Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Serway CP 7 3 00 0.0 points A.2 cm length of tungsten filament in a small lightbulb has a resistance of 0.02 Ω. Find its diameter. (The resistivity is 5.6 0 8 Ω m). Correct answer: 0.206835 mm. Let : L.2 cm 0.02 m, R 0.02 Ω, and ρ 5.6 0 8 Ω m. Resistance is R ρl A ( ρl ) 2 4ρL d πd 2, π 2 4ρL d πr 4(5.6 0 8 Ω m)(0.02 m) π(0.02 Ω) ( ) 000mm m 0.206835 mm. Serway CP 7 57 002 0.0 points Alengthofmetalwirehasaradiusof0.003m andaresistanceof0.07ω. Whenthepotential difference acrossthe wireis 6 V, the electron drift speed is found to be 0.000293 m/s. Based on these data, calculate the density of free electron in the wire. Correct answer:.72442 0 29 m 3. Let : r 0.003 m, R 0.07 Ω, V 6 V, and v d 0.000293 m/s. The current in the wire is I V R 6 V 0.07 Ω 228.57 A. From v d I, the density of free electron nqa is I n v d q e (πr 2 ) 228.57 A (0.000293 m/s)(.6 0 9 C) π(0.003 m) 2.72442 0 29 m 3. Serway CP 7 04 003 0.0 points In a particular television picture tube, the measured beam current is 93.8 µa. How many electrons strike the tube screen every 9 s? Correct answer:.248 0 6 electrons. Let : I 93.8 µa 9.38 0 5 C/s and t 9 s. The current is Thus n I t q e I q t nq e t. (9.38 0 5 C/s)(9 s).602 0 9 C/electron.248 0 6 electrons.
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 2 Serway CP 7 34 004 (part of 2) 0.0 points A high voltage transmission line with a resistanceof0.75ω/km carriesacurrent of 636A. The line is at a potential of 800 kv at the power stationand carriesthe current toacity located 30 km from the power station. What is the power loss due to resistance in the line? Correct answer: 39.4384 MW. Let : I 636 A, ρ 0.75 Ω/km, and l 30 km. The power loss is P l I 2 R I 2 ρ l (636 A) 2 (0.75 Ω/km)(30 km) MW 0 6 W 39.4384 MW. 005 (part 2 of 2) 0.0 points What fraction of the transmitted power does this loss represent? Correct answer: 7.7525%. Let : V 8 0 5 V and I 636 A. The transmitted power is P T V I (8 0 5 V)(636 A) 5.088 0 8 W, sotheratioofpowerlosstopowertransmitted is P l 3.94384 07 W P T 5.088 0 8 W 00% 7.7525%. Serway CP 7 20 006 0.0 points A certain lightbulb has a tungsten filament witharesistance of23ωwhen coldand 48Ω when hot. If the equation R R 0 [+α T] can be used over the large temperature range involved here, find the temperature of the filament when itishot. Assumethat α, thetemperature coefficient of resistivity of tungsten, is 0.0045 ( C) and that the temperature of the cold filament is 32 C. Correct answer: 239.73 C. Resistance is Let : R 0 23 Ω, R 48 Ω, T 0 32 C, and α 0.0045 ( C). R R 0 [+ α T] R 0 +R 0 α(t f T 0 ) T f T 0 R R 0 R 0 α T f R R 0 R 0 α +T 0 48 Ω 23 Ω (23 Ω) [0.0045 ( C) ] +32 C 239.73 C. Serway CP 7 64 007 (part of 2) 0.0 points A 66 g sample of a conducting material is all that is available. The resistivity of the material is measured to be 8 0 8 Ω m,
