Current and Resistance

Transcription

1 Chapter 17 Current and esistance Quick Quizzes 1. (d. Negative charges moving in one direction are equivalent to positive charges moving in the opposite direction. Thus, Ia, Ib, Ic, and Id are equivalent to the movement of,, 4, and charges respectively, giving I < I < I <. d b c. (b. Under steady-state conditions, the current is the same in all parts of the wire. Thus, the drift velocity, given by vd = I nqa, is inversely proportional to the cross-sectional area.. (c, (d. Neither circuit (a nor circuit (b applies a difference in potential across the bulb. Circuit (a has both lead wires connected to the same battery terminal. Circuit (b has a low resistance path (a short between the two battery terminals as well as between the bulb terminals. 4. (b. The slope of the line tangent to the curve at a point is the reciprocal of the resistance at that point. Note that as increases, the slope (and hence 1 increases. Thus, the resistance decreases. l l. (b. Consider the expression for resistance: = ρ = ρ. Doubling all linear A π r dimensions increases the numerator of this expression by a factor of, but increases the denominator by a factor of 4. Thus, the net result is that the resistance will be reduced to one-half of its original value. 6. (a. The resistance of the shorter wire is half that of the longer wire. The power dissipated, ( P =, (and hence the rate of heating will be greater for the shorter wire. Consideration of the expression P = I I a might initially lead one to think that the reverse would be true. However, one must realize that the currents will not be the same in the two wires. 69

2 7 CHAPTE (b. Ia = Ib > Ic = Id > Ie = I f. Charges constituting the current I a leave the positive terminal of the battery and then split to flow through the two bulbs; thus, Ia = Ic + I e. Because the potential difference is the same across the two bulbs and because the power delivered to a device is P = I, the 6 W bulb with the higher power rating must carry the d ( greater current, meaning that Ic > I e. Because charge does not accumulate in the bulbs, all the charge flowing into a bulb from the left has to flow out on the right; consequently I = I and I = I. The two currents leaving the bulbs recombine to form the current back c e into the battery, I f + I d = Ib. f 8. (a B, (b B. Because the voltage across each resistor is the same, and the rate of energy delivered to a resistor is P = (, the resistor with the lower resistance (that is, B dissipates more power. From Ohm s law, I = V. Since the potential difference is the same for the two resistors, B (having the smaller resistance will carry the greater current.

3 Current and esistance 71 Answers to Even Numbered Conceptual Questions. In the electrostatic case in which charges are stationary, the internal electric field must be zero. A nonzero field would produce a current (by interacting with the free electrons in the conductor, which would violate the condition of static equilibrium. In this chapter we deal with conductors that carry current, a nonelectrostatic situation. The current arises because of a potential difference applied between the ends of the conductor, which produces an internal electric field. 4. The number of cars would correspond to charge Q. The rate of flow of cars past a point would correspond to current. 6. The W bulb has the higher resistance. Because ( V = P, and both operate from 1 V, the bulb dissipating the least power has the higher resistance. The 1 W bulb carries more current, because the current is proportional to the power rating of the bulb. 8. An electrical shock occurs when your body serves as a conductor between two points having a difference in potential. The concept behind the admonition is to avoid simultaneously touching points that are at different potentials. 1. The knob is connected to a variable resistor. As you increase the magnitude of the resistance in the circuit, the current is reduced, and the bulb dims. 1. Superconducting devices are expensive to operate primarily because they must be kept at very low temperatures. As the onset temperature for superconductivity is increased toward room temperature, it becomes easier to accomplish this reduction in temperature. In fact, if room temperature superconductors could be achieved, this requirement would disappear altogether. 14. The amplitude of atomic vibrations increases with temperature, thereby scattering electrons more efficiently.

4 7 CHAPTE 17 Answers to Even Numbered Problems m s electrons s ma m s 1. ma 1. (a 1.8 m (b.8 mm Ω 16. ~ $1 (Assumes a 4 W dryer used for about 1 min/d with electric energy costing about 1 cents/kwh (a.8 1 A (b A C. 9.7 Ω C 4. ( 1 6. (a.6 mv (b 1.1 mv m C. (a $.9 (b $.6 4. (a MW (b.71 or 7.1% 6. (a. 1 J (b 18 min 8. (a 184 W (b 461 C 4. (a. cents (b 71% 4. (a $1.61 (b.8 cents (c 41.6 cents h d

