 2 dR (r + 2R)2 (r + 2R)3 (r + 2R)3 (a) Since E2 = E3 and R2 = 2R1 , from symmetry we know that the currents through E2 and E3 are the same: i2 = i3 = i. Therefore the current through E1 is i1 = 2i. Then from Vb Va = E2 iR2 = E1 + (2R1 )(2i) we get 4 2 0V 1 i = 4E2 +ER = 4(1::00V + :2:0 = 0:33 A : R1 2 ) So the current through E1 is i1 = 2i = 0:67 A, owing downward; the current through either E2 and E3 is 0:33 A, owing upward. (b) Va Vb = iR2 + E2 = (0:333 A)(2:0 ) + 4:0 V = 3:3 V: By symmetry, when the two batteries are connected in parallel the current i going through either one is the same. So from E = ir + (2i)R we get iR = 2i = 2E =(r + 2R). When connected in series 2E iR r iR r iR R = 0, or iR = 2E =(2r + R). (b) In series, since R > r. (c) In parallel, since R < r. When all the batteries are connected in parallel, each supplies a current i and therefore iR = Ni. Then from E = ir + iR R = ir + Nir we get iR = N E =[(N + 1)r]. When all the batteries are connected in series ir = iR and Etotal = N E = Nir r + iR R = NiR r + iR r; so iR = N E =[(N + 1)r]. (a) Since P = E 2 =Req , the higher the power rating the smaller the value of Req . To achieve this, we can let the low position connect to the larger resistance (R1 ), middle position connect to the smaller resistance (R2 ), and the high position connect to both of them in parallel.
46P 45P 44P 43P 762 CHAPTER 28 CIRCUITS (b) For P = 100 W, Req = R1 = E 2 =P = (120 V)2 =100 W = 144 ; for P = 300 W, Req = R1 R2 =(R1 + R2 ) = (144 )R2 =(144 + R2 ) = (120 V)2 =300 W: Solve for R2 : R2 = 72 : (a) R2 , R3 and R4 are in parallel with an equivalent resistance of )(50 )(75 R2 3 R = R R + RRRR4 R R = (50 )(50 )(50(50 )(75 ) )(50 )(75 ) = 19 : + + 2 3 2 4+ 3 4 Thus Req = R1 + R = 100 + 19 = 1:2 102 : (b) i1 = E =Req = 6:0 V=(1:1875 102 ) = 5:1 10 2 A; i2 = (E V1 )=R2 = (E i1 R1 )=R2 = [6:0 V (5:05 10 2 A)(100 )]=50 = 1:9 10 2 A; i3 = (E V1 )=R3 == i2 R2 =R3 = (1:9 10 2 A)(50 =50 ) = 1:9 10 2 A; i4 = i1 i2 i3 = 5:0 10 2 A 2(1:895 10 2 A) = 1:2 10 2 A:
48P 47P (a) First nd the currents. Let i1 be the current in R1 and take it to be positive if it is upward. Let i2 be the current in R2 and take it to be positive if it is to the left. Let i3 be the current in R3 and take it to be positive if it is to the right. The junction rule produces i1 + i2 + i3 = 0 :
The loop rule applied to the lefthand loop produces E1 i3 R3 + i1 R1 = 0
and applied to the righthand loop produces E2 i2 R2 + i1 R1 = 0 :
Substitute i1 = i2 i3 , from the rst equation, into the other two to obtain E1 i3 R3 i2 R1 i3 R1 = 0