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 3 and the density is 7.35 g/cm 3. The material is to be shaped into a solid cylindrical wire that has a total resistance of.4 Ω. What length is required? Correct answer: 2.5357 m. Let: W 66 g 0.066 kg, ρ 8 0 8 Ω m, n 7.35 g/cm 3 7350 kg/m 3, and R.4 Ω. The volume of the material is V mass density 0.066 kg 7350 kg/m 3 The volume is and the resistance is R ρl A ρl V L L RV ρ 8.97959 0 6 m 3. V AL A V L, ρl2 V (.4 Ω) (8.97959 0 6 m 3 ) 8 0 8 Ω m 2.5357 m. 008 (part 2 of 2) 0.0 points What must be the diameter of the wire? Correct answer: 0.95504 mm. The cross-sectional area of the wire is ( ) 2 d A π V 2 L. Thus, the diameter is 4V d πl 4(8.97959 0 6 m 3 ) π(2.5357 m) 03 mm m 0.95504 mm. Serway CP 7 23 009 0.0 points At 29 C the carbon resistor in an electric circuit, connected to a 4.4 V battery, has a resistance of 8 Ω. What is the current in the circuit when the temperature of the carbon rises to 33 C? Use a temperature coefficient of resistivity of 0.0005( C). Correct answer: 37.3629 ma. Let : T 0 29 C, V 4.4 V, R 0 8 Ω, T 33 C, and α 0.0005( C). At 33 C the resistance has increased to R R 0 (+α T) (8 Ω) { + [ 0.0005( C) ] (4 C) } 7.764 Ω, so the current is I V R 4.4 V 7.764 Ω 03 ma A 37.3629 ma. Serway CP 7 54 00 (part of 2) 0.0 points
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 4 Thecurrent inaconductor variesovertimeas shown in the figure. Current(A) 6 6 5 5 4 4 3 3 2 2 0 0 0 2 3 4 5 6 7 0 2 3 4 5 6 7 Time(s) How much charge passes through a cross section of the conductor in the time interval t 0 s to t 4 s? Correct answer: 5 C. Let : t 4 s. The charge is the area under the I vs t graph, represented here by two rectangles and two triangles: Q (2 A)(4 s)+(6 A 2 A)( s) + (6 A 2 A)( s) 2 + (6 A 2 A)(0.5 s) 2 5 C. 0 (part 2 of 2) 0.0 points What constant current would transport the same total charge during the 4 s interval as does the actual current? Correct answer: 3.75 A. I Q t 5 C 3.75 A. 4 s Serway CP 7 30 02 0.0 points A platinum resistance thermometer has a resistance of 200 Ω when placed in an ice 0 C bathand07.6ωwhenimmersedinacrucible containing a melting substance. What is the melting point of the substance? The temperature coefficient of the platinum is 0.00392( C). The resistor was calibrated at a temperature of 20 C. Correct answer: 08.67 C. and Let : T 0 20 C, R 200 Ω, T 0 C, R 07.6 Ω, and α 0.00392( C). The resistance at 20 C is R R 0 +α(t T 0 ) 200 Ω +[0.00392( C) ] (0 C 20 C) 27.04 Ω, R R 0 [+α(t T 0 )] R 0 +R 0 α(t T 0 ) T T 0 R R 0 αr 0 T T 0 + R R 0 αr 0 T 20 C 07.6 Ω 27.04 Ω + [0.00392( C) ](27.04 Ω) 08.67 C. Serway CP 7 46 03 0.0 points An electric resistance heater is to deliver 70 kcal/h to a room using 2 V electricity. If fuses come in 0 A, 20 A, and 30 A sizes, whatisthesmallestcurrent thatcansafelybe used in the heater circuit? Correct answer:.2434 A.
