Unformatted text preview: CHAPTER 27 CURRENT AND RESISTANCE 733 CHAPTER 27 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 8 A, rightward (a) { (c) rightward (a) and (c) tie, then (b) device 2 (a) and (b) tie, then (d), then (c) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. (a), (b), and (c) tie, then (d) (b), (a), (c) increase tie of A, B , and C , then tie of A + B and B + C , then A + B + C (a) { (d) top{bottom, front{back, left{right (a) { (c) 1 and 2 tie, then 3 C , then A and B tie, then D C , A, B (a) and (c) tie, then (b) (b), (a), (c) lower (for xed V , smaller R gives larger P ) a, b, and c tie, then d (zero) 734 CHAPTER 27 CURRENT AND RESISTANCE 13. (a) conductors: 1 and 4; semiconductors: 2 and 3; (b) 2 and 3; (c) all four Solutions to Exercises & Problems The number of electrons is given by 10 6 n = q = it = (20060 10 A)(1 s) = 1:25 1015 : 19 C e e 1:
2E 1E (a) The charge that passes through any cross section is the product of the current and time. Since 4:0 min = (4:0 min)(60 s/min) = 240 s, q = it = (5:0 A)(240 s) = 1200 C. (b) The number of electrons N is given by q = Ne, where e is the magnitude of the charge on an electron. Thus N = q=e = (1200 C)=(1:60 10 19 C) = 7:5 1021 . Use i = vl:
3P 100 10 6 A i = = vl (30 m/s)(50 10 2 m) = 6:7 10 6 C/m2 : 4P where r is the radius of the sphere. This means Suppose the charge on the sphere increases by q in time t. Then in that time its potential increases by V = 4q r ;
0 q = 40 r V : Now q = (iin iout ) t, where iin is the current entering the sphere and iout is the current leaving. Thus t = q = 40 r V iin iout (0:10 m)(1000 V) = (8:99 109 F/m)(1:0000020 A 1:0000000 A) = 5:6 10 3 s : iin iout CHAPTER 27 CURRENT AND RESISTANCE 735 Use J = i=A = 4i=d2 / i=d2 , where i is the safe current and d is the diameter of the wire. You can easily verify that the 14gauge wire has the highest value of i=d2 so it has the maximum safe current density.
5000 5E 4000 J (A/in.2)
3000 2000 0 50 100 150 200 250 diameter d (mil) 6E (a) The magnitude of the current density is given by J = nqvd , where n is the number of particles per unit volume, q is the charge on each particle, and vd is the drift speed of the particles. The particle concentration is n = 2:0 108 cm 3 = 2:0 1014 m 3 , the charge is q = 2e = 2(1:60 10 19 C) = 3:20 10 19 C, and the drift speed is 1:0 105 m/s. Thus J = (2 1014 m 3 )(3:2 10 19 C)(1:0 105 m/s) = 6:4 A/m2 :
Since the particles are positively charged the current density is in the same direction as their motion, to the north. (b) The current cannot be calculated unless the crosssectional area of the beam is known. Then i = JA can be used. (a (b
7E 4(1 2 10 10 A) i J = A = di =4 = (2::5 10 3 m)2 = 2:4 10 5 A/m2 : 2
2 J 2 4 10 5 vd = ne = (8:47 10:28m3 )(1A/m 10 19 C) = 1:8 10 15 m/s : = :60 736 CHAPTER 27 CURRENT AND RESISTANCE The crosssectional area of wire is given by A = r2 , where r is its radius. The magnitude of the current density is J = i=A = i=r2 , so 8E i 0:50 A 4 r = J = 4 A/m2 ) = 1:9 10 m : (440 10 The diameter is d = 2r = 2(1:9 10 4 m) = 3:8 10 4 m.