and The rst of these yields E2 i2 R2 i2 R1 i3 R1 = 0 :
R 1 i3 = ER +i2R 1 : 1 3 Substitute this into the second equation and solve for i2 . You should obtain E( 3 i2 = R 2RR1 + RR) +E1 R1 1 2 + R1 3 R2 R3 00 ) (3 00 V)(5 00 ) = (5:00(1:00 :V)(5:00 + 4:
)(4:00 ):+ (2:00 :
)(4:00 ) = 0:158 A : )(2 00 ) + (5:00 CHAPTER 28 CIRCUITS 763 Substitute into the expression for i3 to obtain ( 0 158 A)(5 R 1 i3 = ER +i2R 1 = 3:00 V 5:00 :+ 4:00 :00 ) = 0:421 A : 1 3 i1 = i2 i3 = ( 0:158 A) (0:421 A) = 0:263 A : Note that the current in R1 is actually downward and the current in R2 is to the right. The current in R3 is also to the right. The power dissipated in R1 is P1 = i2 R1 = ( 0:263 A)2 (5:00 ) = 0:346 W, the power 1 dissipated in R2 is P2 = i2 R2 = ( 0:158 A)2 (2:00 ) = 0:0499 W, and the power dissipated 2 in R3 is P3 = i2 R3 = (0:421 A)2 (4:00 ) = 0:709 W. 3 (b) The power supplied by E1 is i3 E1 = (0:421 A)(3:00 V) = 1:26 W and the power supplied by E2 is i2 E2 = ( 0:158 A)(1:00 V) = 0:158 W. The negative sign indicates that E2 is
actually absorbing energy from the circuit. (a) Use P = E 2 =Req , where (12 00 ) Req = 7:00 + (12:0 )(4:0 ):0 )(4::0 )RR (4:00 )R : + (12 + Put P = 60:0 W and E = 24:0 V and solve for R: R = 19:5 . (b) Since P / Req , we must minimize Req , which means R = 0. (c) Now we must maximize Req , or set R = 1. (d) Since Req, max = 7:00 + (12:0 )(4:00 )=(12:0 + 4:00 ) = 10:0 ; Pmin = E 2 =Req, max = (24:0 V)2 =10:0 = 57:6 W. Since Req, min = 7:00 ; Pmax = E 2 =Req, min = (24:0 V)2 =7:00 = 82:3 W. The voltage dierence across R is VR = E R0 =(R0 +2:00 ), where R0 = (5:00 R)=(5:00 + R). So
2 1 2 VR = 1 E R0 E PR = R R R0 + 2:00 = R 1 + 2:00 =R0 E 2 1 + (2:00 )(5:00 + R) 2 E 2 : =R (5:00 )R f (R) 2
50P 49P Finally, To maximize PR we need to minimize the expression f (R). Set df (R) = 4:00 2 + 49 = 0 dR R2 25 764 CHAPTER 28 CIRCUITS to obtain R = (4:00 2 )(25)=49 = 1:43 : (a) The copper wire and the aluminum sheath are connected in parallel, so the potential dierence is the same for them. Since the potential dierence is the product of the current and the resistance, iC RC = iA RA , where iC is the current in the copper, IA is the current in the aluminum, RC is the resistance of the copper, and RA is the resistance of the aluminum. The resistance of either component is given by R = L=A, where is the resistivity, L is the length, and A is the crosssectional area. The resistance of the copper wire is RC = C L=a2 and the resistance of the aluminum sheath is RA = A L=(b2 a2 ). Substitute these expressions into iC RC = iA RA , and cancel the common factors L and to obtain iC C = iA A : a2 b2 a2 Solve this equation simultaneously with i = iC + iA , where i is the total current. You should get 2 iC = (r2 rrC C i+ r2 2 A C) C C A and 2 r2 iA = (r2 (rA 2 )C )+Cri2 : r
