Current and Resistance

Transcription

1 27 Current and Resistance CHAPTER OUTLINE 27.1 Electric Current 27.2 Resistance 27.3 A Model for Electrical Conduction 27.4 Resistance and Temperature 27.5 Superconductors 27.6 Electrical Power * An asterisk indicates a question or problem new to this edition. ANSWERS TO OBJECTIVE QUESTIONS OQ27.1 Answer (d). One ampere hour is (1 C/s)(3 600 s) coulombs. The ampere hour rating is the quantity of charge that the battery can lift though its nominal potential difference. OQ27.2 (i) Answer (e). We require ρl/a A 3ρL/A B. Then A A /A B 1/3. (ii) Answer (d). π r A 2 /π r B 2 1/3 gives r A /r B 1/ 3. OQ27.3 The ranking is c > a > b > d > e. Because (a) I ΔV /R, so the current becomes 3 times larger. P IΔV I 2 R, so the current is 3 times larger. (c) R is 1/4 as large, so the current is 4 times larger. (d) R is 2 times larger, so the current is 1/2 as large. (e) R increases by a small percentage, so the current has a small decrease. OQ27.4 (i) Answer (a). The cross-sectional area decreases, so the current density increases, thus the drift speed must increase. 200

2 Chapter OQ27.5 (ii) Answer (a). The cross-sectional area decreases, so the resistance per unit length, R/L ρ/a, increases. Answer (c). I ΔV /R 1.00 V/10.0 Ω A C/s. Because current is constant, I dq/dt Δq/ Δt, and we find that Δq IΔt ( C/s) ( 20.0 s) 2.00 C OQ27.6 Answer (c). The resistances are: R 1 ρl A ρl πr 2, R 2 ρl π ( 2r) 2 ( 1 4)ρL πr 2, R 3 ρ( 2L) π ( 3r) 2 ( 2 / 9)ρL πr 2. OQ27.7 OQ27.8 Answer (a). The new cross-sectional area is three times the original. Originally, R ρl A. Finally, R f ρ( L 3) 3A ρl 9A R 9. Answer. Using R Ω at T 20.0 C, we have R R 0 ( 1+ αδt ) or α R R 0 1 ΔT At T 20.0 C, we have R R 0 ( 1+ α ΔT) 10.0 Ω ( 90.0 C 20.0 C) C C C 20.0 C OQ27.9 Answer (a). R V/I 2 V/2 A 1 Ω. OQ27.10 Answer (c). Compare resistances: R A R B ρl A π(d A / 2) 2 ρl B π(d B / 2) 2 L A L B 2 d B d A 2 2L B Compare powers: P A P B ΔV 2 R A ΔV 2 R B R B R A 2. 2 d B L B d B 9.66 Ω OQ27.11 Answer (e). R A ρ L A ( A 2ρ B )L A 2R. Therefore, B P A P B ΔV 2 R A ΔV 2 R B R B R A 1 2 OQ27.12 (i) Answer (a). P V 2 /R, and V is the same for both bulbs, so the 25 W bulb must have higher resistance so that it will have lower power. (ii) Answer. V is the same for both bulbs, so the 100 W bulb must have lower resistance so that it will have more current.

3 202 Current and Resistance OQ27.13 Answer (d). Because wire B has twice the radius, it has four times the cross-sectional area of wire A. For wire A, R A R ρl/a. For wire B, R B ρ(2l)/(4a) (1/2)ρL/A R/2. ANSWERS TO CONCEPTUAL QUESTIONS CQ27.1 CQ27.2 CQ27.3 CQ27.4 CQ27.5 CQ27.6 Choose the voltage of the power supply you will use to drive the heater. Next calculate the required resistance R as ΔV 2 P. Knowing the resistivity ρ of the material, choose a combination of wire length and cross-sectional area to make A R ρ. You will have to pay for less material if you make both and A smaller, but if you go too far the wire will have too little surface area to radiate away the energy; then the resistor will melt. Geometry and resistivity. In turn, the resistivity of the material depends on the temperature. The conductor does not follow Ohm s law, and must have a resistivity that is current-dependent, or more likely temperaturedependent. In a normal metal, suppose that we could proceed to a limit of zero resistance by lengthening the average time between collisions. The classical model of conduction then suggests that a constant applied voltage would cause constant acceleration of the free electrons. The drift speed and the current would increase steadily in time. It is not the situation envisioned in the question, but we can actually switch to zero resistance by substituting a superconducting wire for the normal metal. In this case, the drift velocity of electrons is established by vibrations of atoms in the crystal lattice; the maximum current is limited; and it becomes impossible to establish a potential difference across the superconductor. The resistance of copper increases with temperature, while the resistance of silicon decreases with increasing temperature. The conduction electrons are scattered more by vibrating atoms when copper heats up. Silicon s charge carrier density increases as temperature increases and more atomic electrons are promoted to become conduction electrons. The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons more efficiently.

