PROBLEMAS RESUELTOS (45) DEL CAPÍTULO II DE LABORATORIO DE FÍSICA II - TIPPENS

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Estos son los problemas resueltos (45) del Capítulo II de Laboratorio de Física II tomados del Tippens

Transcript of PROBLEMAS RESUELTOS (45) DEL CAPÍTULO II DE LABORATORIO DE FÍSICA II - TIPPENS

  • 1.Chapter 28. Direct-Current Circuits Physics, 6th EditionChapter 28. Direct-Current CircuitsResistors in Series and Parallel (Ignore internal resistances for batteries in this section.)28-1. A 5- resistor is connected in series with a 3- resistor and a 16-V battery. What is the effective resistance and what is the current in the circuit?3 5Re = R1 + R2 = 3 +5 ; Re = 8.00 V 16 V I== I = 2.00 AR 8 16 V 28-2. A 15- resistor is connected in parallel with a 30- resistor and a 30-V source of emf. What is the effective resistance and what total current is delivered?R1 R2 (15 )(30 ) Re ==; Re = 10.0 R1 + R2 15 + 30 15 30 30 VV 30 VI== ; I = 3.00 A R 10 28-3. In Problem 28-2, what is the current in 15 and 30- resistors? 30 V For Parallel: V15 = V30 = 30 V;I15 =;I15 = 2.00 A15 30 V I 30 =;I30 = 1.00 A Note: I15 + I30 = IT = 3.00 A30 28-4. What is the equivalent resistance of 2, 4, and 6- resistors connected in parallel? 24 6 Re = 2 + 4 + 6 ;Re = 12.0 129

2. Chapter 28. Direct-Current CircuitsPhysics, 6th Edition 28-5. An 18- resistor and a 9- resistor are first connected in parallel and then in series with a 24-V battery. What is the effective resistance for each connection? Neglecting internal resistance, what is the total current delivered by the battery in each case? R1 R2 (18 )(9 )30 VRe == ;Re = 6.00 R1 + R2 18 + 9 918 V 24 V I= =;I = 4.00 AR 6.00 Re = R1 + R2 = 18 +9 ;Re = 27.0 18 9V 24 VI== I = 0.889 A 24 V R 27 28-6. A 12- resistor and an 8- resistor are first connected in parallel and then in series with a 28-V source of emf. What is the effective resistance and total current in each case?R1 R2 (12 )(8 ) Re == ; Re = 4.80 12 8R1 + R2 12 + 8 28 V8 12 28 V V 28 VI= =; I = 5.83 A R 4.80 V 28 VRe = R1 + R2 = 12 +8 ;Re = 20.0 I== I = 1.40 A R 20 28-7. An 8- resistor and a 3- resistor are first connected in parallel and then in series with a 12-V source. Find the effective resistance and total current for each connection?(3 )(8 )V 12 VRe =;Re = 2.18 I= =;I = 5.50 A 3+8 R 2.18 V 12 V Re = R1 + R2 = 3 +8 ; Re = 11.0 I== I = 1.09 AR 11 130 3. Chapter 28. Direct-Current Circuits Physics, 6th Edition 28-8. Given three resistors of 80, 60, and 40 , find their effective resistance when connected in series and when connected in parallel.Series:Re = 80 + 60 + 40 ;Re = 180 11111Parallel: = = ++ ; Re = 18.5 Re Ri 80 60 40 28-9. Three resistances of 4, 9, and 11 are connected first in series and then in parallel. Find the effective resistance for each connection. Series:Re = 4 + 9 + 11 ; Re = 24.0 11 1 11 Parallel:= =+ + ;Re = 2.21 Re Ri 4 9 11 *28-10. A 9- resistor is connected in series with two parallel resistors of 6 and 12 . What isthe terminal potential difference if the total current from the battery is 4 A?