A C C C A
51P p The denominators are the same and each has the value (b2 a2 )C + a2 A = (0:380 10 3 m)2 (0:250 10 3 m)2 (1:69 10 8 m) + (0:250 10 3 m)2 (2:75 10 8 m) = 3:10 10 15 m3 : Thus and 3 2 (2:75 10 8 iC = (0:250 10 3:m) 10 15 m3 m)(2:00 A) = 1:11 A 10 (0:380 10 3 m)2 (0:250 10 3 m)2 (1:69 10 8 m)(2:00 A) iA = 3:10 10 15 m3 = 0:893 A : (b) Consider the copper wire. If V is the potential dierence, then the current is given by V = iC RC = iC C L=a2 , so
2 10 3 m)2 0 V) L = aV = :(0:250 69 10 8(12:m) = 126 m : iC C (1 11 A)(1: CHAPTER 28 CIRCUITS 765 The part of R0 connected in parallel with R is given by R1 = R0 x=L, where L = 10 cm. The voltage dierence across R is then VR = E R0 =Req , where R0 = RR1 =(R + R1 ) and Req = R0 (1 x=L) + R0 . Thus
2 2 VR = 1 E RR1 =(R + R1 ) PR = R R R (1 x=L) + RR =(R + R ) : 0 1 1 52P Some simple algebra then leads to
2 100R PR = (100R=R (E x=R0 ) x2 )2 ; + 10x 0 where x is measured in cm. Since i = E =(r + Rext ) and imax = E =r, we have Rext = R(imax =i 1:50 V=1:00 mA = 1:50 103 : Thus (a) Rext = (1:5 103 )(1=10% 1) = 1:35 104 ; (b) Rext = (1:5 103 )(1=50% 1) = 1:50 103 ; (c) Rext = (1:5 103 )(1=90% 1) = 167 : (d) Since r = 20:0 + R, R = 1:50 103 20:0 = 1:48 103 .
54P 53P 1) where r = (a) Put r roughly in the middle of its range; adjust current roughly with B ; make ne adjustment with A. (b) Relatively large percentage changes in A causes only small percentage charges in the resistance of the parallel combination, thus permitting ne adjustment; any change in A causes half as much change in this combination. (a) The current in R1 is given by V i1 = R + R R E=(R + R ) = 2:0 + (4:0 )(65::00
)=(4:0 + 6:0 ) = 1:14 A : 1 2 3 2 3 Thus
55P (b) All we need to do is to interchange the subscripts 1 and 3 in the equation above. Now i3 = E =[R3 + R2 R1 =(R2 + R1 )] = (5:0 V)=[6:0 +(2:0 )(4:0 )=(2:0 +4:0 )] = 0:6818 A and i1 = [(5:0 V) (0:6818 A)(6:0 )]=2:0 = 0:45 A, the same as before. i3 = E R V1 = E Ri1 R1 = 5:0 V (1::14 A)(2:0 ) = 0:45 A : 6 0 3 3 766 CHAPTER 28 CIRCUITS From 56P i 1+i 2 a 2R R iA i2 d i2 iA R R Vab = E = 2Ri1 + R(i1 + iA ) = Ri2 + R(i2 iA ) Vac = 2i1 R = Vad = i2 R ; we solve for iA to obtain iA = E =7R.