5 Current and esistance d l d = ( Any diameter and length related by m l, such as l= 1. m and d =. mm. Yes. 48. (a 76 Ω, 144 Ω (b 4.8 s, lower potential energy (c.4 s, converted to internal energy and light 8 (d $1.6, energy, $ J. (a J (b 1.9 cents. 89 µ V 4. (a 18 C (b.6 A 7 6. (a Ω (b Ω 8..6 k Ω (nichrome, 4.4 k Ω (carbon 6. No. The fuse should be rated at.87 A or less. 6. (a V I = V I 1. V. 1 A 1. V. 1 A. V.1 1 A. 1 Ω. 1 Ω. 1 Ω +.4 V +.1 A 4 Ω +. V +. A Ω +. V +.4 A 14 Ω +.7 V +.7 A 9.7 Ω +.7 V +.1 A 7. Ω (b The resistance of the diode is very large when the applied potential difference has one polarity and is rather small when the potential difference has the opposite polarity. 64. (a 9. m (b.9 mm 66. (b 1.4 Ω versus Ω for the more precise equation

6 74 CHAPTE 17 Problem Solutions 17.1 The charge that moves past the cross section is Q= I( t, and the number of electrons is ( t Q I n = = e e ( 8. 1 C s ( 1. min( 6. s min = = C The negatively charged electrons move in the direction opposite to the conventional current flow. 1 electrons ur v ur I 17. The drift speed in a conductor of cross-sectional area A and carrying current I is v = I nqa, where n is the number of free charge carriers per unit volume and q is the d charge of each of those charge carriers. In the given conductor, v d = I. C s nqa = = ( 7. 1 m ( C( 4. 1 m.1 1 m s 17. The current is Q I = = t =, giving. Thus, the change that passes is Q ( t 1. V Q= ( t = (.1 A(. s =. C 1. Ω 17.4 Q= I( t and the number of electrons is ( t ( C s( 1. s Q I n = = = = -19 e e C electrons

7 Current and esistance The period of the electron in its orbit is T = π r v, and the current represented by the orbiting electron is Q e v e I = = = t T π r 6 19 ( ( 11 π (.9 1 m.19 1 m s C 1. 1 C s 1. ma = = = 17.6 The mass of a single gold atom is m atom M 197 g mol = = = = N A 6. 1 atoms mol g.7 1 kg The number of atoms deposited, and hence the number of ions moving to the negative electrode, is. 1 kg m n = = = m atom kg 1 Thus, the current in the cell is 1 19 ( ( C Q ne I = = = =.19 A = 19 ma t t (.78 h( 6 s 1 h 17.7 The drift speed of electrons in the line is v d I I = =, or nqa ne d ( π 4 v d ( 41 A = = 8 19 ( 8. 1 m ( C π (. m The time to travel the length of the -km line is then m s L 1 m 1 yr t = = -4 7 = v.4 1 m s.16 1 s d 7 yr

8 76 CHAPTE 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 density N = = =, ρ A ρ n mass per atom M N M A or ( 6. 1 mol (.7 g cm ( 1 6 cm 1 m n = = 6. 1 m 6.98 g mol 8 The drift speed of the electrons in the wire is then v d = I. C s nea = = ( 6. 1 m ( C( 4. 1 m m s 17.9 (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 V I = = =. A = ma 4 Ω ( V = Imax= ( max 8 1 A Thus, if = 4. 1 Ω, ( V max = V and if = Ω, ( V =.16 V max 17.1 The volume of the copper is 1. 1 kg m V = = = m density kg m 7 7 Since, V = A L, this gives A L = m (1

9 Current and esistance 77 (a From ρl =, we find that A 8 ρ Ω m A = L= L=. Ω ( 8 (.4 1 m Inserting this expression for A into equation (1 gives.4 1 m m 8 L = 7, which yields L = 1.8 m L (b From equation (1, 7 π d m A = = 4 L, or ( ( m m d = = πl 7 7 π( 1.8 m m.8 mm = = 17.1 From ρl =, we obtain A π d ρl A = =, or 4 8 ( Ω ( 4 ρ L m. 1 m d = = = = π π(. Ω m.17 mm ( Ω ( ρl ρl m 1 m = = = =.1 Ω A π d 4 π m ( 17.1 (a 1 V = = = Ω I.4 A (b From, ρl =, A ( (.4 1 m A π Ω ρ = = = Ω L. m m