4 Chapter CQ27.7 CQ27.8 Because there are so many electrons in a conductor (approximately electrons/m 3 ) the average velocity of charges is very slow. When you connect a wire to a potential difference, you establish an electric field everywhere in the wire nearly instantaneously, to make electrons start drifting everywhere all at once. Voltage is a measure of potential difference, not of current. Surge implies a flow and only charge, in coulombs, can flow through a system. It would also be correct to say that the victim carried a certain current, in amperes. SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 21.1 *P27.1 Electric Current The drift speed of electrons in the line is v d I nqa I n e πd 2 / 4 The time to travel the 200-km length of the line is then 4I Δt L Ln e πd2 v d Substituting numerical values, ( m 3 ) C Δt m s π 0.02 m A 1 yr s 27.1 yr 2 *P27.2 The period of revolution for the sphere is T 2π, and the average ω current represented by this revolving charge is I q T qω 2π. P27.3 We use I nqav d, where n is the number of charge carriers per unit volume, and is identical to the number of atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationship above. For aluminum, which has a molar mass of 27, we know that Avogadro s number of atoms, N A, has a mass of 27.0 g. Thus, the mass per atom is m 27.0 g 27.0 g N A g atom

5 204 Current and Resistance Thus, n ρ m Therefore, density of aluminum mass per atom 2.70 g cm g atom n atoms cm atoms m 3 v d I nqa m s or, v d mm s A ( m 3 ) C ( m 2 ) P27.4 The period of the electron in its orbit is T 2πr/v, and the current represented by the orbiting electron is I ΔQ Δt e T v e 2πr ( C) m s 2π m C s 1.05 ma P27.5 If N is the number of protons, each with charge e, that hit the target in time t, the average current in the beam is I ΔQ/ Δt Ne/ Δt, giving N I ( Δt) e 23.0 s C/s C/proton protons P27.6 (a) From Example 27.1 in the textbook, the density of charge carriers (electrons) in a copper wire is n electrons/m 3. With A πr 2 and q e, the drift speed of electrons in this wire is v d I n q A I ne πr C s m 3 π m C m s 2 The drift speed is smaller because more electrons are being conducted. To create the same current, therefore, the drift speed need not be as great.

6 Chapter P27.7 From I dq, we have dq I dt. dt From this, we derive the general integral: Q dq I dt In all three cases, define an end-time, T: Q I 0 e t/τ dt T Integrating from time t 0 to time t T: Q ( I 0 τ )e t/τ dt 0 τ We perform the integral and set Q 0 at t 0 to obtain I 0 τ ( 1 e ) T/τ Q I 0 τ e T/τ e 0 (a) If T τ: Q( τ ) I 0 τ ( 1 e 1 ) ( 0.632)I 0 τ If T 10 τ: Q( 10τ ) I 0 τ ( 1 e 10 ) ( )I 0 τ (c) If T : Q( ) I 0 τ ( 1 e ) I 0 τ P27.8 (a) J I A 5.00 A π ka m2 ( m) T 0 (c) Current is the same. The cross-sectional area is greater; therefore the current density is smaller. (d) A 2 4A 1 or πr 2 2 4πr 1 2 so r 2 2r cm. (e) I 5.00 A (f) J J 1 1 ( A/m 2 ) A/m 2 P27.9 We are given q 4t 3 + 5t + 6. The area is (a) A 2.00 cm m 100 cm m 2 I ( 1.00 s) dq ( 12t 2 + 5) 17.0 A dt t1.00 s t1.00 s J I A 17.0 A 85.0 ka m2 2 m

7 206 Current and Resistance P27.10 (a) We obtain the speed of each deuteron from K 1 2 mv2 : v 2K m ( J) kg m/s The time between deuterons passing a stationary point is t in I q t, so t q I C C s s So the distance between individual deuterons is vt ( m/s)( s) m One nucleus will put its nearest neighbor at potential ( C) V k eq r N m 2 C m V This is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect. P27.11 (a) J I A A π A m2 ( m) From J nev d, we have (c) n J ev d From I ΔQ, we have Δt Δt ΔQ I 2.55 A m 2 ( C) m s N A e I s m 3 ( ) C A (This is about 382 years!)

8 Chapter P27.12 To find the total charge passing a point in a given amount of time, we use I dq, from which we can write dt s q dq I dt ( 100 A)sin 120π t s dt q 100 C 120π 0 cos π 2 cos C 120π C P27.13 The molar mass of silver g/mole and the volume V is V ( area) ( thickness ) m m 3 The mass of silver deposited is m Ag ρv kg m 3 ( m) ( m 3 ) kg And the number of silver atoms deposited is The current is then atoms N kg atoms g I ΔV R 12.0 V 6.67 A 6.67 C s 1.80 Ω The time interval required for the silver coating is Δt ΔQ Ne ( ) C I I 6.67 C s s 3.64 h g 1 kg Section 27.2 Resistance P27.14 From Equation 27.7, we obtain I ΔV R 120 V A 500 ma 240 Ω

9 208 Current and Resistance *P27.15 From Ohm s law, R ΔV / I, and from Equation 27.10, R ρ / A ρ / ( πd 2 / 4) Solving for the resistivity gives ρ 4 πd2 R πd Ω m ΔV I 2 π m 4( 50.0 m) Then, from Table 27.2, we see that the wire is made of silver. P27.16 ΔV IR and R ρ. The area is A A mm m mm m V 36.0 A From the potential difference, we can solve for the current, which gives ΔV Iρ A I 6.43 A P27.17 From the definition of resistance, R ΔV I ( m 2 ) 1.50 m I ΔVA ρ V Ω m 120 V 13.5 A 8.89 Ω P27.18 Using R ρl A and data from Table 27.2, we have which yields L ρ Cu Cu πr Cu πr Al 2 ρ Al L Al r 2 Al ρ Al 2 2 r Cu ρ Cu r Al r Cu ρ Al Ω m ρ Cu Ω m 1.29 P27.19 (a) Given total mass m ρ m V ρ m A A m ρ m, where ρ m mass density. Taking ρ resistivity, R ρ A ρ m ρ m ρρ m 2 m.