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 5 Let : P 70 kcal/h and V 2 V. The current is I P V 70 kcal/h 2 V.2434 A. The 20 amp fuse can be used. 486 J kcal h 3600 s Serway CP 7 4 04 0.0 points We estimate that there are 262 million plug-in electric clocks in the United States, approximately one clock for each person. The clocks convert energy at the average rate of.8 W. To supply this energy, how many metric tons of coal are burned per hour in coal-fired electric-generating plants that are, on average, 8.5% efficient? The heat of combustion for coal is 33 MJ/kg. Correct answer: 278.093 metric ton. Let : R.8 W/clock, L 33 MJ/kg 3.3 0 7 J/kg, N 2.62 0 8 clocks, and t h. The total energy to run the clocks for one hour is E tot P t (.8 W/clock)(2.62 0 8 clocks) ( ) 3600 s ( h) h.69776 0 2 J. The plant operates at 8.5% efficiency, so E tot 0.85E in 0.85(mL), m E tot 0.85 L.69776 0 2 J 0.85(3.3 0 7 J/kg) ( ) metric ton 000 kg 278.093 metric ton. Serway CP 8 54 05 0.0 points 3.3 0 6 2.4 µf Ω 0 V An emf of 0 V is connected to a series RC circuit consisting of a 3.3 0 6 Ω resistor and a 2.4 µf capacitor. Find the time required for the charge on the capacitor to reach 86% of its final value. Correct answer: 5.576 s. Let : R 3.3 0 6 Ω C 2.4 µf 2.4 0 6 F E 0 V, and p 0.86. From Q Q max ( e t/τ ), where τ RC, we get p Q Q max e t/τ e t/τ p t τ ln( p) t τ ln( p) RCln( p) ( 3.3 0 6 Ω )( 2.4 0 6 F ) ln( 0.86) 5.576 s. S
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 6 keywords: Serway CP 8 27 06 (part of 3) 0.0 points 45.7 V 22.85 V 39.4 Ω 5.53 Ω 5 Ω Find the current through the 5 Ω (lower) resistor. Correct answer:.29286 A. E r E 2 I r 2 I 2 R I 3 Let : E 45.7 V, E 22.85 V, r 39.4 Ω, r 2 5.53 Ω, and R 5 Ω. Assuming currents I, I 2, and I 3 in the direction show, we get I 3 I +I 2. Applying Kirchhoff s loop rule, we can get two equations. E I r +I 3 R () E 2 I 2 r 2 +I 3 R (I 3 I )r 2 +I 3 R I r 2 +I 3 (R+r 2 ), (2) Multiplying Eq. () by r 2, Eq. (2) by r, E r 2 I r r 2 +r 2 I 3 R E 2 r I r r 2 +I 3 r (R+r 2 ) Adding, E r 2 +E 2 r I 3 [r 2 R+r (R+r 2 )] E r 2 +E 2 r I 3 r 2 R+r (R+r 2 ) (45.7 V)(5.53 Ω)+(22.85 V)(39.4 Ω) (5.53 Ω)(5 Ω)+(39.4 Ω)(5 Ω+5.53 Ω).29286 A. 07 (part 2 of 3) 0.0 points Determine the current in the 39.4 Ω (upper) resistor. Correct answer: 0.667694 A. From (), get I E I 3 R r 45.7 V (.29286 A)(5 Ω) 39.4 Ω 0.667694 A. 08 (part 3 of 3) 0.0 points Determine the current in the 5.53 Ω (middle) resistor. Correct answer: 0.62563 A. From (2), get I 2 E 2 I 3 R r 2 22.85 V (.29286 A)(5 Ω) 5.53 Ω 0.62563 A.
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 7 Serway CP 8 22 09 (part of 2) 0.0 points Consider the following circuit 34 V.8 Ω 7.4 Ω 6.4 Ω What is the current through the lower resistor? Correct answer: 4.3324 A. I I 2 R R 2 R 3 and R 2 are in series, so R 23 R 2 +R 3.44783Ω+6.4Ω 7.84783Ω. The current I 3 in the circuit is I 3 E R 23 34 V 7.84783 Ω 4.3324 A. Alternate Solution: From Kirchhoff s junction rule, I 3 I +I 2. Applying Kirchhoff s loop rule, we obtain two equations: E I 3 R 3 +I R () E I 3 R 3 +I 2 R 2 I 3 R 3 +(I 3 I )R 2 (2) Multiplying Eq. () by R 2 and Eq. (2) by R, we obtain E R 2 I 3 R 3 R 2 +I R R 2 () E R I 3 R 3 R +(I 3 I )R 2 R (2) Adding, E R 2 +E R I 3 (R 3 R 2 +R 3 R +R 2 R ). Since R 3 R 2 +R 3 R +R 2 R I 3 E R 3 (6.4 Ω)(7.4 Ω)+(6.4 Ω)(.8 Ω)+(7.4 Ω)(.8 Ω 72.2 Ω, then Let : R.8 Ω, R 2 7.4 Ω, R 3 6.4 Ω, and E 34 V. R and R 2 are in parallel, so ( R 2 + ) R R 2 (.8 Ω + 7.4 Ω.44783 Ω. I 3 E R 2 R 3 ) E R 2 +E R I 3 R 3 R 2 +R 3 R +R 2 R (34 V)(7.4 Ω)+(34 V)(.8 Ω) 72.2 Ω 4.3324 A. 020 (part 2 of 2) 0.0 points What is the current through the middle (7.4 Ω) resistor? Correct answer: 0.847645 A. For resistors R and R 2, V V 2 I 3 R 2. I 2 V 2 R 2 I 3R 2 R 2 (4.3324 A)(.44783 Ω) 7.4 Ω 0.847645 A.