10 78 CHAPTE We assume that your hair dryer will use about 4 W of power for 1 minutes each day of the year. The estimate of the total energy used each year is min 1 hr E= P( t = (.4 kw 1 ( 6 d = 4 kwh d 6 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 = ( 4 kwh.1 = $.4 or ~$1 kwh The resistance is 9.11 V = = =. Ω, so the resistivity of the metal is I 6. A ( πd 4 (. π(. 1 m A Ω ρ = = = = Ω L L 4. m Thus, the metal is seen to be silver. ( m With different orientations of the block, three different values of the ratio LA are possible. These are: L 1 cm 1 1 = = =, A cm 4 cm 8 cm.8 m 1 ( ( L cm 1 1 = = =, A 1 cm 4 cm cm. m ( ( and L 4 cm 1 1 = = = A 1 cm cm. cm. m ( ( (a (b I = = ( L A max 8 min ρ min I = = ( L A min 8 max ρ max ( 6. V(.8 m 8 = =.8 1 A Ω m ( 6. V(. m 7 = = A Ω m

11 Current and esistance The volume of material, = L = ( π r L stretched to decrease its radius, the length increases such that ( π f f = ( π V A, in the wire is constant. Thus, as the wire is r r Lf = L = L = ( 4. L = 16L r f.r The new resistance is then r L r L giving f L L 16L L = = = = = = 6 1. Ω = 6 Ω f f ρ ρ ρ 16 ( 4 ρ 6 ( Af πrf π ( r 4 πr 17. Solving = + α ( 1 T T for the final temperature gives 14 Ω 19 Ω T = T + = + = 4. 1 ( C ( 19 Ω C C - 1 α 17.1 From Ohm s law, V = Iii = I ff, so the current in Antarctica is I f i = Ii = f 1 + α Ii 1 + α ( T T i ( Tf T ( C ( 8. C. C ( = A = 1.98 A 1 1 ( C ( 88. C. C The expression for the temperature variation of resistance, = 1 + α ( TT T = C, gives the temperature coefficient of resistivity of this material as α Then, at T 1. Ω 1. Ω T T = = = ( ( 1. Ω ( 9 C C = C, the resistance will be ( ( C ( ( C with = 1 + α C C = 1. Ω 1 C = 9. 7 Ω

12 8 CHAPTE At 8 C,. V I = = = α ( TT ( Ω 1 + (. 1 C ( 8 C C or I = =.6 1 A 6 ma 17.4 If 41. at T C and 41.4 at T 9. C, then gives the temperature coefficient of resistivity of the material making up this wire as = Ω = = Ω = = 1 + α ( TT α 41.4 Ω 41. Ω T T = = = ( ( 41. Ω ( 9. C C ( C ( Ω ( ρl m 1. m = = =.67 1 A. 1 m 6 Ω (a At =. C, = 1+ α ( T T T gives a resistance of 1 ( (.9 1 ( C ( = Ω +. C. C =.89 1 Ω (b At = 1. C, = 1 + α ( T T T yields 1 ( (.9 1 ( C ( = Ω + 1. C. C =.4 1 Ω 17.6 (a At C, the resistance of this copper wire is = ρ( L A = ρ( L πr potential difference required to produce a current of I =. A is and the L V = I = I ρ = Ω π r 8 (. A( m ( 1. m (. 1 m π or = = 4 V 6. 1 V.6 mv

13 Current and esistance 81 (b At T = C, the potential difference required to maintain the same current in the copper wire is V ( α( I I 1 α T T V 1 T T = = + = + -1 or V (. 6 mv 1 + (.9 1 C ( C C = = 1.1 mv 17.7 (a The resistance at. C is 8 ( Ω ( L m 4. m = ρ = =. Ω A π (. 1 m and the current will be (b At. C, 9. V I = = =. Ω. A ( = 1 + α T T 1 ( ( ( ( =. Ω C. C. C =.1 Ω Thus, the current is 9. V I = = =.1 Ω.9 A 17.8 The resistance of the heating element when at its operating temperature is ( ( 1 V = = = 1.7 Ω P 1 W ρ L 1 α TT A, the cross-sectional area is From = 1+ α( T T = + ( ρ L ( A= 1 + α TT 8 ( 1 1 m( 4. m Ω = Ω 1 ( 1 ( C ( C. C A = m 7