10 Chapter Thus, mr ρρ m 1.82 m ( kg) Ω Ω m ( kg/m 3 ) V m ρ m, or π r 2 m ρ m Thus, r m πρ m kg 1.82 m π kg/m m The diameter is twice this distance: diameter 280 µm P27.20 (a) Given total mass m ρ m V ρ m A A m ρ m, where ρ m mass density. Taking ρ resistivity, R ρ A ρ m ρ m ρρ m 2 m. Thus, mr ρρ m. Volume V m ρ m, or 1 4 π d2 m ρ m d 4 m 1 π ρ m m π ρ m ρρ m mr π m 2 ρ m 2 ρρ m mr ρm π ρ m R 1 4 P27.21 (a) From the definition of resistance, R ΔV I 120 V 9.25 A 13.0 Ω The resistivity of Nichrome (from Table 27.2) is Ω m.

11 210 Current and Resistance We find the length of wire from R ρ A ρ πr 2 solving for the length gives Rπr 2 ρ 2 ( 13.0 Ω)π m Ω m 170 m Section 27.3 A Model for Electrical Conduction *P27.22 (a) n is unaffected. J I I so it doubles. A (c) J nev d so v d doubles. (d) τ mσ is unchanged as long as σ does not change due to a 2 nq temperature change in the conductor. *P27.23 J σe so σ J E A m V m P27.24 (a) From Appendix C, the molar mass of iron is M Fe g mol g mol kg mol ( Ω m) 1. ( 1 kg 10 3 g) From Table 14.1, the density of iron is ρ Fe kg m 3, so the molar density is ( molar density ) Fe ρ Fe kg m 3 M Fe kg mol mol m 3 (c) The density of iron atoms is density of atoms N A ( molar density ) atoms mol atoms m 3 mol m 3

12 Chapter (d) With two conduction electrons per iron atom, the density of charge carriers is n ( charge carriers atom) ( density of atoms) 2 electrons atom electrons m 3 atoms m 3 (e) With a current of I 30.0 A and cross-sectional area A m 2, the drift speed of the conduction electrons in this wire is v d I nqa 30.0 C s ( m 3 ) C m s ( m 3 ) P27.25 From Equations and 27.13, the resistivity and drift velocity can be related to the electric field within the copper wire: and ρ m ne 2 τ τ v d ee m τ ee m m ρne 2 m ρne 2 E ρne E ρnev d where n is the electron density. From Example 27.1, n N Aρ Cu M The electric field is then E ρnev d E Ω m ( kg/m 3 ) mol kg/mol ( m 3 ) ( C) ( m/s) 0.18 V/m m 3

13 212 Current and Resistance Section 27.4 P27.26 R R α ( ΔT) Solving, Resistance and Temperature [ ] gives 140 Ω 19.0 Ω ΔT C ΔT C T 20.0 C And the final temperature is T C P27.27 If we ignore thermal expansion, the change in the material s resistivity with temperature ρ ρ 0 [1 + αδt] implies that the change in resistance *P27.28 is R R 0 R 0 αδt. The fractional change in resistance is defined by f (R R 0 )/R 0. Therefore, f R 0αΔT R 0 At the low temperature T C we write αδt ( C 1 )( 50.0 C 25.0 C) 0.12 [ ] R C ΔV R I α T C T 0 C where T C. At the high temperature T h, R h ΔV ΔV I h 1 A R 1 + α T 0 [ ( T h 0 )] Then, and 58.0 C 20.0 C 88.0 C 20.0 C ( ΔV) ( 1.00 A) ( C) 1 ( ΔV) I C ( C) 1 I C ( 1.00 A) 1.98 A P27.29 We use Equation and refer to Table 27.2: R R α T T Ω 34.0 C 20.0 C ( C) Ω P27.30 (a) From R ρl/a, the initial resistance of the mercury is R i ρl i A i ρl i πd i 2 4 ( m) Ω m π m 1.22 Ω

14 Chapter Since the volume of mercury is constant, V A f L f A i L i gives the final cross-sectional area as A f A i ( L i L f ). Thus, the final resistance is given by R f ρ L f A f the resistance is then ΔR R R f R i R i ΔR R R f R i cm cm 1 ρ L 2 f 2 ρ L 2 f. The fractional change in A i L i ( A i L i ) 1 L f ρ L i A i L i increase 2 1 *P27.31 (a) The resistance at 20.0 C is ( 34.5 m) 2.99 Ω 2 R 0 ρ A Ω m π m and the current is I ΔV R V 3.00 Ω 3.01 A At 30.0 C, from Equation 27.20, R R 0 [ 1+ α ( ΔT) ] 2.99 Ω 30.0 C 20.0 C ( C) 1 The current is then I ΔV R V 3.10 Ω 2.90 A P27.32 (a) We require two conditions: 3.10 Ω R ρ 1 1 πr 2 + ρ 2 2 πr 2 [1] where carbon 1 and Nichrome 2, and for any ΔT R ρ 1 1 πr 2 ( 1 + α 1 ΔT) + ρ 2 2 πr 2 ( 1 + α 2 ΔT) [2] Setting equations [1] and [2] equal to each other, we have ρ 1 1 πr + ρ πr ρ πr 2 ( 1+ α 1 ΔT) + ρ 2 2 πr 2 ( 1+ α 2 ΔT)