The magnitude of the current is i = (np + ne )e = (1:1 1018 =s + 3:1 1018 =s)(1:60 10 19 C) = 0:67 A : The current ows toward the negative terminal. (a) i = (nh + ne )e = (2:25 1015 =s + 3:50 1015 =s)(1:60 10 19 C) = 9:20 10 4 A: (b) J = i=A = 9:20 10 4 A=[(0:165 10 3 m)2 ] = 1:08 104 A/m2 : (a) J = np evp = (8:70=10 6 m3 )(1:60 10 19 C)(470 105 m/s) = 6:54 10 7 A/m2 : 2 (b) i = JA = Re J = (6:37 106 m)2 (6:54 10 7 A/m2 ) = 8:34 107 A:
12P 11P 10E 9E r s (b) Now let N be the number of particles in a length L of the beam. They will all pass through the beam cross section at one end in time t = L=v, where v is the particle speed. The current is the charge that moves through the cross section per unit time. That is, i = 2eN=t = 2eNv=L. Thus N = iL=2ev. Now nd the particle speed. The kinetic energy of a particle is K = 20 MeV = (20 106 eV)(1:60 10 19 J/eV) = 3:2 10 12 J : p Since K = 1 mv2 , v = 2K=m. The mass of an alpha particle is 4 times the mass of a 2 proton or m = 4(1:67 10 27 kg) = 6:68 10 27 kg, so (a) The charge that strikes the surface in time t is given by q = i t, where i is the current. Since each particle carries charge 2e, the number of particles that strike the surface is 6 q 25 N = e = i2et = (0:2(1 10 10 A)(3:0 s) = 2:3 1012 : 2 :6 19 C) 10 12 J) v = 2(3:2 10 27 kg = 3:1 107 m/s 6:68 s CHAPTER 27 CURRENT AND RESISTANCE 737 and (c) Use conservation of energy. The initial kinetic energy is zero, the nal kinetic energy is 20 MeV = 3:2 10 12 J, the initial potential energy is qV = 2eV , and the nal potential energy is zero. Here V is the electric potential through which the particles are accelerated. Conservation of energy leads to Kf = Ui = 2eV , so
12 32 f V = Ke = 2(1::60 10 19JC) = 10 106 V : 2 10 6 )(20 10 2 N = 2iL = 2(1(0:25 1019 C)(3: 107m) = 5:0 103 : ev :60 10 1 m/s) Use vd = J=ne = i=Ane: Thus 13P L L t = v = i=Ane = LAne i d 4 2 28 3 19 = (0:85 m)(0:21 10 m )(8:47 10 = m )(1:60 10 C)
= 8:1 102 s = 13 min : 300 A The current density is the sum of the contribution from the electrons and the particles: 14P J = J + Je = q n v + ene ve = (2e)n v + e(2n )ve = 2en (v + ve ) = 2(1:60 10 19 C)(2:8 1015 =cm3 )(25 m/s + 88 m/s) = 10 A/cm2 :
The direction of the current is to the east. (a)
15P (b) Now 1 1 r J dA = J0 1 R 2rdr = 3 R2 J0 = 3 AJ0 : i= cylinder 0
Z Z R The result is dierent from that in part (a) because the current density in (b) is lower near the center of the cylinder (where the area is smaller for the same radial interval) and J0 Z R r 2rdr = 2 R2 J = 2 AJ : i= J dA = R 0 3 0 3 cylinder 0
Z 738 CHAPTER 27 CURRENT AND RESISTANCE higher outward, resulting in a higher average current density over the cross section and consequently a greater current than that in part (a).
16E 7 3 : R = L = (3:00 10 :0 m)(10m02 10 m) = 0:536 : A 56 10 4 The resistance of the wire is given by R = L=A, where is the resistivity of the material, L is the length of the wire, and A is the crosssectional area of the wire. The crosssectional area is A = r2 = (0:50 10 3 m)2 = 7:85 10 7 m2 . Here r = 0:50 mm = 0:50 10 3 m is the radius of the wire. Thus 3 7 2 85 = RA = (50 10 )(7:m 10 m ) = 2:0 10 8 m : L 2:0
18E 17E 1 L L Li (1:0 0 = = RA = (V=i)A = V A = (2:0 V)(1m)(4:10A) m2 ) = 2:0 106 ( m) 1 : :0 6 19E Since the potential dierence V and current i are related by V = iR, where R is the resistance of the electrician, the fatal voltage is V = (50 10 3 A)(2000 ) = 100 V. The resistance of the coil is given by R = L=A, where L is the length of the wire, is the resistivity of copper, and A is the crosssectional area of the wire. Since each turn of wire has length 2r, where r is the radius of the coil, L = (250)2r = (250)(2)(0:12 m) = 188:5 m. 2 If rw is the radius of the wire, its crosssectional area is A = rw = (0:65 10 3 m)2 = 1:33 10 6 m2 . According to Table 27{1, the resistivity of copper is 1:69 10 8 m. Thus 20E R = L = (1:69 :10 m)(188:5 m) = 2:4 : A 1 33 10 6 m2
(a) i = V=R = 23:0 V=15:0 10 3 = 1:53 103 A: (b) J = i=A = 4i=d2 = 4(1:53 10 3 A)=[(6:00 10 3 m)2 ] = 5:41 107 A/m2 : (c) = RA=L = (15:0 10 3 )()(6:00 10 3 m)2 =[4(4:00 m)] = 10:6 10 8 m:
21E 8 CHAPTER 27 CURRENT AND RESISTANCE 739 The material is platinum. Use Eq. 2717: 0 = (T T0 ), and solve for T : 1 T = T0 + 22E 0 1 = 20 C + 58 1 3 =K 50 1 = 57 C : 4:3 10 Here we noted that =0 = R=R0 . In Eq. 2717, let (T ) = 20 where 0 is the resistivity at 20 C: 0 = 20 0 (T T0 ), and solve for T : 1 1 T = T0 + = 20 C + 4:3 10 3 =K = 2:5 102 C : This agrees well with Fig. 2710, from which you can deduce T = 5:2 102 K; or 2:5 102 C. Use R = dV=di = local slope of the V vs i curve.