(6 )(12 )9Re = = 4 ;Re = 4 + 9 = 13 6 6 + 12 12 VT = IR = (4 A)(13 ); VT = 52.0 V VT 4A *28-11. For the circuit described in Problem 28-10, what is the voltage across the 9- resistorand what is the current through the 6- resistor?V9 = (4 A)(9 ) = 36 V; V9 = 36.0 V The rest of the 52 V drops across each of the parallel resistors: V6 = V7 = 52 V 36 V; V6 = 16 VV6 16 VI6 = = ; I6 = 2.67 A R6 6 131 4. Chapter 28. Direct-Current Circuits Physics, 6th Edition *28-12. Find the equivalent resistance of the circuit drawn in Fig. 28-19. Start at far right and reduce circuit in steps: R = 1 + 3 + 2 = 6 ; (6 )(3 ) R '' == 2 ; Re = 2 + 4 + 2 ; Re = 8 6+34144 3 336 2822 2 2 *28-13. Find the equivalent resistance of the circuit shown in Fig. 28-20.Start at far right and reduce circuit in steps: R = 1 + 2 = 3 ; (6 )(3 ) R' == 2 ; R = 2 + 3 = 5 6+3 (5 )(4 ) Re == 2.22 ; Re = 2.22 5+41 444 63 4 5Re6 2 32 3 3 *28-14. If a potential difference of 24 V is applied to the circuit drawn in Fig. 28-19, what is the414 current and voltage across the 1- resistor? 24 V24 V24 V 3 336 Re = 8.00 ; I == 3.00 A;8222 The voltage across the 3 and 6- parallel 4connection is found from It and the 2- combination resistance:24 V 26V V3 = V6 = (2 )(3.00 A);V6 = 6.00 V;I6 == 1A6 2Thus, I1 = I6 = 1.00 A, and V1 = (1 A)(1 ) = 1 V;V1 = 1 V; I1 = 1 A 132 5. Chapter 28. Direct-Current CircuitsPhysics, 6th Edition *28-15. If a potential difference of 12-V is applied to the free ends in Fig. 28-20, what is thecurrent and voltage across the 2- resistor? 112 V12 V4 Re = 2.22 ; I == 5.40 A;62.22 3 2 12 V Note that V5 = 12 V; I5 == 2.40 A5 12 V4 4.8 V36 V3,6 = (2.4 A)(2 ) = 4.80 V;I3 = = 1.6 A33 4 5 I2 = I1 = 1.60 A; V2 = (1.6 A)(2 ) = 3.20 V 12 V 412 V I2 = 1.60 A; V2 = 3.20 V23EMF and Terminal Potential Difference 28-16. A load resistance of 8 is connected in series with a 18-V battery whose internalresistance is 1.0 . What current is delivered and what is the terminal voltage?E 18 V I= =; I = 2.00 Ar + RL 1.0 + 8 28-17. A resistance of 6 is placed across a 12-V battery whose internal resistance is 0.3 .What is the current delivered to the circuit? What is the terminal potential difference?E12 V I= =; I = 1.90 Ar + RL 0.3 + 6 VT = E Ir = 12 V (1.90 A)(0.3 );VT = 11.4 V133 6. Chapter 28. Direct-Current Circuits Physics, 6th Edition 28-18. Two resistors of 7 and 14 are connected in parallel with a 16-V battery whose internalresistance is 0.25 . What is the terminal potential