15 214 Current and Resistance simplifying, ρ 1 1 πr 2 + ρ 2 2 πr 2 ρ 1 1 πr 2 + ρ 1 1 πr 2 α 1ΔT + ρ 2 2 πr 2 + ρ 2 2 πr 2 α 2ΔT or ρ 2 2 πr α ΔT ρ πr α 2 1ΔT, which gives ρ 2 2 α 2 ρ 1 1 α 1 [3] The two equations [1] and [3] are just sufficient to determine 1 and 2. The design goal can be met. From Table 27.2, α ( C) 1 and α ( C) 1. Use equation [3] to solve for 2 in terms of 1 : 2 ρ 1 ρ 2 α 1 α 2 1 then substitute this into equation [1]: and so R ρ 1 1 πr + ρ 2 ρ 1 α 1 2 πr 2 1 ρ 1 1 α 1 ρ 2 α 2 πr 2 α Ω Ω m π( m) m 2 ρ 1 α 1 1 ρ 2 α 2 1 ( Ω m) Ω m Therefore, m and m m 1 P27.33 (a) The resistivity is computed from ρ ρ α ( T T 0 ) : 1+ ( C 1 )( 30.0 C) ρ Ω m Ω m The current density is J σe E ρ V/m Ω m 1 Ω A V A/m 2 (c) The current density is related to the current by J I A I πr 2. I J( πr 2 ) ( A/m 2 ) π ( m) ma

16 Chapter (d) The mass density gives the number-density of free electrons; we assume that each atom donates one conduction electron: (e) n P27.34 For aluminum, kg m e /m 3 1 mol g Now J nqv d gives the drift speed as 10 3 g kg v d nq J A/m e /m free e 1 mol ( C/e ) m/s The sign indicates that the electrons drift opposite to the field and current. The applied voltage is ΔV E ( V/m) ( 2.00 m) V. α E C 1 (Table 27.2) and α C 1 (Table 19.1) The resistance is then R ρ A ρ 0 ( 1+ α E ΔT) 1+ αδt A 1+ αδt 1.23 Ω 1.71 Ω 1+ α R E ΔT 2 0 ( 1+ αδt ) ( ( C) 1 ) 120 C 20.0 C C C 20.0 C P27.35 Room temperature is T From Equation 27.19, ρ Al ρ 0 Al 1+ α Al ( T T 0 ) Cu 3 ρ 0 Then, substituting numerical values from Table 27.2 gives T T ρ 0 α Al ρ 0 Cu 1 Al C Ω m Ω m 1

17 216 Current and Resistance and solving for the temperature gives T 20.0 C 207 C T 227 C where we have assumed three significant figures throughout. Section 27.6 Electrical Power *P27.36 (a) P ( ΔV)I ( J C) ( C s) W A large electric generating station, fed by a trainload of coal each day, converts energy faster. I P A P πr 2 *P27.37 P I ( π r 2 ) ( W m 2 )[ π( m) 2 ] W Terrestrial solar power is immense compared to lightning and compared to all human energy conversions. P 0.800( hp) ( 746 W hp) W Then, from P IΔV, P27.38 From Equation 27.21, I P ΔV W V 448 A P27.39 (a) From Equation 27.21, P IΔV A V 7.50 W 120 V P IΔV I P ΔV W From Equation 27.23, P27.40 From Equation 27.21, 8.33 A 14.4 Ω P ΔV 2 R R ΔV 2 P ( 120 V) W ( V) P IΔV A W 15.0 µw

18 Chapter P27.41 From Equation 27.21, 6.00 V P IΔV A 2.10 W P27.42 If the tank has good insulation, essentially all of the energy electrically transmitted to the heating element becomes internal energy in the water: ΔE ( internal) E ( electrical). Our symbol E (electrical) represents the same thing as the textbook s T ET, namely electrically transmitted energy. Since ΔE ( internal) mcδt and E ( electrical) PΔt ( ΔV) 2 Δt/R where c J/kg C the resistance is R (ΔV)2 Δt cmδt 240 V J/kg C P27.43 From P ( ΔV) 2 /R, we find that ( R ΔV i ) 2 P The final current is I f ΔV f R 2 ( s) 6.53 Ω ( 109 kg) 29.0 C (120 V)2 100 W 144 Ω 140 V 144 Ω A The power during the surge is ( P ΔV f ) 2 R So the percentage increase is 136 W 100 W 100 W ( 140 V)2 144 Ω 136 V % P27.44 You pay the electric company for energy transferred in the amount E PΔt. (a) P Δt ( 40 W) ( 2 weeks) $ d 1 week 24 h 1 d k $ kwh P Δt ( 970 W) ( 3 min) 1 h k 60 min $ kwh $

19 218 Current and Resistance (c) P Δt ( W) ( 40 min) 1 h k 60 min P27.45 (a) The total energy stored in the battery is ΔU E q( ΔV) It( ΔV) ( 55.0 A h) ( 12.0 V) 660 W h kwh The value of the electricity is Cost kwh $ kwh 1 C 1 J 1 A s 1 V C $ kwh $ P27.46 (a) The resistance of 1.00 m of 12-gauge copper wire is R ρ A Ω ρ π ( d 2) 4ρ 2 π d Ω m 2 π m The rate of internal energy production is $ W s 1 J 1.00 m 2 (c) 2.1 W P IΔV I 2 R ( 20.0 A) Ω ( 1.00 m) Ω 2 R 4ρ Ω m 2 π d π m P IΔV I 2 R ( 20.0 A) 2 ( Ω) 3.42 W It would not be as safe. If surrounded by thermal insulation, it would get much hotter than a copper wire. P27.47 The power of the lamp is P IΔV U / Δt, where U is the energy transformed. Then the energy you buy, in standard units, is U ΔVIΔt ( 110 V) ( 1.70 A) 1 day 16.2 MJ 24 h 1 day s h 1 J 1 C V C A s