Resistance (102 )
24E 23E 0 = 20 15 10 5 0 +1 +2 +3 +4 potential difference (V) (a) 25E 2 10 8 m)(4:0 L V = iR = i A = (12 A)(1:69 (5:2 10 3 m=2)2 10 m) = 3:8 10 4 V : 740 CHAPTER 27 CURRENT AND RESISTANCE (b) Since it moves in the direction of the electron drift which is against the direction of the current, its tail is negative compared to its head. (c) Use the result of 13P:
2 L t = v = lAne = Ldi ne i 4 d 2 m)(5:2 10 3 m)2 (8:47 1028 =m3 )(1:60 10 19 C) = (1:0 10 = 238 s = 3 min 58 s :
26E 4(12 A) (a) The new area is A0 = AL=L0 = A=2. (b) The new resistance is R0 = R(A=A0 )(L0 =L) = 4R: Since the mass and density of the material do not change, the volume remains the same. If L0 is the original length, L is the new length, A0 is the original crosssectional area, and A is the new crosssectional area, then L0 A0 = LA and A = L0 A0 =L = L0 A0 =3L0 = A0 =3. The new resistance is 0 R = L = A3L3 = 9 L0 = 9R0 ; A A0 0= where R0 is the original resistance. Thus R = 9(6:0 ) = 54 . Use R / L=A and note that A = 1 d2 / d2 . The resistance of the second wire is given by 4
28E 27E A R2 = R A1 2 29P L2 = R = d1 2 L2 = R(2)2 (1=2) = 2R : L1 d2 L1 The resistance of conductor A is given by L RA = r2 ;
where rA is the radius of the conductor. If ro is the outside diameter of conductor B and 2 ri is its inside diameter, then its crosssectional area is (ro ri2 ) and its resistance is
A RB = (r2L r2 ) :
o i CHAPTER 27 CURRENT AND RESISTANCE 741 The ratio is 2 RA = ro ri2 = (1:0 mm)2 (0:50 mm)2 = 3:0 : 2 RB rA (0:50 mm)2 (a) Denote the copper wire with subscript c and the iron wire with subscript i. Let 2 ic = V=Rc = ii = V=Ri , or Rc = Ri , we get c Lc =rc = i Li =ri2 , i.e., 30P in (a). So this is impossible.