and i1 c i1+i A i 1+i 2 b (a) E = V + ir = 12 V + (10 A)(0:050 ) = 12:5 V. (b) Now E = V 0 + (imotor + 8:0 A)r, where V 0 = i0A R light = (8:0 A)(12 V=10 A) = 9:6 V. Solve for imotor : imotor = (E V 0 )=r 8:0 A = (12:5 V 9:6 V)=0:050 8:0 A = 50 A. The current in R2 is i. Let i1 be the current in R1 and take it to be downward. According to the junction rule the current in the voltmeter is i i1 and it is downward. Apply the loop rule to the lefthand loop to obtain
58P 57P E iR2 i1 R1 ir = 0 :
Apply the loop rule to the righthand loop to obtain i1 R1 (i i1 )RV = 0 :
Solve these equations for i1 . The second equation yields + i = R1 R RV i1 :
V Substitute this into the rst equation to obtain E (R2 + r)(R1 + RV ) i1 + R1 i1 = 0 : R
V This has the solution R i1 = (R + r)(R E+ V ) + R R : 2 1 RV 1 V CHAPTER 28 CIRCUITS 767 The reading on the voltmeter is E R i1 R1 = (R + r)(R RV R1 ) + R R 2 1+ V 1 V (3:0 V)(5:0 103 )(250 ) = (300 + 100 )(250 + 5:0 103 ) + (250 )(5:0 103 ) = 1:12 V :
The current in the absence of the voltmeter can be obtained by taking the limit as RV becomes innitely large. Then (3:0 E i1 R1 = R + R1 + r = 250 + V)(250 ) = 1:15 V : 300 + 100 1 R2 The fractional error is (1:15 1:12)=(1:15) = 0:030: This is 3:0%. The current in the ammeter is given by iA = E =(r + R1 + R2 + RA ). The current in R1 and R2 without the ammeter is i = E =(r + R1 + R2 ). The percent error is then
59P RA r + R1 + R2 = i = i iA = 1 i i r + R1 + R2 + RA r + R1 + R2 + RA 0 10 = 2:0 + 5:0 :+ 4
0 + 0:10 = 0:90% : :
The currents in R and RV are i and i0 i, respectively. Since V = iR = (i0 i)RV we have, by dividing both sides with V , 1 = (i0 =V i=V )RV = (1=R0 1=R)RV , i.e. 1= 1 R R0
61P 60P 1 : R
V Let the current in the ammeter be i0 . We have V = i0 (R + RA ), or R = V=i0 RA = R0 RA , where R0 = V=i0 is the apparent reading of the resistance. (a) In the rst case i0 = E =Req = E =[RA + R0 + RV R=(R + RV )] = 12:0 V=[3:00 + 100 + (300 )(85:0 )=(300 + 85:0 )] = 7:09 10 2 A, and V = E i0 (RA + R0 ) = 12:0 V (0:0709 A)(103:00 ) = 4:70 V. In the second case V = E R0 =(R0 + R0 ), where R0 = RV (R + RA )=(RV + R + RA ) = (300 )(300 + 85:0 )=(300 + 85:0 + 3:00 ) = 68:0 :
62P 768 CHAPTER 28 CIRCUITS So V = (12:0 V)(68:0 )=(68:0 + 100 ) = 4:86 V, and i0 = V (R + RA ) = 4:86 V=(300 + 85:0 ) = 5:52 10 2 A. (b) In the rst case R0 = V=i0 = 4:70 V=(7:09 10 2 A) = 66:3 ; and in the second case R0 = V=i0 = 4:86 V=(5:52 10 2 A) = 88:0 : Let i1 be the current in R1 and R2 and take it to be positive if it is toward point a in R1 . Let i2 be the current in Rs and Rx and take it to be positive if it is toward b in Rs . The loop rule yields (R1 + R2 )i1 (Rx + Rs )i2 = 0. Since points a and b are at the same potential i1 R1 = i2 Rs . The second equation gives i2 = i1 R1 =Rs . This expression is substituted into the rst equation to obtain (R1 + R2 )i1 = (Rx + Rs ) R1 i1 : R
s
63P Solve for Rx . You should get Rx = R2 Rs =R1 . (a) Refer to the diagram to the right. From
64P a R i1 c Rs i1 +i2 i2 b i r i 1i R d i2+ i Rx E = Vc Vd = i1 R + (i1 i)R
= i2 Rs + (i2 + i)Rx and Va Vb = (Va Vc ) (Vb Vc ) = i1 R + i2 Rs = ir ;
we solve for i: (b) The condition for i = 0 is Rs = Rx . Since R1 = R2 = R, this is equivalent to Rx = Rs (R2 =R1 ), consistent with the result of Problem 63. Use q = q0 e t= , or t = ln(q0 =q). (a) t1=3 = ln[q0 =(2q0 =3)] = ln(3=2) = 0:41: (b) t2=3 = ln[q0 =(q0 =3)] = ln 3 = 1:1:
65E E( x i = (R + 2r)(RRs+ RR) ) 2R R : +
s x s x CHAPTER 28 CIRCUITS 769 (a) = RC = (1:40 106 )(1:80 10 6 F) = 2:52 s: (b) qo = E C = (12:0 V)(1:80 F) = 21:6 C: (c) The time t satises q = q0 (1 e t=RC ), or 66E t = RC ln q 0 q0 21 = (2:52 s) ln 21:6 C:6 C:0 C = 3:40 s : q 16 During charging the charge on the positive plate of the capacitor is given by 67E q = C E (1 e t= ) ;
where C is the capacitance, E is applied emf, and is the time constant. The equilibrium charge is qeq = C E . You want q = 0:99qeq = 0:99C E , so 0:99 = 1 e t= : Thus e t= = 0:01 : Take the natural logarithm of both sides to obtain t= = ln 0:01 = 4:6 and t = 4:6 .