20 Chapter In kilowatt hours, the energy is U ΔVIΔt ( 110 V) ( 1.70 A) 1 day 4.49 kwh 24 h 1 J 1 C 1 day V C A s W s J So operating the lamp costs (4.49 kwh)($0.110/kwh) $0.494 /day. P27.48 The energy taken in by electric transmission for the fluorescent bulb is PΔt 11 J s 100 h cost J For the incandescent bulb, PΔt 40 W 100 h s 1 h $0.110 kwh s 1 h J k W s J J $0.110 cost J J $0.440 savings $0.440 $0.121 $0.319 P27.49 First, we compute the resistance of the wire: R ρ ( Ω m)25.0 m A π m Ω The potential drop across the wire is then (a) ΔV IR ( A) ( 298 Ω) 149 V The magnitude of the electric field in the wire is E ΔV 149 V 25.0 m 5.97 V m The power delivered to the wire is P ( ΔV)I ( 149 V) ( A) 74.6 W (c) We use Equation and Table 27.2: R R 0 1+ α T T Ω 337 Ω C h s $ C

21 220 Current and Resistance To find the power delivered, we first compute the current flowing through the wire: then, I ΔV R 149 V 337 Ω A P ( ΔV)I ( 149 V) ( A) 66.1 W P27.50 The battery takes in energy by electric transmission: PΔt ( ΔV)I( Δt) ( 2.3 J C) ( C s) 4.2 h 469 J It puts out energy by electric transmission: (a) (c) ( ΔV)I( Δt) ( 1.6 J C) ( C s) 2.4 h efficiency useful output total input 249 J 469 J s 1 h The only place for the missing energy to go is into internal energy: 469 J 249 J + ΔE int ΔE int 221 J s 1 h 249 J We imagine toasting the battery over a fire with 221 J of heat input: Q mcδt ΔT Q mc 221 J kg ( 975 J/kg C) 15.1 C P27.51 We compute the resistance of the wire from P ( ΔV)2 R R ( ΔV)2 P ( 110 V)2 500 W 24.2 Ω (a) Then, Equation 27.10, R ρ, gives us the length of wire used: A 2 RA ( ρ 24.2 Ω)π m Ω m 3.17 m

22 Chapter From Equation 27.20, the resistance of the wire at this temperature is ( ) R R 0 [ 1+ αδt ] 24.2 Ω Ω The power delivered to the coil is then P ( ΔV)2 R ( 110 V) Ω 340 W P27.52 We find the energy transferred into a number N of these clocks in one year: T ET P total Δt NP one clock Δt 2.50 W/clock ( 365 d/yr) 24 h/d clocks kwh ( 1 kw/1000 W) Divide this energy into the total cost claimed by the politician to find the cost of the electricity: cost $ kwh $0.017 / kwh This is significantly lower than the average cost of electricity in the United States. While the situation is not actually impossible, the politician would have a better argument by using the actual average cost of electricity in the United States, which would raise his estimate of the total cost to operate the clocks to about $650 million every year. P27.53 At operating temperature, (a) P IΔV ( 1.53 A) ( 120 V) 184 W Use the change in resistance to find the final operating temperature of the toaster. R R 0 ( 1 + αδt ) 120 V 1.53 A 120 V 1.80 A 1 + ( ( C) 1 )ΔT which gives ΔT 441 C and T 20.0 C C 461 C

23 222 Current and Resistance P27.54 Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the dryer is ( 600 s d) 365 d P Δt 400 J s 1 kwh J J 20 kwh We suppose that electrically transmitted energy costs on the order of ten cents per kilowatt-hour. Then the cost of using the dryer for a year is on the order of Cost ( 20 kwh) ( $0.10 kwh) $2 ~ $1 P27.55 We first compute the power delivered to the resistor: P IΔV ( 2.00 A) ( 120 V) 240 W The change in internal energy of the water as it is heated from 23.0 C to 100 C is ( J kg C) ( 77.0 C) 161 kj ΔE int kg The time interval required to heat the water is then Δt ΔE int P P27.56 (a) We know that J 240 W 672 s mechanical power output efficiency total power input ( 2.50 hp)(746 W/1 hp) (120 V) I from which, we calculate the current as I J/s J/s 0.9(120 V) 120 V 17.3 A The energy delivered to the motor in 3.00 h is energy input P input Δt J/s [ ] 3.00( s) J 22.4 MJ (c) At $0.110/kWh, the cost of running the motor for 3.00 h is cost J $ kwh k J 10 3 W s h s $0.684

24 Chapter Additional Problems *P27.57 From Equation 27.22, P ( ΔV)2, we find that the total resistance R needed in the wire is R ( ΔV)2 P ( 20 V)2 48 W 8.3 Ω We then solve for the length of the wire from Equation 27.10: m 2 RA ρ 8.3 Ω Ω m P27.58 At T , R R 0. Then, from Equation 27.20, R R 0 1+ α ( T T 0 ) 2R 0 Solving for the change in temperature gives T T 0 1 α ( C) 1 T 20.0 C 256 C T 276 C m 1.1 km P27.59 We find the amount of current each headlight draws: P IΔV I P ΔV 36.0 W 12.0 V 3.00 A For two headlights, the total current from battery is 6.00 A. The battery rating is the total amount of charge the battery can deliver, without being recharged, over a time interval Δt at a rate (current) I: ΔQ IΔt 90.0 A h The total time interval to discharge the battery is then Δt ΔQ I P27.60 (a) P IΔV ( ΔV)2 R Lightbulb A: Lightbulb B: 90.0 A h 6.00 A 15.0 h R ( ΔV)2 P R ( ΔV)2 P R ( ΔV)2 P I Q Δt P QΔV Δt ΔV P ( 120 V) W 576 Ω ( 120 V)2 100 W 144 Ω ( 120 V) 1.00 C 25.0 W 4.80 s