31P ri = r i = 9:68 108 = 2:39 : rc c 1:69 108 (b) If Ji = i=Ai = Jc = i=Ac then Ai = Ac , or ri = rc , which is inconsistent with the result r (b) Let R = c L=(d2 =4) and solve for the diameter d of the copper wire: Denote the copper wire with subscript c and the aluminum wire with subscript a. (a) L 10 8 R = a A = (2:75(5:2 10 3m)(1:3 m) = 1:3 10 3 : m)2 d= r 4c L = s R 4(1:69 10 8 m)(1:3 m) = 4:6 10 3 m : (1:3 10 3 ) (a) Since = RA=L = Rd2 =4L = (1:09 10 3 )(5:50 10 3 m)2 =[4(1:60 m)] = 1:62 10 8 m, the material is silver. (b) 3 10 8 : L R = A = 4L = 4(1:62 (2:00 m)(1200 10 m) = 5:16 10 8 : d2 10 m)2 (a) The current in each strand is i = 0:750 A=125 = 6:00 10 3 A: (b) The potential dierence is V = iR = (6:00 10 3 A)(2:65 10 6 ) = 1:59 10 8 V: (c) The resistance is Rtotal = 2:65 10 6 =125 = 2:12 10 8 :
34P 33P 32P Use J = E=, where E is the magnitude of the electric eld in the wire, J is the magnitude of the current density, and is the resistivity of the material. The electric eld is given 742 CHAPTER 27 CURRENT AND RESISTANCE by E = V=L, where V is the potential dierence along the wire and L is the length of the wire. Thus J = V=L and V 115 V = LJ = = 8:2 10 4 m : (10 m)(1:4 104 A/m2 ) The resistance at operating temperature T is R = V=i = 2:9 V=0:30 A = 9:67 . Thus from
35P R R0 = R0 (T T0 ) we nd 1 R 1 T = T0 + R 0 = 20 C + 1 4:5 10 3 =K 9:67 1 = 1:8 103 C : 1:1 (a) i = V=R = 35:8 V=935 = 3:83 10 2 A. (b) J = i=A = 3:83 10 2 A=(3:50 10 4 m2 ) = 109 A/m2 : (c) vd = J=ne = (109 A/m2 )=[(5:33 1022 =m3 )(1:60 10 19 C)] = 1:28 10 2 m/s. (d) E = V=L = 35:8 V=0:158 m = 227 V/m: Use R=L = =A = 0:150 =km. (a) For copper J = i=A = (60:0 A)(0:150 =km)=(1:69 10 8 m) = 5:32 105 A/m2 ; and for aluminum J = (60:0 A)(0:150 =km)=(2:75 10 8 m) = 3:27 105 A/m2 . (b) Denote the mass densities as m . For copper (m=L)c = (m A)c = (8960 kg/m3 ) (1:69 10 8 m)=(0:150 =km) = 1:01 kg/m; and for aluminum (m=L)a = (m A)a = (2700 kg/m3 )(2:75 10 8 m)=(0:150 =km) = 0:495 kg/m: Use J = E = (n+ + n )evd . (a) 14 = (2: vd = (n En )e = [(62070 10 =cm3 m)(120 V/m) = 1:73 cm/s : + + 550) ](1:60 10 19 C) (b) J = E = (2:70 10 14= m)(120 V/m) = 3:24 10 12 A/m2 :
39P 38P 37P 36P + Use R / L=A. The diameter of a 22gauge wire is 1=4 that of a 10gauge wire. Thus from R = L=A we nd the resistance of 25 ft of 22gauge copper wire to be R = (1:00 )(25 ft=1000 ft)(4)2 = 0:40 : CHAPTER 27 CURRENT AND RESISTANCE 743 (a) Let T be the change in temperature and be the coecient of linear expansion for copper. Then L = L T and L = T = (1:7 10 5 =C )(1:0 C) = 1:7 10 5 : This is 0:0017%. The fractional change in area is A = 2 T = 2(1:7 10 5 =C )(1:0 C) = 3:4 10 5 : A 40P L This is 0:0034%. For small changes in the resistivity , length L, and area A of a wire, the change in the resistance is given by R = @R + @R L + @R A : @ @L @A Since R = L=A, @R=@ = L=A = R=, @[email protected] = =A = R=L, and @[email protected] = L=A2 = R=A. Furthermore, = = T , where is the temperature coecient of resistivity for copper (4:3 10 3 =C , according to Table 27{1). Thus R = + L A = ( + 2 ) T = ( ) T R L A 3 =C 1:7 10 5 =C ) (1:0 C ) = 4:3 10 3 : = (4:3 10 This is 0:43%. (b) The fractional change in resistivity is much larger than the fractional change in length and area. Changes in length and area aect the resistance much less than changes in resistivity.