68E (a) The voltage dierence V across the capacitor varies with time as V (t) = E (1 e t=RC ). At t = 1:30 s we have V (t) = 5:00 V, so 5:00 V = (12:0 V)(1 e 1:30 s=RC ), which gives = (1:30 s)= ln(12=7) = 2:41 s. (b) C = =R = 2:41 s=15:0 k = 161 pF. (a) The charge on the positive plate of the capacitor is given by
69P q = C E (1 e t= ) ;
where E is the emf of the battery, C is the capacitance, and is the time constant. The value of is = RC = (3:00 106 )(1:00 10 6 F) = 3:00 s. At t = 1:00 s, t= = (1:00 s)=(3:00 s) = 0:333 and the rate at which the charge is increasing is dq = C E e t= = (1:00 10 6 )(4:00 V) e 0:333 = 9:55 10 7 C/s : dt 3:00 s 770 CHAPTER 28 CIRCUITS (b) The energy stored in the capacitor is given by q2 UC = 2C :
and its rate of change is Now so dUC = q dq : dt C dt q = C E (1 e t= ) = (1:00 10 6 )(4:00 V)(1 e 0:333 ) = 1:13 10 6 C ; dUC = 1:13 10 6 C (9:55 10 7 C/s) = 1:08 10 6 W : dt 1:00 10 6 F P = (9:55 10 7 A)2 (3:00 106 ) = 2:74 10 6 W :
(d) The rate at which energy is delivered by the battery is (c) The rate at which energy is being dissipated in the resistor is given by P = i2 R. The current is 9:55 10 7 A, so iE = (9:55 10 7 A)(4:00 V) = 3:82 10 6 W :
The energy delivered by the battery is either stored in the capacitor or dissipated in the resistor. Conservation of energy requires that iE = (q=C ) (dq=dt) + i2 R. Except for some roundo error the numerical results support the conservation principle. (a) The potential dierence V across the plates of a capacitor is related to the charge q on the positive plate by V = q=C , where C is capacitance. Since the charge on a discharging capacitor is given by q = q0 e t= , where q0 is the charge at time t = 0 and is the time constant, this means V = V0 e t= ; where q0 =C was replaced by V0 , the initial potential dierence. Solve for by dividing the equation by V0 and taking the natural logarithm of both sides. The result is:
70P t = ln (V=V ) = ln [(1:0010:0=s(100 V)] = 2:17 s : V) 0
(b) At t = 17:0 s, t= = (17:0 s)=(2:17 s) = 7:83, so V = V0 e t= = (100 V) e 7:83 = 3:96 10 2 V : CHAPTER 28 CIRCUITS 771 The time it takes for the voltage dierence across the capacitor to reach VL is given by VL = E (1 e t=RC ). Solve for R: 0:500 s R = C ln[E =(tE V )] = (0:150 10 6 F) ln[95:0 V=(95:0 V 72:0 V)] L 6 ; = 2:35 10 where we used t = 0:500 s given in the problem. (a) The initial energy stored in a capacitor is given by