difference and the current indelivered to the circuit?16 V 7 14 (7 )(14 ) 0.25 R' = = 4.67 ;Re = 0.25 + 4.67 7 + 14 E16 VI==; I = 3.25 A VT = E Ir = 16 V (3.25 A)(0.25 ); r + R ' 4.917 VT = 15.2 V; I = 3.25 A 28-19. The open-circuit potential difference of a battery is 6 V. The current delivered to a 4-resistor is 1.40 A. What is the internal resistance?E = IRL + Ir;Ir = E - IRL E IRL 6 V - (1.40 A)(4 ) r==;r = 0.286 I 1.40 A 28-20. A dc motor draws 20 A from a 120-V dc line. If the internal resistance is 0.2 , what isthe terminal voltage of the motor? VT = E Ir = 120 V (20A)(0.2 );VT = 116 V 28-21. For the motor in Problem 28-21, what is the electric power drawn from the line? Whatportion of this power is dissipated because of heat losses? What power is delivered bythe motor?Pi = EI = (120 V)(20 A); Pi = 2400 W PL = I2r = (20 A)2(0.2); PL = 80 W Po = VTI = (116 V)(20 A); Po = 2320 W;Note: Pi = PL + Po; 2400 W = 80 W + 2320 W134 7. Chapter 28. Direct-Current Circuits Physics, 6th Edition 28-22. A 2- and a 6- resistor are connected in series with a 24-V battery of internalresistance 0.5 . What is the terminal voltage and the power lost to internal resistance? E 24 VRe = 2 + 6 + 0.5 = 8.50 ; I= == 2.82 ARe 8.5 VT = E Ir = 24 V (2.82 A)(0.5 ); VT = 22.6 VPL = I2 r = (2.82 A)2 (0.5 );PL = 3.99 W*28-23. Determine the total current and the current through each resistor for Fig. 28-21 whenE = 24 V, R1 = 6 , R2 = 3 , R3 = 1 , R4 = 2 ,and r = 0.4 .R3 = 1 24 V(3 )(6 ) R1,2 == 2 ; R1,2,3 = 2 + 1 = 3 R1R2 2 3 + 6 6 R430.4 (3 )(2 ) R3 = 1 Re == 1.20 ;24 V 24 V 3 + 2 32 2 2 R4 R4 0.4 0.4 Re = 1.20 + 0.4 = 1.60 24 V24 V24 V IT =;IT = 15.0 A1.60 1.2 1.6 0.4 18 V V4 = V1.2= (1.2 )(15 A) = 18 VI4 =2I4 = 9.0 A;I3 = 15 A 9 A = 6 A; V3 = (6 A)(1 ) = 6 V; V1 = V2 = 18 V 6 V;12 V 12 V V1 = V2 = 12 V;I2 == 4 A; I1 ==2A;3 6IT = 15 A, I1 = 2 A, I2 = 4 , I3 = 6 , I4 = 9 A.The solution is easier using Kirchhoffs laws, developed later in this chapter.135 8. Chapter 28. Direct-Current Circuits Physics, 6th Edition *28-24. Find the total current and the current through each resistor for Fig. 28-21 when E = 50 V,R1 = 12 , R2 = 6 , R3 = 6 , R4 = 8 ,and r = 0.4 . R3 = 6 50 V (12 )(6 )R1 R2 8R1,2 = = 4 ; R1,2,3 = 4 + 6 = 10 12 R412 + 6 6 0.4 R3 = 6 (10 )(8 ) 50 V10 50 V R= = 4.44 10 + 8 484R4R4 0.4 0.4 Re = 4.44 + 0.4 = 4.84 24 V50 V 50 V1.6 IT =;IT = 10.3 A2.86 4.84 0.4 45.9 V V4 = Vp= (4.44 )(10.3 A) = 45.9 VI4 = 8I4 = 5.73 A; I3 = 10.3 A 5.73 A = 4.59 A; V3 = (4.59 A)(6 ) = 27.5 V; 18.4 V 18.4 V V1 = V2 = 45.9 V 27.5 V = 18.4 V; I2 == 3.06 A; I1 = = 1.53 A ; 6 12 IT = 10.3 A, I1 = 1.53 A, I2 = 3.06 , I3 = 4.59 , I4 = 5.73 A.Kirchhoffs Laws28-25. Apply Kirchhoffs second rule to the current loop in Fig. 28-22. What is the net voltagearound the loop? What is the