41P (a) Assume, as in Fig. 2725, that the cone has current i, from bottom to top. The current is the same through every crosssection. We can nd an expression for the electric eld at every cross section, in terms of the current and then use this expression to nd the potential dierence V from bottom to top of the cone. The resistance of the cone is given by R = V=i. Consider any cross section of the cone. Let J denote the current density at that cross section and assume it is uniform over the cross section. Then the current through the cross R section is given by i = J dA = r2 J , where r is the radius of the cross section. Now J = E=, where is the resistivity and E is the magnitude of the electric eld at the cross section. Thus i = r2 E= and E = i=r2 . The current density and electric eld have dierent values on dierent cross sections because dierent cross sections have dierent radii. 744 CHAPTER 27 CURRENT AND RESISTANCE Let x measure distance downward from the upper surface of the cone. The radius increases linearly with x, so we may write r = a + b L a x:
The coecients in this function have been chosen so r = a when x = 0 and r = b when x = L. Thus i a + b a x 2 : E= L The potential dierence between the upper and lower surfaces of the cone is given by i Z L a + b a x 2 dx V= E dx = L 0 0 L i L a + b a x 1 = i L 1 1 = b a L 0 b a a b iL = i b L a b ab a = ab : Z L L R = V = ab : i (b) If b = a then R = L=a2 = L=A, where A = a2 is the crosssectional area of the
cylinder.
42P The resistance is p p From Eq. 2720 / T 1 / v. But according to Chapter 20 v / T . Thus / T . Since P = iV , q = it = Pt=V = (7:0 W)(5:0 h)(3600 s=h)=9:0 V = 1:4 104 C:
44E 43E The power dissipated is given by the product of the current and the potential dierence: P = iV = (7:0 10 3 A)(80 103 V) = 560 W : R = P=i2 = 100 W=(3:00 A)2 = 11:1 : 45E CHAPTER 27 CURRENT AND RESISTANCE 745 The horse power required is 46E iV (10 A)(12 V) P = 80% = (80%)(746 W/hp) = 0:20 hp :
47E (a) Electrical energy is transferred to heat at a rate given by P=V ; R
where V is the potential dierence across the heater and R is the resistance of the heater. Thus V)2 P = (120 = 1:0 103 W = 1:0 kW : 14 (b) The cost is given by 2 C = (1:0 kW)(5:0 h)(5:0 cents/kW h) = 25 cents :
48E (a) R = V 2 =P = (120 V)2 =500 W = 28:8 : (b) The rate is given by i=e = P=eV = 500 W=[1:60 10 19 C(120 V)] = 2:60 1019 =s: Use P = V 2 =R. The power dissipated in the second case is
49E P = (1:50 V=3:00 V)2 (0:540 W) = 0:135 W : A) i J = A = [(0:10 in.)(24(25 10 2 m/in.)]2 = 4:9 106 A/m2 : :54 (b) E = J= = J = (1:69 10 8 m)(4:9 106 A/m2 ) = 8:3 10 2 V/m: (c) V = EL = (8:3 10 2 V/m)(1000 ft)(0:3048 m/ft) = 25 V: (d) P = V i = (25 V)(25 A) = 6:3 102 W: (a) 50E 746 CHAPTER 27 CURRENT AND RESISTANCE (a) 51E 2 V i = V = L=A = V d R 4L 2 m/in.)]2 = 1:74 A : = (1:20 V)[(0:0400 in.)(2:54 100 m) 4(1:69 10 8 m)(33: (b) 4(1 74 A) i 4 J = A = di2 = [(0:0400 in.)(2::54 10 2 m/in.)]2 = 2:15 106 A/m2 : (c) E = V=L = 1:20 V=33:0 m = 3:63 10 2 V/m: (d) P = V i = (1:20 V)(1:74 A) = 2:09 W: Use P = i2 R = i2 L=A, or L=A = P=i2 . So the new values of L and A satisfy P 30 P 30 L L A new = i2 new = 42 i2 old = 16 A old ; i.e., (L=A)new = 1:875(L=A)old : Also note that (LA)new = (LA)old . Solve the above two p p equations for Lnew and Anew : Lnew = 1:875Lold = 1:369Lold ; Anew = 1=1:875Aold = 0:730Aold : (a) Since P = i2 R = J 2 A2 R, the current density is 1 P 1 J = A P = A L=A = R
s r s s
53P 52P P LA = 1:0 W 5 2 5 m)(2:0 10 2 m)(5:0 10 3 m)2 = 1:3 10 A/m : (3:5 10 (b) From P = iV = JAV we get P 1 W P V = JA = r2 J = (5:0 10 3 m):0(1:3 105 A/m2 ) = 9:4 10 2 V : 2
(a) From P = V 2 =R = AV 2 =L we solve for L: 2 (2:60 10 6 m2 )(75:0 V)2 L = AV = (5:00 10 7 m)(5000 W) = 5:85 m : P
54P CHAPTER 27 CURRENT AND RESISTANCE 747 (b) Since L / V 2 the new length should be L0 = L(V 0 =V )2 = (5:85 m)(100 V=75:0 V)2 = 10:4 m :
55P (a) The monthly cost is (100 W)(24 h/d)(31 d/month)(6 cents/kW h) = 446 cents = $4:46, assuming a 31day month. (b) R = V 2 =P = (120 V)2 =100 W = 144 . (c) i = P=V = 100 W=120 V = 0:833 A.