2 q0 UC = 2C ;
72P 71P where C is the capacitance and q0 is the initial charge on one plate. Thus q0 = 2CUC = 2(1:0 10 6 F)(0:50 J) = 1:0 10 3 C :
(b) The charge as a function of time is given by q = q0 e t= , where g is the time constant. The current is the derivative of the charge: and the initial current is given by this function evaluated for t = 0: i0 = q0 = . The time constant is = RC = (1:0 10 6 F)(1:0 106 ) = 1:0 s. Thus i0 = (1:0 10 3 C)=(1:0 s) = 1:0 10 3 A. (c) Substitute q = q0 e t= into VC = q=C to obtain
3 VC = q0 e t= = 1::0 10 6 C e t=1:0 s = (1:0 103 V) e 1:0t ; C 1 0 10 F p p i = dq = q0 e t= dt where t is measured in seconds. Substitute i = (q0 = ) e t= into VR = iR to obtain q0 R e t= = (1:0 10 3 C)(1:0 106 ) e t=1:0 s = (1:0 103 V) e 1:0t ; VR = 1:0 s where t is measured in seconds. (d) Substitute i = (q0 = ) e t= into P = i2 R to obtain
2 3 C)2 (1 6 0R P = q 2 e 2t= = (1:0 10 (1:0 s)2:0 10 ) e 2t=1:0 s = (1:0 W) e 2:0t ; where t is again measured in seconds. 772 CHAPTER 28 CIRCUITS The potential dierence across the capacitor varies as a function of time t as V (t) = V0 e t=RC , which gives t 2: R = C ln(V =V ) = (2:0 100 s F) ln 4 = 7:2 105 : 6 0 Here we used V = V0 =4 at t = 2:0 s. (a) The charge q on the capacitor as a function of time is q(t) = (E C )(1 e t=RC ), so the charging current is i(t) = dq=dt = (E =R)e t=RC . The energy supplied by the emf is then
74P 73P U= Z 1 where UC = 1 C E 2 is the energy stored in the capacitor. 2 (b) Z 1 2Z 1 2 Rdt = E i e 2t=RC dt = 1 C E 2 : UR = R 2
0 0 0 E2 Z 1 e E idt =
R
0 t=RC dt = C E 2 = 2UC ; Use the result of 73P: R = t=[C ln(V0 =V )]. Then for tmin = 10:0 s 10:0 Rmin = (0:220 F) ln(5:s00=0:800) = 24:8 ; and for tmax = 6:00 ms 6:00 ms (24:8 ) = 1:49 104 ; Rmax = 10:0 s where in the last equation we used = RC . When S is open for a long time, the charge on C is qi = E2 C . When S is closed for a long time, the current i in R1 and R2 is i = (E2 E1 )=(R1 + R2 ) = (3:0 V 1:0 V)=(0:20 + 0:40 ) = 3:33 A. The voltage dierence V across the capacitor is then V = E2 iR2 = 3:0 V (3:33 A)(0:40 ) = 1:67 V. Thus the nal charge on C is qf = V C . So the change in the change on the capacitor is q = qf qi = (V E2 )C = (1:67 V 3:0 V)(10 F) = 13 C.