14 8 CHAPTE (a From = ρl A, the initial resistance of the mercury is i 7 ( Ω m( 1. m π ( 1. 1 m 4 ρli i = = = 1. Ω A (b Since the volume of mercury is constant, V = Af Lf = Ai L i gives the final crosssectional area as Af Ai ( Li Lf f f i i =. Thus, the final resistance is given by ρlf ρlf = =. The fractional change in the resistance is then A A L ( f i f ρ Lf Ai Li Lf = = 1= 1= 1 i i ρ Li Ai Li 1.4 = 1= or a.8% increase 17. The resistance at. C is. Ω = = = 17 Ω α ( T T ( C ( C. C 1 Solving = 1 + α ( TT for T gives the temperature of the melting potassium as.8 Ω 17 Ω T = T + =. C + = 6. C 1 α.9 1 ( C ( 17 Ω W I = P. A = 1 V = and 1 V = = = 4. Ω I. A 17. (a The energy used by a 1-W bulb in 4 h is ( ( ( ( E= P t= 1 W 4 h =.1 kw 4 h =.4 kwh and the cost of this energy, at a rate of $.1 per kilowatt-hour is ( ( cost = E rate =.4 kwh $.1 kwh = $.9

15 Current and esistance 8 (b The energy used by the oven in. h is 1 kw E= P t= I( t= (. C s( J C (. h = kwh 1 J s and the cost of this energy, at a rate of $.1 per kilowatt-hour is ( ( cost = E rate = kwh $.1 kwh = $ The maximum power that can be dissipated in the circuit is P ( V I ( ( = = = max max 1 V 1 A W Thus, one can operate at most 18 bulbs rated at 1 W per bulb (a The power loss in the line is ( ( ( 7 P loss = I= 1 A.1 Ω km 16 km =. 1 W = MW (b The total power transmitted is P input ( V I ( ( = = = = V 1 A 7. 1 W 7 MW Thus, the fraction of the total transmitted power represented by the line losses is Ploss MW fraction loss = = =.71 P 7 MW input or 7.1% 17. The energy required to bring the water to the boiling point is ( ( ( ( E= mc T = = The power input by the heating element is P input. kg J kg C 1 C. C J ( V I ( ( = = 1 V. A = 4 W = 4 J s Therefore, the time required is E J 1 min t = = = 67 s = 11. min P 4 J s 6 s input

16 84 CHAPTE 17 E= P t = 9 W 1 h = 9 J s 6 s =. 1 J 17.6 (a ( ( ( ( (b The power consumption of the color set is ( V I ( ( P = = 1 V. A = W Therefore, the time required to consume the energy found in (a is E. 1 J 1 min t = = = s = 18 min P J s 6 s 17.7 The energy input required is ( ( ( ( E= mc T = = 1. kg J kg C. C 1. C.1 1 J and, if this is to be added in t = 1. min = 6 s, the power input needed is E.1 1 J P = = = 419 W t 6 s The power input to the heater may be expressed as resistance is ( P =, so the needed ( ( 1 V = = = 4.4 Ω P 419 W 17.8 (a At the operating temperature, ( V I ( ( P = = 1 V 1. A = 184 W 1 T T, the temperature T is given by T = T +. The α resistances are given by Ohm s law as (b From = + α ( ( 1 V = =, and I 1. A ( = = I 1 V 1.8 A Therefore, the operating temperature is ( 1 1. ( ( 1 ( C 1 ( T =.C +.4 = 461C

17 Current and esistance The resistance per unit length of the cable is P I P L = = = = Ω L L I. W m. 1 m ( A From = ρl A, the resistance per unit length is also given by L= ρ A. Hence, the ρ cross-sectional area is π r = A=, and the required radius is L 8 ρ Ω m r = = =.16 m = 1.6 cm π - ( L π (. 1 Ω m 17.4 (a The power input to the motor is P input ( V I ( ( = = 1 V 1.7 A = 1 W =.1 kw At a rate of $.6/kWh, the cost of operating this motor for 4. h is ( ( P input cost = Energy used rate = t rate (b The efficiency is ( ( ( =.1 kw 4. h 6. cents kwh =. cents (. hp(.746 kw hp Poutput Eff = = =.71 or 71% P.1 kw input