56P (a) Let P be the power dissipated, i be the current in the heater, and V be the potential dierence across the heater. They are related by P = iV . Solve for i: W P i = V = 1250 V = 10:9 A : 115 (b) According to Ohm's law V = iR, where R is the resistance of the heater. Solve for R: 115 R = V = 10:9V = 10:6 : i A (c) The thermal energy E generated by the heater in time t (= 1:0 h = 3600 s) is E = Pt = (1250 W)(3600 s) = 4:5 106 J :
57P Let RH be the resistance at the higher temperature (800 C) and let RL be the resistance at the lower temperature (200 C). Since the potential dierence is the same for the two temperatures, the power dissipated at the lower temperature is PL = V 2 =RL , and the power dissipated at the higher temperature is PH = V 2 =RH , so PL = (RH =RL )PH . Now RL = RH + RH T , where T is the temperature dierence TL TH = 600 C . Thus 500 H PL = R +RH T PH = 1 +P T = 1 + (4:0 10 4 =W )( 600 C ) = 660 W : C H RH
58P (a) The rate n is given by n = i=e = (15 10 6 A)=(1:60 10 19 C) = 9:4 1013 =s. (b) The rate is P = (9:4 1013 =s)(16 106 eV)(1:60 10 19 J/eV) = 2:4 102 W. 748 CHAPTER 27 CURRENT AND RESISTANCE (a) The charge q that ows past any cross section of the beam in time t is given by q = i t and the number of electrons is N = q=e = (i=e) t. This is the number of electrons that are accelerated. Thus 10 6 N = (0:50 :A)(0:10 19 C s) = 3:1 1011 : 1 60 10 (b) Over a long time t the total charge is Q = nqt, where n is the number of pulses per unit time and q is the charge in one pulse. The average current is given by = Q=t = nq. i Now q = i t = (0:50 A)(0:10 10 6 s) = 5:0 10 8 C, so = (500 s 1 )(5:0 10 8 C) = 2:5 10 5 A : i (c) The accelerating potential dierence is V = K=e, where K is the nal kinetic energy of an electron. Since K = 50 MeV, the accelerating potential is V = 50 kV = 5:0 107 V. During a pulse the power output is 59P P = iV = (0:50 A)(5:0 107 V) = 2:5 107 W :
This is the peak power. The average power is i P = V = (2:5 10 5 A)(5:0 107 V) = 1:3 103 W : Let the heat of vaporization be Lv . Then mLv = P = iV , where m = 21 mg/s: So (5:2 V) Lv = iV = 21 A)(12kg/s = 3:0 106 J/kg : 6 m 10
61P 60P The rate of change of mechanical energy of the pistonEarth system, mgv, must be equal to the rate at which heat is generated from the coil, i2 R: mgv = i2 R. Thus
2 R (0: 2 (550 ) v = img = (12240 A):80 m/s2 ) = 0:27 m/s : kg)(9 Use P = V 2 =R / V 2 , which gives P / V 2 2V V . The percentage change is roughly P=P = 2V=V = 2(110 115)=115 = 8:6%: 62P CHAPTER 27 CURRENT AND RESISTANCE 749 (b) A drop in V causes a drop in P , which in turn lowers the temperature of the resistor in the coil. At a lower temperature R is also decreased. Since P / R 1 a decrease in R will result in an increase in P , which partially osets the decrease in P due to the drop in V . Thus the actual drop in P will be smaller when the temperature dependency of the resistance is taken into consideration. (a) 120 ; (b) 107 ; (c) 5:3 10 (d) 5:9 10 (e) 276 63P 3 =K; 3 =K; ...
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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
 GROUPTEST
 Current, Resistance

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