77P 76P 75P (a) At t = 0 the capacitor is completely uncharged and the current in the capacitor branch is as it would be if the capacitor were replaced by a wire. Let i1 be the current in R1 and CHAPTER 28 CIRCUITS 773 take it to be positive if it is to the right. Let i2 be the current in R2 and take it to be positive if it is downward. Let i3 be the current in R3 and take it to be positive if it is downward. The junction rule produces i1 = i2 + i3 ;
the loop rule applied to the lefthand loop produces E i1 R1 i2 R2 = 0 ;
and the loop rule applied to the righthand loop produces i2 R2 i3 R3 = 0 :
Since the resistances are all the same you can simplify the mathematics by replacing R1 , R2 , and R3 with R. The solution to the three simultaneous equations is 2E = 2(1:2 103 V) = 1:1 10 3 A i1 = 3R 3(0:73 106 ) and 3V 1 i2 = i3 = 3E = 3(0::2 10 6 ) = 5:5 10 4 A : R 73 10 At t = 1 the capacitor is fully charged and the current in the capacitor branch is 0. Then i1 = i2 and the loop rule yields E i1 R1 i1 R2 = 0 :
The solution is (b) Take the upper plate of the capacitor to be positive. This is consistent with current into that plate. The junction and loop equations are E = 1:2 103 V = 8:2 10 4 A : i1 = i2 = 2R 2(0:73 106 )
i1 = i2 + i3 ; E i1 R i2 R = 0 ; q i R + i R = 0: 2 C 3 Use the rst equation to substitute for i1 in the second and obtain E 2i2 R i3 R = 0. Thus i2 = (E i3 R)=2R. Substitute this expression into the third equation above to obtain (q=C ) (i3 R) + (E =2) (i3 R=2) = 0. Now replace i3 with dq=dt and obtain 3R dq + q = E : 2 dt C 2 and 774 CHAPTER 28 CIRCUITS This is just like the equation for an RC series circuit, except that the time constant is = 3RC=2 and the impressed potential dierence is E =2. The solution is
i h q = C2E 1 e 2t=3RC : The current in the capacitor branch is i3 = dq = 3E e 2t=3RC : dt R
The current in the center branch is
3 i2 = 2E i2 = 2E 6E e Rh R iR = E 3 e 2t=3RC 2t=3RC
V2 / 2 / 3 6R and the potential dierence across R2 is
h V2 = i2 R = E 3 e 6 2t=3RC i / 6 : t The graph is shown to the right. (c) For t = 0, e 2t=3RC is 1 and VR = E =3 = (1:2 103 V)=3 = 400 V. For t = 1, e 2t=3RC is 0 and VR = E =2 = (1:2 203 V)=2 = 600 V. (d) \A long time" means after several time constants. Then the current in the capacitor branch is very small and can be approximated by 0. (a) i = 0 : current in E1 = 0:665 A up, current in E2 = 1:26 A up; VAB = 6:29 V; i = 4:0 A : current in E1 = 0:248 A up, current in E2 = 3:21 A up; VAB = 14:2 V; i = 8:0 A : current in E1 = 0:169 A down, current in E2 = 5:17 A up; VAB = 34:6 V; i = 12 A : current in E1 = 0:587 A down, current in E2 = 7:13 A up; VAB = 55:1 V; E1 is discharging for i = 0 and 4:0 A; it is charging for i = 6:0 and 12 A; E2 is discharging for all values of i; (b) the emf is 6:29 V and the resistance is 2:56 (a) VT = ri + E ; (b) 13:6 V; (c) 0:060 79P 78P CHAPTER 28 CIRCUITS 775 (a) i2 = i1 + i4 + i5 , i3 + i4 + i5 = 0, 16 V+(7 )i1 +(5 )i2 +4 V = 0, 10 V+(8 )i3 (9 )i4 4 V (5 )i2 = 0, 12 V (4 )i5 + (9 )i4 = 0; (b) 1 1 0 1 0 1 1 6 0 6 [A] = 6 7 5 0 0 4 0 5 8 9 0 0 0 9 3 2 0 0 7 6 6 [C ] = 6 12 V 7 7 4 6V 5 12 V
2 80P 1 3 1 7 0 7; 7 0 5 4 (c) fi1 ; i2 ; i3 ; i4 ; i5 g = f0:717 A; 1:40 A; 0:680 A; 0:714 A; 1:39 Ag (a) 6:4 V; (b) 3:6 W; (c) 17 W; (d) 5:6 W; (e) a
82P 81P (a) Vc = E0 e t= ; (b) 12 V; (c) 0:77 s; (d) 3:8 F ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
 Fall '07
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