net IR drop? What is the current in the loop?Indicate output directions of emfs, assume direction of20 V+2current, and trace in a clockwise direction for loop rule:IE = IR; 20 V 4 V = I(6 ) + I(2 ); 64V16 V(8 )I = 16 V; I=;I = 2.00 A 8 Net voltage drop = E = 16 V;IR = (8 )(2 A) = 16 V136 9. Chapter 28. Direct-Current CircuitsPhysics, 6th Edition 28-26. Answer the same questions for Problem 28-25 where the polarity of the 20-V battery ischanged, that is, its output direction is now to the left? ( Refer to Fig. in Prob. 28-25. ) E = -20 V 4 V = -24 V; IR = I(2 ) + I(6 = (8)IE = IR; -24 V = (8 )I; I = -4.00 A; IR = (8 )(-4 A) = -24 VThe minus sign means the current is counterclockwise (against the assume direction) *28-27. Use Kirchhoffs laws to solve for the currents through the circuit shown as Fig. 28-23.4 First law at point P: I1 + I2 = I3 Current rule I1 nd Loop A (2 law): E = IRLoop rule2 P A 5V 6I 5 V 4 V = (4 )I1 + (2 )I1 (6 )I22 4V I2Simplifying we obtain:(1) 6I1 6I2 = 1 A3 BI3 1 Loop B: 4 V 3 V = (6 )I2 + (3 )I3 + (1 )I3 Simplifying: (2) 6I2 + 4I3 = 1 A, but I3 = I1 + I2 3VSubstituting we have: 6I2 + 4(I1 + I2) = 1 Aor (3) 4I1 + 10I2 = 1 AFrom which,I1 = 0.25 A 2.5 I2 ; Substituting into (1): 6(0.25 A 2.5I2 ) 6 I2 = 1 A1.5 A 15I2 6I2 = 1 A; -21I2 = -0.5 A; I2 = 0.00238 A;I2 = 23.8 mAPutting this into (1), we have: 6I1 6(0.0238 A) = 1 A, and I1 = 190 mA Now, I1 + I2 = I3so thatI3 = 23.8 mA + 190 mAor I3 = 214 mA3 *28-28. Use Kirchhoffs laws to solve for the currents in Fig. 28-24.I1Current rule:I1 + I3 = I2or (1) I3 = I2 I1 20 V A 5V 4 I2 Loop A: 20 V = (3 )I1 + (4 )I2 ; (2) 3I1 + 4I2 = 20 A P4V I2 Loop B: 8 V = (6 )I3 + (4 )I2; (3) 3I3 + 2I2 = 4 A2B4I3Outside Loop: 20 V 8 V = (3 )1 (6 )3 or I1 2 I3 = 4 A 8V 137 10. Chapter 28. Direct-Current Circuits Physics, 6th Edition *28-28. (Cont.) (3) 3I3 + 2I2 = 4 A and I3 = I2 I1 3 3(I2 I1) + 2I2 = 4 A; 3I1 = 5I2 4 A I1(2) 3I1 + 4I2 = 20 A; (5I2 4 A) + 4I2 = 20; 20 V A5V4 I2PI2 = 2.67 A; 3I1 = 5(2.67 A) 4 A; I1 = 3.11 A4V I2I3 = I2 I1 = 2.67 A 3.11 A = -0.444 A 2B 4 I3Note: I3 goes in opposite direction to that assumed.8V I1 = 3.11 A, I2 = 2.67 A, I3 = 0.444 A *28-29. Apply Kirchhoffs laws to the circuit of Fig. 28-25. Find the currents in each branch.Current rule: (1)I1 + I4 = I2 + I3 1.5 3VApplying loop rule gives six possible equations: 3I1 (2) 1.5I1 + 3I2 = 3 A; (3) 3I2 5I3 = 0 I2 5 I3(4) 5I3 + 6I4 = 6A; (5) 1.5I1 6I4 = -3A I4 6V6(6) 6I4 + 3I2 = 6 A (7) 1.5I1 + 5I3 = 3A Put I4 = I2 + I3 I1; into (4): 5I3 + 6(I2 + I3 I1) = 6 A -6I1 + 6I2 + 11I3 = 6 A Now, solving (2) for I1 gives: I1 = 2 A 2I2, which can be used in the above equation. -6(2 A 2I2) + 6I2 + 11I3 = 6 A, which gives: 18I2 + 11I3 = 18 A But, from (3), we put I2 = 5 3 I 3 into above equation to find that:I3 = 0.439 