18 86 CHAPTE The total power converted by the clocks is ( ( P= = 6 8. W W and the energy used in one hour is ( ( 8 1 E= P t= W 6 s =.4 1 J The energy input required from the coal is E coal 1 E.4 1 J = = = efficiency. 1 J The required mass of coal is thus E coal J. 6 1 J kg m = = heat of combustion =.9 1 kg or 1 metric ton m = (.9 1 kg 9 me = tric tons 1 kg E= P t = 4. W 14. d 4. h d = Wh = 1.4 kwh 17.4 (a ( ( ( 4 ( ( ( cost = E rate = 1.4 kwh $.1 kwh = $1.61 E= P t =.97 kw. min 1 h 6 min = kwh (b ( ( ( ( cost = E rate - ( ( = kwh $.1 kwh = $. 8 =.8 cents (c E P t ( ( ( = =. kw 4. min 1 h 6 min =.47 kwh ( ( ( cost = E rate =.47 kwh $.1 kwh = $.416 = 41.6 cents 17.4 E= P t= (.18 kw( 1 h and cost= E rate= ( P t( $.7 kwh so ( ( ( cost =.18 kw 1 h $.7 kwh = $.6 = 6 cents

19 Current and esistance The energy used was cost $ E = = = rate $.8 kwh. 1 kwh E. 1 kwh The total time the furnace operated was t = = = 14 h, and since P 4. kw January has 1 days, the average time per day was 14 h average daily operation = = 1. d.6 h d 17.4 The energy saved is ( Phigh Plow t ( ( E= = = = and the monetary savings is 4 W 11 W 1 h.9 1 Wh.9 kwh ( ( savings = E rate =.9 kwh $.8 kwh = $. = cents The power required to warm the water to 1 C in 4. min is ( T (. kg( J kg C( 1 C C t ( 4. min( 6 s 1 min Q mc P = = = = t The required resistance (at 1 C of the heating element is then ( ( 1 V = = = 41 Ω P. 1 W so the resistance at C would be 41 Ω = = = 4 Ω C 1 C C ( T T α - 1 ( (. 1 W We find the needed dimensions of a Nichrome wire for this heating element from ( A d = ρl = ρl π 4, where l is the length of the wire and d is its diameter. This 4ρ l gives 4 Ω= π d or d ρ 1 1 Ω m = = = ( m π( 1 Ω l π( 1 Ω l l 8 8

20 88 CHAPTE 17 Thus, ( 8 any combination of length and diameter satisfying the relation d m be suitable. A typical combination might be = l will l= 1. m and d = = = m. 1 m. mm Yes, such heating elements could easily be made from less than. cm of Nichrome. The volume of material required for the typical wire given above is 6 1 cm ( π l π( 4 ( ( 7 V = Al= d =. 1 m 1. m = 1. 1 m =.1 cm 1 m The energy that must be added to the water is J 1 kwh E= mc( T = ( kg ( 8 C 1 C 1 kwh 6 = kg C.6 1 J and the cost is cos t E rate ( ( = = 1 kwh $.8 kwh = $ (a From ( P =, the resistance of each bulb is dim bright ( ( 1 V = = = P. W dim ( ( 1 V = = = P 1 W bright 76 Ω and 144 Ω (b The current in the dim bulb is Pdim. W I = = =.8 A 1 V so the time for 1. C to pass through the bulb is Q 1. C t = = = I.8 A 4.8 s When the charge emerges from the bulb, it has lower potential energy

21 Current and esistance 89 (c The time for the dim bulb to dissipate 1. J of energy is E 1. J t = = = P. W dim.4 s Electrical potential energy is transformed into internal energy and light (d In. days, the energy used by the dim bulb is ( ( ( 4 E= P t=. W. d 4. h d = Wh = 18. kwh dim and the cost is cos t E rate ( ( = = 18. kwh $.7 kwh = $1.6 The electric company sells energy, and the unit cost is $.7 1 kwh unit cost = 6 kwh =.6 1 J 8 $ Joule From ( P =, the total resistance needed is ( ( V = = = 8. Ω P 48 W Thus, from = ρl A, the length of wire required is A L = = ρ ( 8. Ω ( m 8. 1 m Ω m 1.1 km = = 17. (a The power required by the iron is ( V I ( ( = = = 1 V 6. A 7. 1 W and the energy transformed in minutes is J 6 s E= t= ( = s 1 min 7. 1 min J

