Chapter 17 Current and Resistance Problem Solutions 17.1 The charge that moves past the cross section is ∆Q = I ( ∆t ) , and the number of electrons is n= ∆Q I ( ∆t ) = e e ( 80.0 × 10 = C s) ( 10.0 min) ( 60.0 s min) = 3.00× 1020 electrons 1.60 × 10−19 C The negatively charged electrons move in the direction opposite to the conventional current flow. −3 17.3 The current is I = ∆Q ∆V ∆V = . Thus, the change that passes is ∆Q = ( ∆t ) , giving ∆t R R 1.00 V ∆Q = ( ∆t ) = ( 0.100 A) ( 20.0 s) = 2.00 C 10.0 Ω 69 70 CHAPTER 17 17.8 Assuming that, on average, each aluminum atom contributes one electron, the density of charge carriers is the same as the number of atoms per cubic meter. This is n= or density ρ N ρ = = A , mass per atom M NA M ( 6.02 × 10 n= 23 mol ) ( 2 .7 g cm3 ) ( 106 cm3 1 m3 ) 26.98 g mol = 6.0 × 1028 m3 The drift speed of the electrons in the wire is then vd = 17.9 I 5.0 C s = = 1.3 × 10 − 4 m s 28 3 n e A ( 6.0 × 10 m ) ( 1.60× 10−19 C) ( 4.0× 10−6 m2 ) (a) The carrier density is determined by the physical characteristics of the wire, not the current in the wire. Hence, n is unaffected . (b) The drift velocity of the electrons is vd = I nqA . Thus, the drift velocity is doubled when the current is doubled. 17.11 ( ∆V ) max = I max R = ( 80 × 10−6 A ) R Thus, if R = 4.0 × 105 Ω , ( ∆V ) max = 32 V and if R = 2 000 Ω , ( ∆V ) max = 0.16 V 17.13 From R = d= ρL π d2 ρ L , we obtain A = , or = A 4 R 4ρ L = πR ( )( ) 4 5.6 × 10 −8 Ω ⋅ m 2 .0 × 10−2 m π ( 0.050 Ω ) = 1.7 × 10 −4 m = 0.17 mm Current and Resistance 17.16 71 We assume that your hair dryer will use about 400 W of power for 10 minutes each day of the year. The estimate of the total energy used each year is E= ( ∆t ) = ( 0.400 kW) 10 min 1 hr ( 365 d) = 24 kWh d 60 min If your cost for electrical energy is approximately ten cents per kilowatt-hour, the cost of using the hair dryer for a year is on the order of $ cost = E × rate = ( 24 kWh ) 0.10 = $2.4 kWh 17.19 or ~$1 The volume of material, V = AL0 = ( π r02 ) L0 , in the wire is constant. Thus, as the wire is ( ) 2 2 stretched to decrease its radius, the length increases such that π rf L f = ( π r0 ) L0 giving 2 2 r r 2 L f = 0 L0 = 0 L0 = ( 4.0 ) L0 = 16 L0 rf r 0.25 0 The new resistance is then Rf = ρ 17.20 Lf Af Lf πr 2 f =ρ 16 L0 π ( r0 4 ) 2 L 2 = 16 ( 4 ) ρ 0 2 = 256R0 = 256( 1.00 Ω ) = 256 Ω π r0 Solving R = R0 1 + α ( T − T0 ) for the final temperature gives T = T0 + 17.23 =ρ R − R0 140 Ω − 19 Ω = 20 °C + = 1.4 × 103 ° C 4.5 × 10 -3 ( ° C) −1 ( 19 Ω ) α R0 At 80°C, I= or 5.0 V ∆V ∆V = = R R0 1 + α ( T − T0 ) ( 200 Ω ) 1 + ( − 0.5× 10−3 ° C−1 ) ( 80° C− 20° C) I = 2.6 × 10−2 A = 26 mA 72 CHAPTER 17 17.31 I= ∆V and 17.33 = 600 W = 5.00 A 120 V R= ∆V 120 V = = 24.0 Ω I 5.00 A The maximum power that can be dissipated in the circuit is max = ( ∆V ) I max = ( 120 V ) ( 15 A ) = 1.8 × 103 W Thus, one can operate at most 18 bulbs rated at 100 W per bulb. 17.39 The resistance per unit length of the cable is I2 L 2 .00 W m = 2 = = 2 .22 × 10−5 Ω m 2 L I ( 300 A) R = L From R = ρ L A , the resistance per unit length is also given by R L = ρ A . Hence, the ρ 2 cross-sectional area is π r = A = , and the required radius is RL r= 17.45 ρ 1.7 × 10−8 Ω ⋅ m = = 0.016 m = 1.6 cm π ( R L) π ( 2.22 × 10-5 Ω m ) The energy saved is ∆E = ( high − low ) ⋅ t = ( 40 W − 11 W ) ( 100 h ) = 2.9 × 10 3 Wh = 2.9 kWh and the monetary savings is savings = ∆E ⋅ rate = ( 2.9 kWh ) ( $0.080 kWh ) = $0.23 = 23 cents Current and Resistance 17.52 The resistance of the 4.0 cm length of wire between the feet is −8 ρ L ( 1.7 × 10 Ω ⋅ m ) ( 0.040 m) R= = = 1.79 × 10−6 Ω , 2 A π ( 0.011 m) so the potential difference is ∆V = IR = ( 50 A ) ( 1.79 × 10 −6 Ω ) = 8.9 × 10 −5 V = 89 µ V 17.60 Each speaker has a resistance of R = 4.00 Ω and can handle 60.0 W of power. From = I 2 R , the maximum safe current is Imax = R = 60.0 W = 3.87 A 4.00 Ω Thus, the system is not adequately protected by a 4.00 A fuse. 73