25 224 Current and Resistance (c) (d) (e) (f) The charge is the same. It is at a location that is lower in potential. P ΔU Δt Δt ΔU P 1.00 J 25.0 W s Because of energy conservation, the energy entering and leaving the lightbulb is the same. Energy enters the lightbulb by electric transmission and leaves by heat and electromagnetic radiation. ΔU PΔt ( 25.0 J s) ( s d) ( 30.0 d) J Cost J P27.61 The resistance of one wire is $ kwh k Ω mi W s J 100 mi 50.0 Ω. h s $1.98 The whole wire is at nominal 700 kv away from ground potential, but the potential difference between its two ends is 50.0 Ω IR A Then it radiates as heat power 50.0 kv P IΔV ( A) V P27.62 (a) From ρ RA ( ΔV) A I (m) R (Ω) ρ (Ω m) MW we compute (c) ρ Ω m The average value is within 1% of the tabulated value of Ω m given in Table 27.2.

26 Chapter P27.63 The original stored energy is U E,i 1 2 QΔV i 1 2 (a) When the switch is closed, charge Q distributes itself over the plates of C and 3C in parallel, presenting equivalent capacitance Q 2 C. 4C. Then the final potential difference is ΔV f Q 4C for both. The smaller capacitor then carries charge CΔV f Q 4C C Q 4. The larger capacitor carries charge 3C Q 4C 3Q 4. (c) The smaller capacitor stores final energy 1 2 C ΔV f Q 2 32C 1 2 3C Q 4C. The larger capacitor possesses energy 2 3Q2 32C C Q 4C 2 (d) The total final energy is Q2 32C + 3Q2 32C Q2. The loss of potential 8C energy is the energy appearing as internal energy in the resistor: Q 2 2C Q2 8C + ΔE int so ΔE int 3Q2 8C P27.64 (a) The heater should put out constant power P Q Δt mc T f T i Δt kg ( J/kg C) 100 C 20 C 349 J s ( 4 min) Then its resistance should be described by ( P ΔV)2 R ( R ΔV)2 P ( 120 J C)2 349 J s 41.3 Ω 1 min 60 s

27 226 Current and Resistance Its resistivity at 100 C is given by ρ ρ 0 1+ α T T Ω m Ω m Then for a wire of circular cross section, from Equation 27.10, R ρ A ρ π r ρ 4 2 π d Ω ( Ω m) 4 π d 2 d m or d 2 ( m) One possible choice is m and d m. If and d are made too small, the surface area will be inadequate to transfer heat into the water fast enough to prevent overheating of the filament. To make the volume less than 0.5 cm 3, we want and d less than those described by π d m 3. Substituting d 2 ( m) gives π ( m) m 3, 3.65 m and d m. Thus our answer is: Any diameter d and length related by d 2 ( ), where d and are in meters. Yes; for V cm 3 of Nichrome, 3.65 m and d mm. *P27.65 The power the beam delivers to the target is ( V) W P IΔV A The mass of cooling water that must flow through the tube each second if the rise in the water temperature is not to exceed 50 C is found from Therefore, Q PΔt ( Δm)cΔT Δm Δt P cδt J/s J/kg C 50.0 C kg/s

28 Chapter P27.66 (a) Since P IΔV, we have I P ΔV W 667 A 12.0 V From P U / Δt, the time the car runs is Δt ΔU P J W s So it moves a distance of Δx vδt ( 20.0 m s) ( s) 50.0 km P27.67 (a) Assuming the change in V is uniform: E x dv ( x) E x ΔV dx Δx ( V) m V/m or 8.00 V/m in the positive x direction. From Equation 27.10, we have R ρ A (c) From Equation 27.7, ( m) Ω Ω m π m I ΔV R 4.00 V Ω 6.28 A (d) From Equation 27.5, the current density is given by (e) J I A 6.28 A A m MA m 2 π m The field and the current are both in the x direction. We intend to derive the equivalent of Equation We start with the definition of current density, J I/A, and, using Equations 27.7 and 27.10, note that the current is given by Then, I ΔV R E R EA ρ J I A EA/ ρ A E ρ

29 228 Current and Resistance so E ρj ( Ω m) ( A m 2 ) 8.00 V/m P27.68 (a) Assuming the change in V is uniform: E x dv ( x) E x ΔV dx Δx 0 V L 0 + V L Therefore, the electric field is V/L in the positive x direction. From Equation 27.10, we have R ρ A ρl 4ρL π π d 2 d2 4 (c) From Equation 27.7, I ΔV R Vπ d 2 4ρ L (d) From Equation 27.5, the current density is given by (e) J I Vπ d2 4ρ L V ρl in the positive x direction A π d 2 4 The field and the current both have the same direction. We intend to derive the equivalent of Equation We start with the definition of current density, J I/A, and, using Equations 27.7 and 27.10, note that the current is given by Then, I ΔV R E R EA ρ J I A EA/ ρ A E ρ so E ρ J ρ V ρl V L P27.69 Since there are 2 wires, the total length is 100 m. The resistance of the wires is (a) R Ω 300 m 100 m Ω We find the potential difference at the customer s house from ( ΔV) home ( ΔV) line IR 120 V ( 110 A) ( Ω) 116 V