22 9 CHAPTE 17 (b The cost of operating the iron for minutes is cost = E rate 1 kwh = ( J 6 ( $.8 kwh = $.19 = 1.9 cents.6 1 J 17.1 Ohm s law gives the resistance as ( V I ( A L =. From = ρl A, the resistivity is given by ρ =. The results of these calculations for each of the three wires are summarized in the table below. L ( m ( Ω ρ ( Ω m The average value found for the resistivity is ρ Σ ρ i 6 av = = Ω m which differs from the value of by.%. 8 ρ = 1 1 Ω m = Ω m given in Table The resistance of the 4. cm length of wire between the feet is 8 ( Ω ( π (.11 m ρl m.4 m = = = A 6 Ω, so the potential difference is ( ( V = I = Ω = = µ 6 A V 89 V

23 Current and esistance (a The total power you now use while cooking breakfast is P = ( 1 + W = 1.7 kw The cost to use this power for. h each day for. days is h cost = P ( t rate = ( 1.7 kw. (. days ( $.1 kwh = $.6 day (b If you upgraded, the new power requirement would be: P = ( 4 + W= 9 W and the required current would be 9 W I = P 6.4 A A = 11 V = > No, your present circuit breaker cannot handle the upgrade (a The charge passing through the conductor in the interval t. s is represented by the area under the I vs t graph given in Figure P17.4. This area consists of two rectangles and two triangles. Thus, Q= A + A + A + A rectangle 1 rectangle triangle 1 triangle Current (A 6 (. s (. A ( 4. s. s( 6. A. A = (. s. s( 6. A. A (. s 4. s( 6. A. A 1 4 Time(s Q = 18 C (b The constant current that would pass the same charge in. s is Q 18 C I = = = t. s.6 A

24 9 CHAPTE (a From = ( I, the current is P I = = = 1. V (b The time before the stored energy is depleted is 8. 1 W 667 A E storage 7. 1 J t = = = P 8. 1 J s. 1 s Thus, the distance traveled is ( ( d= v t= = = 4. m s. 1 s. 1 m. km 17.6 The volume of aluminum available is 11 1 kg mass V = = = m density.7 1 kg m (a For a cylinder whose height equals the diameter, the volume is π d π d V = d= 4 4 and the diameter is ( 1 1 4V m d = = =.7 8 m π π The resistance between ends is then ρl A (b For a cube, V ρd ( π d 4 4 ρ 8 ( Ω π(.7 8 m m = = = = = Ω π d = L, so the length of an edge is 1 ( V ( 4. 1 L= = 6 1 m =.4 9 m The resistance between opposite faces is Ω m ρl ρl ρ = = = = = Ω A L L.4 9 m

25 Current and esistance The current in the wire is 1. V I = = = 1 A.1 Ω Then, from vd d = I nqa, the density of free electrons is I n = = ve r ( π 1 A (.17 1 m s( C π (. 1 m 4 19 or 8 n =.77 1 m 17.8 At temperature T, the resistance of the carbon wire is c = c 1 + αc( T T of the nichrome wire is = + ( T T end, the total resistance is n n 1 αn ( ( α α ( = + = T c n c n c c n n, and that. When the wires are connected end to If this is to have a constant value of 1. kω as the temperature changes, it is necessary that T and + = 1. kω c n cαc + nα n = (1 ( From equation (1, = 1. kω c n, and substituting into equation ( gives ( Ω ( ( k n. 1 C + n.4 1 C = Solving this equation gives n =.6 k Ω nichro ( me wire ( Then, = 1. kω.6 kω= 4.4 k Ω carbon wire c 17.9 (a From = (, the resistance is ( ( 1 V = = = 1 W 144 Ω (b Solving ρl = for the length gives A ( ( A 144 Ω.1 mm 1 m L = = = ρ.6 1 m 1 mm -8 6 Ω 6 m

26 94 CHAPTE 17 (c The filament is tightly coiled to fit the required length into a small space (d From L = L 1 + α ( TT, where 6 α 4. 1 ( C 1 =, the length at T = C is L = L 1 = 6 m C = α ( T T 6 ( ( C 1 ( 6 C m 17.6 Each speaker has a resistance of = 4. Ω and can handle 6. W of power. From P = I, the maximum safe current is I max = = 6. W 4. Ω =.87 A Thus, the system is not adequately protected by a 4. A fuse The cross-sectional area of the conducting material is A= π ( router rinner Thus, (. 1 Ω m( 4. 1 m ( 1. 1 m (. 1 m ρl 7 = = =.7 1 Ω= 7 MΩ A π 17.6 (a V I = V I 1. V. 1 A 1. V. 1 A. V.1 1 A. 1 Ω. 1 Ω. 1 Ω +.4 V +.1 A 4 Ω +. V +. A Ω +. V +.4 A 14 Ω +.7 V +.7 A 9.7 Ω +.7 V +.1 A 7. Ω (b The resistance of the diode is very large when the applied potential difference has one polarity and is rather small when the potential difference has the opposite polarity.