AFrom (2): 1.5I1 + 3(0.439 A) = 3 A; and I1 = 0.536 AFrom (3): 3I2 5(0.439 A) = 0; and I2 = 0.736 AFrom (4): 5(0.439 A) + 6I4 = 6 A; and I4 = 0.634 A Currents in each branch are: I1 = 536 mA, I2 = 732 mA, I3 = 439 mA, I4 = 634 mA Note: Not all of the equations are independent. Elimination of two may yield another. It is best to start with the current rule, and use it to eliminate one of the currents quickly.138 11. Chapter 28. Direct-Current CircuitsPhysics, 6th Edition The Wheatstone Bridge28-30. A Wheatstone bridge is used to measure the resistance Rx of a coil of wire. Theresistance box is adjusted for 6 , and the contact key is positioned at the 45 cm markwhen measured from point A of Fig. 28-13. Find Rx.( Note: l1 + l2 = 100 cm ) R3l2 (6 )(55 cm) Rx = = ; Rx = 7.33 l1 (45 cm)28-31. Commercially available Wheatstone bridges are portable and have a self-containedgalvanometer. The ratio R2/R1 can be set at any integral power of ten between 0.001 and1000 by a single dual switch. When this ratio is set to 100 and the known resistance R isadjusted to 46.7 , the galvanometer current is zero. What is the unknown resistance? R2Rx = R3= (46.7 )(100) ;Rx = 4670 R1 28-32. In a commercial Wheatstone bridge, R1 and R2 have the resistances of 20 and 40 ,respectively. If the resistance Rx is 14 , what must be the known resistance R3 for zerogalvanometer deflection?R220 Rx = R3= (14 ); Rx = 7.00 R140 Challenge Problems:28-33. Resistances of 3, 6, and 9 are first connected in series and then in parallel with an 36-Vsource of potential difference. Neglecting internal resistance, what is the current leavingthe positive terminal of the battery? 36 VRe = Ri = 3 + 6 + 9 = 18 ; I=; I = 2.00 A18 139 12. Chapter 28. Direct-Current Circuits Physics, 6th Edition11 1 1 136 V 28-33. (Cont.)= =+ +; Re = 1.64 ; I =; I = 22.0 ARe Ri 3 6 9 1.64 28-34. Three 3- resistors are connected in parallel. This combination is then placed in serieswith another 3- resistor. What is the equivalent resistance? 31 1 1 1= + += 1 ; Re = R + 3 R' 3 3 3 V 3 33Re = 1 + 3 ; Re = 4 *28-35. Three resistors of 4, 8, and 12 are connected in series with a battery. A switch allowsthe battery to be connected or disconnected from the circuit? When the switch is open, avoltmeter across the terminals of the battery reads 50 V. When the switch is closed, thevoltmeter reads 48 V. What is the internal resistance in the battery?RL = 4 + 8 + 12 = 24 ; E = 50 V; VT = 48 V = IRL 48 V I== 2.00 A ; E VT = Ir; 50 V 48 V = Ir24 V50 V - 48 Vr= ;r = 1.00 2.00 A*28-36. The generator in Fig. 28-26 develops an emf of E1 = 24 V and has an internal resistanceof 0.2 . The generator is used to charge a battery E2 = 12 V whose internal resistanceis 0.3 . Assume that R1 = 4 and R2 = 6 . What is the terminal voltage across thegenerator? What is the terminal voltage across the battery? 