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 8 Serway CP 8 03 02 0.0 points A lightbulb marked 75 W [at] 07 V is screwed into a socket at one end of a long extension cord in which each of the two conductorshasaresistance of0.877ω. Theother end of the extension cord is plugged into a 07 V outlet. Find the actual power of the bulb in this circuit. Correct answer: 73.3057 W. V r r R bulb Let : V 07 V, P 75 W, and r 0.877 Ω. The total resistance is R eq R bulb +2r For the bulb P ( V)2 R bulb R bulb ( V)2 P (07 V)2 75 W 52.653 Ω. Now, presuming the bulb resistance is unchanged, the current in the circuit shown is I V R eq V R bulb +2r 07 V 52.653 Ω+2(0.877 Ω) 0.692972 A and the actual power dissipated in the bulb is P actual I 2 R bulb (0.692972 A) 2 (52.653 Ω) 73.3057 W. Serway CP 8 44 022 0.0 points Consider an RC circuit in which a capacitor is being charged by a battery connected in a circuit. Initially the capacitor is uncharged. Then the switch is closed. After a time equal to 4.4 time constants, what percent of the final charge is present on the capacitor? Correct answer: 98.7723%. From Q Q max ( e t/τ), when t 4.4τ we have Q Q max e 4.4τ/τ e 4.4 98.7723%. Serway CP 8 49 023 (part of 2) 0.0 points An automobile battery has an emf of 0.5 V and an internal resistance of 0.075 Ω. The headlights have total constant resistance 7.97 Ω. What is the potential difference across the headlight bulbs when they are the only load on the battery? Correct answer: 0.402 V. Let : E 0.5 V, R i 0.075 Ω, and R h 7.97 Ω.
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 9 With the starter off, E I h R i +R h 0.5 V 0.075 Ω+7.97 Ω.3056 A Thus the voltage across the headlight bulbs is V I h R h (.3056 A)(7.97 Ω) 0.402 V. 024 (part 2 of 2) 0.0 points What is the potential difference across the headlight bulbs when the starter motor is operated, requiring an additional 6.5 A from the battery? Correct answer: 9.765 V. R b Let : I s 6.5 A. R h R s The starter requires additional current, so I b I h +I s. Considering the closed loop containing the battery and the headlights, E R i I b R h I h 0 E R i (I h +I s ) R h I h 0 E R i I s (R i +R h )I h I h E R ii s R i +R h Therefore, the voltage across the headlights with the starter on is V I h R h (E R ii s )R h R h +R i [0.5 V (0.075 Ω)(6.5 A)] 7.97 Ω 7.97 Ω+0.075 Ω 9.765 V. Serway CP 8 38 025 (part of 4) 0.0 points A 46 Ω lamp, a 28 Ω electric heater, and a 57 Ω fan are connected in parallel across a 26 V line. What total current is supplied to the circuit? Correct answer: 7.57354 A. Let : R 46 Ω, R 2 28 Ω, R 3 57 Ω, and V 26 V. For the parallel circuit, so and R R + R 2 + R 3 R R 2 +R 2 R 3 +R 3 R R R 2 R 3 R R 2 R 3 R R R 2 +R 2 R 3 +R 3 R R R 2 + R 2 R 3 + R 3 R (46 Ω)(28 Ω) + (28 Ω)(57 Ω) + (57 Ω)(46 Ω) R 4006 Ω 2, (46 Ω)(28 Ω)(57 Ω) 4006 Ω 2 6.6369 Ω I V R 26 V 6.6369 Ω 7.57354 A.