30 Chapter The power delivered to the customer is P I( ΔV) ( 110 A) ( 116 V) 12.8 kw (c) The power dissipated in the wires, or the energy produced in the wires, is P wires I 2 R ( 110 A) 2 ( Ω) 436 W P27.70 The original resistance is R i ρl i /A i. The new length is L L i + δ L i L i (1 + δ ) (a) Constancy of volume implies AL A i L i so A A L i i L A i L i L i (1+ δ ) A i (1+ δ ) The new resistance is R ρl A ρl i (1 + δ) A i /(1 + δ) R (1 + i δ)2 R i (1 + 2δ + δ 2 ) The result is exact if the assumptions are precisely true. Our derivation contains no approximation steps where delta is assumed to be small. P27.71 (a) A thin cylindrical shell of radius r, thickness dr, and length L contributes resistance dr ρd A ρdr ( 2π r)l ρ 2π L The resistance of the whole annulus is the series summation of the contributions of the thin shells: dr r R ρ r b dr 2π L r In this equation ΔV I r c ρ 2π L ln r b r a ρ 2π L ln r b. r a Solving, we get ρ 2π LΔV I ln r b. r a

31 230 Current and Resistance P27.72 The value of 11.4 A is what results from substituting the given voltage and resistance into Equation However, the resistance measured for a lightbulb with an ohmmeter is not the resistance at which it operates because of the change in resistivity with temperature. The higher resistance of the filament at the operating temperature brings the current down significantly. P27.73 Let α be the temperature coefficient at 20.0 C, and α be the temperature coefficient at 0 C. Then ρ ρ α T 20.0 C [ ] and [ ] must both give the correct resistivity at any ρ ρ 1 + α T 0 C temperature T. That is, we must have: ρ 0 [ 1 + α ( T 20.0 C) ] Setting T 0 in equation [1] yields: ρ ρ 0 [ 1 α ( 20.0 C) ] [ ] [1] ρ 1 + α T 0 C and setting T 20.0 C in equation [1] gives: [ ] ρ 0 ρ 1 + α 20.0 C Substitute ρ from the first of these results into the second to obtain: ρ 0 ρ 0 [ 1 α ( 20.0 C) ][ 1 + α ( 20.0 C) ] Therefore, 1 + α 20.0 C which simplifies: Therefore, α 20.0 C α α 1 1 α 20.0 C 1 1 α 20.0 C ( 20.0 C) α ( 20.0 C) 1 α ( 20.0 C) [ ] 1 1 α 20.0 C 1 1 α 20.0 C α α 1 α 20.0 C α [ 1 α ( 20.0 C) ] ( C) ( C) ( C) C

32 Chapter P27.74 (a) We begin from ΔV E or dv E dx. Then, ΔV IR E and the current is P27.75 We begin with I dq dt E R A ρ E A dv E σ A ρ dx σ A dv dx Current flows in the direction of decreasing voltage. Energy flows by heat in the direction of decreasing temperature. R ρ A ρ 0 1+ α T T α T T 0 2 A 0 1+ α ( T T 0 ). ρ α T T 0 A 0 1+ α T T 0 For copper (for T C): ρ Ω m, α C 1, and α C 1. Then, R ρ α T T 0 A α T T 0 ( ) R π R Ω 1 + ( C 1 ) C ( C 1 ) C P27.76 The wire has length, and radius r; its cross-sectional area is A (πr 2, if circular), which is proportional to r 2. Because both and r change with a temperature variation ΔT according to L L 0 ( 1 + α ΔT), the crosssectional area changes according to A A0 ( 1+ α ΔT) 2. Calling R 0 ρ 0 0 A 0 at temperature T 0, we have R 0 ρ 0 0 R ρ 0 1+ α T T α T T 0 2 A 0 A 0 1+ α ( T T 0 ) ρ 0 1+ α T T 0 x A 0 1+ α ( T T 0 )

33 232 Current and Resistance which gives 1+ α T T R R α T T 0 P27.77 (a) Think of the device as two capacitors in parallel. The one on the left has κ 1 1, A x. The equivalent capacitance is κ 1 0 A 1 d + κ 2 0 A 2 d 0 d 0 2d 2 + x + κ 0 d 2 x ( + 2x +κ 2κ x ) The charge on the capacitor is Q CΔV Q 0 ΔV 2d The current is ( + 2x +κ 2κ x) I dq dt dq dx dx dt 0 ΔV ( κ )v 0 ΔVv κ 1 2d d The negative value indicates that the current drains charge from the capacitor. Positive current is clockwise 0 ΔVv d ( κ 1). P27.78 (a) The resistance of the dielectric block is R ρ A d σ A. The capacitance of the capacitor is C κ 0 A. d Then RC only. d κ 0 A κ 0 σ A d σ is a characteristic of the material The resistance between the plates of the capacitor is R κ 0 σ C ρκ 0 C Ω m Ω C 2 N m F

34 Chapter P27.79 The volume of the gram of gold is given by ρ m V. V m ρ Then, A m kg kg m m 3 A m and the resistance is ( m) R ρ A Ω m Ω m 2 P27.80 Evaluate I I 0 exp eδv k B T 1 and R ΔV I with I A, e C, and k B J/K. Parts (a) and : The following includes a partial table of calculated values and a graph for each of the specified temperatures. (i) For T 280 K: I ( A) R( Ω) ΔV V ANS. FIG. P27.80(i)

35 234 Current and Resistance (ii) For T 300 K: I ( A) R( Ω) ΔV V ANS. FIG. P27.80(ii) (iii) For T 320 K: I ( A) R( Ω) ΔV V ANS. FIG. P27.80(iii)