24 V - 12 VI= = 1.14 A 6 + 4 + 0.2 + 0.3 12 V4 V1= E1 Ir = 24 V (1.14 A)(0.2 ) = 23.8 V;24 Vrr 6V2 = 12 V + (1.14 A)(0.3 ) = 12.3 V 140 13. Chapter 28. Direct-Current Circuits Physics, 6th Edition *28-37. What is the power consumed in charging the battery for Problem 28-36. Show that thepower delivered by the generator is equal to the power losses due to resistance and thepower consumed in charging the battery.P = E I = (24 V)(1.143 A)Pe = 27.43 W 12 V 4 PR= I2 Re = (1.143 A)2 (10.5 ); PR = 13.69 W 24 Vrr6 PV = (12 V)(1.143 A) = 13.72 W;Pe = PR + PV; 27.4 W = 13.7 W + 13.7 W*28-38. Assume the following values for the parameters of the circuit illustrated in Fig. 28-8:E1 = 100 V, E2 = 20 V, r1 = 0.3 . r2 = 0.4 , and R = 4 .What are the terminalvoltages V1 and V2? What is the power lost through the 4- resistor? 100 V - 20 V20 VI= = 17.0 A100 V 4 + 0.3 + 0.4 r4r V1= E1 Ir = 100 V (17.0 A)(0.3 ) = 94.9 V;V2 = 20 V + (17.0 A)(0.4 ) = 26.8 V; P = I2 R = (17 A)2(4 ) = 1160 W*28-39. Solve for the currents in each branch for Fig. 28-27.512 VCurrent rule: I1 = I2 + I3 ;Loops: E = IRsI1A (1) 5I1 + 10I2 = 12 A;(2) -10I2 + 20I3 = 6 A10 I2 P (2) -5I2 + 10I3 = 3 A;(3) 5I1 + 20I3 = 18 AI2 4VFrom (1): 5(I2 + I3) + 10 I2 = 12 A 15I2 + 5I3 = 12 A20 BI3Multiplying this equation by 2: -30I2 10I3 = -24 A 6V Now add this to (2): -35I2 + 0 = -21 AandI2 = 0.600 ANow, from (1) and from (2): I1 = 1.20 A andI3 = 0.600 A141 14. Chapter 28. Direct-Current Circuits Physics, 6th Edition *28-40. If the current in the 6- resistor of Fig. 28-28 is 2 A, what is the emf of the battery?Neglect internal resistance. What is the power loss through the 1- resistor? 2A 1 V6 = (2 A)(6 ) = 12 V; V3 = 12 V = I3(3 ) ITE 12 V63I3 == 4 A ; IT = 2 A + 4 A = 6 A; I33 V1 = (1 )(6 A) = 6 V; E = 6 V + 12 V = 18 V; P = IT2 R = (6 A)2(1 ) = 36 W Critical Thinking Problems*28-41. A three-way light bulb uses two resistors, a 50-W filament and a 100-W filament. A three-way switch allows each to be connected in series and provides a third possibility by connecting the two filaments in parallel? Draw a possible arrangement of switches than will accomplish these tasks. Assume that the household voltage is 120 V. What are the resistances of each filament? What is the power of the parallel combination?I1Switch can be set at A, B, or C to give three possibilities:ITV2(120 V) 2 (120 V) 2 R1 R2 P=; R1 = = 288 ; R2 = = 144 I2R 50 W 100 W B 120 V(288 )(144 ) A C For Parallel: Re = = 96 288 + 144 V 2 (120 V) 2 P==; P = 150 W R96 142 15. Chapter 28. Direct-Current CircuitsPhysics, 6th Edition *28-42. The circuit illustrated in Fig. 28-7 consists of a 12-V battery, a 4- resistor, and a switch. When new, the internal