27 Current and esistance The power the beam delivers to the target is 6 ( V I ( ( P = = 4. 1 V 1 A = 1. 1 W The mass of cooling water that must flow through the tube each second if the rise in the water temperature is not to exceed C is found from P = ( m t c( T as 1. 1 J s m = = = t c T ( ( J kg C( C.48 kg s The volume of the material is mass. g 1 m V = = m density 6 = 7.86 g cm 1 cm 6 Since V = A L, the cross-sectional area of the wire is A = VL (a From ρl ρl ρl = = =, the length of the wire is given by A VL V ( 1. Ω ( m V L = = = ρ Ω m 9. m (b The cross-sectional area of the wire is π d V A = =. Thus, the diameter is 4 L 6 ( 4V m d = = = = π L π( 9. m m.9 mm 17.6 (a The cross-sectional area of the copper in the hollow tube is ( ( ( ( A = circumference thickness = = Thus, the resistance of this tube is ρl A ( Ω m(.4 m = = =.6 1 Ω m 4.8 m. 1 m m

28 96 CHAPTE 17 (b The mass may be written as m= ( density Volume= ( density A L From = ρl A, the cross-sectional area is A = ρl, so the expression for the mass becomes 8 ( 1 Ω m( ρ L m m= ( density = ( 8 9 kg m = 76 kg 4. Ω (a At temperature T, the resistance is ρl A =, where ρ = ρ 1 + α( TT, 1 α (, and A= A ( 1+ α TT A 1 α ( T T L= L + TT Thus, ( + α ( + α ( T T + ( T T α ( 1+ α ( TT ρ 1 α 1 T T 1+ α 1+ L T T = = T T + A 1 (b ρ L = = 8 ( Ω ( - π ( m. m = 1.8 Ω A Then = 1 + α TT ( gives ( ( ( = 1.8 Ω 1 C 8. C = 1.4 Ω The more complex formula gives 6 ( 1.4 Ω 1 + ( 17 1 C( 8. C ( 17 1 C( 8. C = = Ω Note: Some rules for handing significant figures have been deliberately violated in this solution in order to illustrate the very small difference in the results obtained with these two expressions.

29 Current and esistance Note that all potential differences in this solution have a value of V = 1 V. First, we shall do a symbolic solution for many parts of the problem, and then enter the specified numeric values for the cases of interest. From the marked specifications on the cleaner, its internal resistance (assumed constant is ( where P 1 i = = W P 1 Equation (1 If each of the two conductors in the extension cord has resistance c, the total resistance in the path of the current (outside of the power source is = + t i c so the current which will exist is is Equation ( I = V t and the power that is delivered to the cleaner P delivered ( V ( V 4 = Ii= i = = t t P P 1 t 1 Equation ( The resistance of a copper conductor of length l and diameter d is l l 4ρCu l c = ρcu = ρcu = A π d 4 π d ( c, max Thus, if is the maximum allowed value of, the minimum acceptable diameter of the conductor is c d min = 4ρCu l π c, max Equation (4 (a If.9, then from Equations ( and (1, = Ω c = + ( t i ( ( 1 V 1.9 Ω = Ω= Ω P W and, from Equation (, the power delivered to the cleaner is P delivered ( 1 V = = ( 1 V Ω ( W W 4 47 W

30 98 CHAPTE 17 (b If the minimum acceptable power delivered to the cleaner is ( and ( give the maximum allowable total resistance as P min, then Equations ( ( 4 t, max = i + c, max = = P P P P min 1 min 1 so c, max ( ( ( ( = = = P P P P P P 1 i P min 1 min 1 1 P min 1 1 When P min = W, then c, max = ( 1 V ( W( W W = Ω and, from Equation (4, d min 8 ( Ω ( π (.18 Ω m 1. m = = 1.6 mm When c, max P min = W, then ( 1 V 1 1 = =.7 9 Ω ( W( W W and, d min 8 ( Ω m( 1. m π (.7 9 Ω 4 = =.9 mm