36 Chapter P27.81 To find the final operating temperature, we begin with R R α T T 0 and solve for the temperature T: T T R 1 T I 0 α R 0 α I 1 In this case, I I 0 10, so T T α ( 9) C C Challenge Problems P27.82 (a) We are given α 1 dρ ρ dt ρ dρ T Separating variables, α dt ρ 0 ρ. We integrate, on both T 0 sides, from the physical situation at temperature T 0 to that at temperature T. Integrating both sides, ln(ρ/ρ 0 ) α(t T 0 ) Thus ρ ρ 0 e α ( T T 0 ) From the series expansion e x 1 + x, with x much less than 1, ρ ρ α T T 0 P27.83 A spherical layer within the shell, with radius r and thickness dr, has resistance dr ρdr 4πr 2 The whole resistance is the absolute value of the quantity b R dr a r b ρdr 4πr ra 2 ρ r 1 r b 4π 1 ra ρ 4π r a r b ρ 1 1 4π r a r b

37 236 Current and Resistance P27.84 Refer to ANS. FIG. P The current flows generally parallel to L. Consider a slice of the material perpendicular to this current, of thickness dx, and at distance x from face A. Then the other dimensions of the slice are w ANS. FIG. P27.84 and y, where by proportion y y 1 x y 2 y 1 L so y y 1 + (y 2 y 1 ) x. The bit of resistance which this slice contributes is L dr ρdx A ρdx wy ρdx w y 1 + y 2 y 1 The whole resistance is that of all the slices: L L x0 0 R dr ρ w L y 2 y 1 With u y 1 + ( y 2 y 1 ) x L ρl R w y 2 y 1 L x0 ( ( x / L) ) ρdx ( ( x / L) ) w y 1 + y 2 y 1 (( y 2 y 1 )/ L)dx y 1 + y 2 y 1 ( x / L) this is of the form ln y 1 + ( y 2 y 1 ) x/l ρl w y 2 y 1 ( ln y 2 ln y 1 ) du, so u ρl L x0 ln y 2 y 1 w y 2 y 1 P27.85 From the geometry of the longitudinal section of the resistor shown in ANS. FIG. P27.85, we see that ( b r) y ( b a) h From this, the radius at a distance y from the base ANS. FIG. P27.85 is r ( a b) y ρdy + b. For a disk-shaped element of volume dr h π r : 2 R ρ π Using the integral formula h 0 dy 2 ( a b) y h + b du ( au + b) 2 1 a( au + b), R ρ h π ab

38 Chapter ANSWERS TO EVEN-NUMBERED PROBLEMS P27.2 qω 2π P ma P27.6 (a) m/s; The drift speed is smaller because more electrons are being conducted. P27.8 (a) 99.5 ka/m 2 ; The current is the same; (c) The current density is smaller; (d) cm; (e) I 5.00 A; (f) A/m 2 P27.10 (a) m; The potential of the nearest neighbor is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect. P C P ma P A P P27.20 (a) mr ρρ m ; 4 ρm π ρ m R 1/4 P27.22 (a) unaffected; doubles; (c) doubles; (d) unchanged P27.24 (a) kg/mol; mol/m 3 ; (c) atoms m 3 ; (d) electrons/m 3 ; (e) m/s P27.26 T C P A P27.30 (a) 1.22 Ω; increase P27.32 (a) The design goal can be met; m and m P Ω P27.36 (a) W; W P W P µw P Ω P27.44 (a) $1.48; $ ; (c) $0.381

39 238 Current and Resistance P27.46 (a) 2.1 W; 3.42 W; (c) It would not be as safe. If surrounded by thermal insulation, it would get much hotter than a copper wire. P27.48 $0.319 P27.50 (a) 0.530; 221 J; (c) 15.1 C P27.52 See P27.52 for full explanation. P27.54 ~ $1 P27.56 (a) 17.3 A; 22.4 MJ; (c) $0.684 P C P27.60 (a) Lightbulb A 576 Ω and Lightbulb B 144 Ω; 4.80 s; (c) The charge is the same. It is at a location that is lower in potential; (d) s; (e) The energy is the same. Energy enters the lightbulb by electric transmission and leaves by heat and electromagnetic radiation; (f) $1.98 P27.62 (a) See the table in P27.62(a); Ω m; (c) The average value is within 1% of the tabulated value of Ω m given in Table P27.64 (a) Any diameter d and length related by d 2 ( ), where d and are in meters; Yes; for V cm 3 of Nichrome, 3.65 m and d mm. P27.66 (a) 667 A; 50.0 km P27.68 (a) V/L in the positive x direction; 4 ρl/π d 2 ; (c)vπd 2 / 4ρL; (d) V/ρL in the positive x direction; (e) ρj ρ V ρl V L E P27.70 See P27.70(a) for the full explanation; The result is exact if the assumptions are precisely true. Our derivation contains no approximation steps where delta is assumed to be small. P27.72 The value of 11.4 A is what results from substituting the given voltage and resistance into Equation However, the resistance measured for a lightbulb with an ohmmeter is not the resistance at which it operates because of the change in resistivity with temperature. The higher resistance of the filament at the operating temperature brings the current down significantly. P27.74 (a) σ A dv ; Current flows in the direction of decreasing voltage. dx Energy flows by heat in the direction of decreasing temperature.

40 Chapter α T T P27.76 R R α T T 0 P27.78 (a) See P27.78 for full explanation; Ω P27.80 (a) See Table P27.80 (i), (ii), and (iii); See ANS. FIG. P27.80 (i), (ii), and (iii). P27.82 (a) ρ ρ 0 e α ( T T ) 0 ; ρ ρ α T T 0 P27.84 ρl ln y 2 w y 2 y 1 y 1