resistance of the battery is 0.4 , and a voltmeter is placed across the terminals of the battery. What will be the reading of the voltmeter when the switch is open and when it is closed? After a long period of time, the experiment is repeated and it is noted that open circuit reading is unchanged, but the terminal voltage has reduced by 10 percent. How do you explain the lower terminal voltage? What is the internal resistance of the old battery?E 12 VWhen new: I ==;I = 2.73 A4R + r 4 + 0.4 VT = E Ir = 12 V (2.73 A)(0.4 );VT = 10.9 V VT reduced by 10% due to increase of rint. V = 10.9 V 0.1(10.9 V); V = 9.81 V 9.81 VE V ' V = IRL = 9.81 V;I== 2.45 A;V ' = E Ir ; r= 4 IE V ' 12 V - 9.81 Vr==;r = 0.893 I2.45 A*28-43. Given three resistors of 3, 9, and 18 ,list all the possible equivalent resistances that can be obtained through various connections?1 1 11All in parallel: = + + ; Re = 2 Re 3 9 18 All in series: Re = R1 + R2 + R3 = 3 + 9 + 18 ; Re = 30 (3 )(9 )Parallel (3,9) in series with (18): Re =+ 18 ;Re = 20.2 3+9(3 )(18 )Parallel (3,18) in series with (9): Re = + 9 ;Re = 11.6 3 + 18 (9 )(18 )Parallel (9,18) in series with (3): Re = + 3 ;Re = 9.00 9 + 18 143 16. Chapter 28. Direct-Current Circuits Physics, 6th Edition(12 )(18 ) *28-43 (Cont.)Series (3 + 9) in parallel with (18): Re =; Re = 20.2 12 + 18 (9 )(21 )Series (3 + 18) in parallel with (9): Re =;Re = 6.30 9 + 21 (3 )(27 )Series (9 + 18) in parallel with (3): Re =;Re = 2.70 3 + 27 *28-44. Refer to Fig. 28-21, assume that E = 24 V, R1 = 8 ,R2 = 3 , R3 = 2 ,R4 = 4 ,and r= 0.5 . What current is delivered to the circuit by the 24-V battery? What are thevoltage and current for the 8- resistor?R3 = 2 24 V (3 )(8 ) 4R1,2 == 2.18 ; R1,2,3 = 2.18 + 2 = 4.18 8R23 +8 R4 0.5 R13 R3 = 2 (4.18 )(4 ) Re = = 2.04 ; 24 V 4.18 24 V 4.18 + 4 R4 44 2.18 R40.5 0.5 Re = 2.04 + 0.5 = 2.54 24 V24 V24 V IT = ; IT = 9.43 A 2.54 2.54 2.04 0.5 V4 = V1,2,3 = (9.43 A) (2.04 ) = 19.3 V19.3 VI4 == 4.82 A ; I3 = 9.43 A 4.82 A = 4.61 A; V3 = (4.61 A)(2 ) = 9.23 V4 10.1 VV1 = V2 = 19.3 V 9.23 V;V1 = V2 = 10.1 V;I1 == 1.26 A ; 8Finally, for the 8- resistor: V1 = 10.1 V and I1 = 1.26 A144 17. Chapter 28. Direct-Current CircuitsPhysics, 6th Edition *28-45. What is the effective resistance of the external circuit for Fig. 28-29 if internal resistanceis neglected. What is the current through the 1- resistor?414 141R1R3R524 V 24 V 24 V636 8 36 2.18 8R2R4(3 )(8 )R4,5 == 2.18 ; R3,4,5 = 2.18 + 1 = 3.18 43 +8 24 V6 3.18 (3.18 )(6 ) R1,2,3,4 = ;R1,2,3,4 = 2.08 3.18 + 6 Re = 2.08 + 4 ;Re = 6.08 4 24 V24 V I3 in the 1- resistor is same as I in R3,4,524 V IT == 3.95 A6.08 V2,3,4,5 = (3.95 A)(2.08 ) = 8.21 V; Also V3,4,5 = 8.21 V 8.21 VI 3,4,5 == 2.58 A;ThereforeI3 = 2.58 A in 1- resistor3.18 Re = 6.08 ; I3 = 2.58 A145