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 0 026 (part 2 of 4) 0.0 points What is the voltage across the fan? Correct answer: 26 V. Because the circuit elements are connected in parallel, the voltage across the fan is the same as that across the whole circuit. 027 (part 3 of 4) 0.0 points What is the current in the lamp? Correct answer: 0.86304 A. The voltage is the same in all 3 appliances, so the current through the lamp is I V R 26 V 46 Ω 0.86304 A. 028 (part 4 of 4) 0.0 points What power is expended in the heater? Correct answer: 567 W. The power consumed by the heater is P 2 V 2 R 2 (26 V)2 28 Ω 567 W. Serway CP 8 57 029 0.0 points A voltage V is applied to a series configuration of 76 resistors, each of value R. The circuit components are reconnected in a parallel configuration, and voltage V is again applied. Calculate the ratio of the series power to the parallel power. Correct answer: 0.000733. Let : R R 2 R 3 R 4 R. When connected in series, the equivalent resistance is R s R +R 2 +...+R n nr. Thus the power consumed is P s ( V)2 R s ( V)2 nr. When connected in parallel, the equivalent resistance is + +...+ n R p R R 2 R n R R p R n Thus the power consumed is and the ratio is P p ( V)2 R p n( V)2. R P s ( V)2 P p nr R n( V) 2 n 2 (76) 2 0.000733. Serway CP 8 08 030 (part of 2) 0.0 points The power supplied to the circuit is.08 W. 0.8 Ω 5.8 Ω 2.27 Ω 2.74 Ω 3.83 Ω Find the equivalent resistance of the circuit. Correct answer: 4.28447 Ω. E
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) Let : R 2.27 Ω, R 2 0.8 Ω, R 3 5.8 Ω, R 4 3.83 Ω, and R 5 2.74 Ω. R 2 R 3 R 4 R R 5 E Correct answer: 6.88999 V. P IV V 2 Let : P.08 W. R eq V PR eq (.08 W)(4.28447 Ω) 6.88999 V. Serway CP 8 33 032 0.0 points Consider a series RC circuit. The capacitor is initially uncharged when the switch is open. R 2 and R 3 are in parallel, so R 23 R 2 + R 3 R 3 +R 2 R 2 R 3 R 23 R 2R 3 R 3 +R 2. R 23 and R 4 are in series, so. MΩ 32 V 4.8 µf S R 234 R 23 +R 4 R 2R 3 +R 4 R 3 +R 2 (0.8 Ω)(5.8 Ω) +3.83 Ω 7.60772 Ω. 5.8 Ω+0.8 Ω R 234 is parallel with R 5, so R 2345 R 234 + R 5 R 5 +R 234 R 234 R 5 R 2345 R 234R 5 R 5 +R 234 (7.60772 Ω)(2.74 Ω) 2.0447 Ω. 2.74 Ω+7.60772 Ω R and R 2345 are in series, so R eq R +R 2345 2.27 Ω+2.0447 Ω 4.28447 Ω. 03 (part 2 of 2) 0.0 points Find the emf across the battery. Find the charge on the capacitor s after the switch is closed. Correct answer: 0.00034475 C. Let : R. MΩ. 0 6 Ω, C 4.8 µf 4.8 0 6 F, E 32 V, and T s. When the switch is closed, the charge on the capacitor will rise exponentially: q Q( e t/τ ) CE [ e t/(rc)] (4.8 0 6 F)(32 V) [ e ] ( s)/(. 06 Ω)/(4.8 0 6 F) 0.00034475 C.
Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) 2 Serway CP 8 6 033 (part of 3) 0.0 points The ammeter in the figure reads.9 A. a Find I. 4.27 Ω I.43 Ω I 3 6.56 Ω.9 A Correct answer:.57752 A. 9.2 V A E 2 Let : R 4.27 Ω, R 2 6.56 Ω, R 3.43 Ω, I.9 A, and E 9.2 V. b Correct answer: 0.322482 A. Applying Kirchhoff s junction rule at point a, I +I 3 I I 3 I I.9 A.57752 A 0.322482 A. 035 (part 3 of 3) 0.0 points Find E 2. Correct answer: 2.925 V. Applying Kirchhoff s loop rule clockwise around the lower loop, E 2 R 3 I 3 R 2 I 0 E 2 R 3 I 3 +R 2 I (.43 Ω)(0.322482 A)+(6.56 Ω)(.9 A) 2.925 V. R E I a R 2 I A b R 3 E 2 I 3 Applying Kirchhoff s loop rule clockwise around the upper loop, I E R 2 I R E R I R 2 I 0 9.2 V (6.56 Ω)(.9 A) 4.27 Ω.57752 A. 034 (part 2 of 3) 0.0 points Find I 3.