SlideShare una empresa de Scribd logo
1 de 11
Descargar para leer sin conexión
Chapter 27. Current and Resistance                                                 Physics, 6th Edition


                       Chapter 27. Current and Resistance

Electric Current and Ohm’s Law

27-1. How many electrons pass a point every second in a wire carrying a current of 20 A?

       How much time is needed to transport 40 C of charge past this point?

                                              C       1e     
        Q = It = (20 C/s)(1 s);      Q = 20              -19 ;   Q = 1.25 x 1020 electrons/s
                                              s  1.6 x 10 C 

                               Q           Q 40 C
                          I=     ;    t=    =     = 2.00 s ;       t = 2.00 s
                               t           I 20 A


27-2. If 600 C of charge pass a given point in 3 s, what is the electric current in amperes?

                                           Q 600 C
                                      I=     =     ;      I = 20 A
                                           t   3s


27-3. Find the current in amperes when 690 C of charge pass a given point in 2 min.

                                          Q 690 C
                                     I=     =       ;    I = 5.75 A
                                          t   120 s


27-4. If a current of 24 A exists for 50 s, how many coulombs of charge have passed through

       the wire?

                               Q = It = (24 A)(50 s);      Q = 1200 C


27-5. What is the potential drop across a 4-Ω resistor with a current of 8 A passing through it?

                                V = IR = (8 A)(4 Ω);       V = 32.0 V


27-6. Find the resistance of a rheostat if the drop in potential is 48 V and the current is 4 A.

                                          V 48 V
                                     R=     =    ;        R = 12.0 Ω
                                          I   4A



                                                  119
Chapter 27. Current and Resistance                                             Physics, 6th Edition


27-7. Determine the current through a 5-Ω resistor that has a 40-V drop in potential across it?

                                       V 40 V
                                  I=    =     ;       I = 8.00 A
                                       R 5Ω


27-8. A 2-A fuse is placed in a circuit with a battery having a terminal voltage of 12 V. What is

       the minimum resistance for a circuit containing this fuse?

                                      V 12 V
                                 R=     =    ;        R = 6.00 Ω
                                      I   2A


7-9.   What emf is required to pass 60 mA through a resistance of 20 kΩ? If this same emf is

       applied to a resistance of 300 Ω, what will be the new current?

                         E = IR = 60 x 10-3 A)(20 x 103 Ω);   E = 1200 V

                                      E 1200 V
                                 I=    =       ;       I = 4.00 A
                                      R 300 Ω



Electric Power and Heat Loss

27-10. A soldering iron draws 0.75 A at 120 V. How much energy will it use in 15 min?

                P = IV = (0.75 A)(120 V); P = 90.0 W;         t = 15 min = 900 s

                        Work
                   P=        ;   Energy = Pt = (90 W)(900 s) ;      E = 81,000 J
                         t


27-11. An electric lamp has an 80-Ω filament connected to a 110-V direct-current line. What is

       the current through the filament? What is the power loss in watts?

                                       V 110 V
                                  I=    =      ;      I = 1.38 A
                                       R 80 Ω




                                               120
Chapter 27. Current and Resistance                                                  Physics, 6th Edition


                                         V 2 (110 V) 2
                                   P=       =          ;    P = 151 W
                                          R    80 Ω

27-12. Assume that the cost of energy in a home is 8 cents per kilowatt-hour. A family goes on

       a 2-week vacation leaving a single 80-W light bulb burning. What is the cost?

              E = Pt = (80 W)(2 wk)(7 day/wk)(24 h/day)(3600 s/h) = 26.9 kW h

                     E = (26.9 kW h)(0.08 c/kw h) =         $2.15    (Rates vary)


27-13. A 120-V, direct-current generator delivers 2.4 kW to an electric furnace. What current is

       supplied? What is the resistance?

                       P 2400W                             V 120 V
                  I=    =      ; I = 20 A;            R=     =      ;    R = 6.00 Ω
                       V 120 V                             I   20 A


27-14. A resistor develops heat at the rate of 250 W when the potential difference across its ends

       is 120 V. What is its resistance?

                                  V2           V 2 (120 V) 2
                          P=         ;    R=      =             R = 57.6 Ω
                                  R             P   250 W;


27-15. A 120-V motor draws a current of 4.0 A. How many joules of electrical energy is used in

       one hour? How many kilowatt-hours?

                                   P = VI = (120 V)(4.0 A) = 480 W

                            E
                       P=     ;     E = Pt = (480 W)(3600 s);       E = 1.73 MJ
                            t

                                         1 kW ⋅ h 
                       E = 1.73 x 106 J           6          E = 0.480 kW h
                                         3.60 x 10 J 


27-16. A household hair dryer is rated at 2000 W and is designed to operate on a 120-V outlet.

       What is the resistance of the device?


                                                    121
Chapter 27. Current and Resistance                                                      Physics, 6th Edition


                                  V2             V 2 (120 V) 2
                            P=       ;      R=      =          ;      R = 7.20 Ω
                                  R               P   2000 W

Resistivity

27-17. What length of copper wire 1/16 in. in diameter is required to construct a 20-Ω resistor at

       200C? What length of nichrome wire is needed?

               Copper: ρ = 1.78 x 10-8 Ω m ;                nichrome: ρ = 100 x 10-8 Ω m

                   1
                    16   ft = 0.0625 in. = 62.5 mil;        A = (62.5 mil) 2 = 3906 cmil

                               ρl          RA (20 Ω)(3906 cmil)
                         R=       ;   l=     =                   ;        l = 7510 ft
                               A           ρ    10.4 Ω ⋅ cmil/ft

                               ρl          RA (20 Ω)(3906 cmil)
                          R=      ;   l=     =                  ;          l = 130 ft
                               A           ρ    600 Ω ⋅ cmil/ft


27-18. A 3.0-m length of copper wire (ρ = 1.78 x 10-8 Ω m) at 200C has a cross section of                 4

       mm2.What is the electrical resistance of this wire?           [ A = 4 mm2 = 4 x 10-6 m2 ]

                                ρ l (1.72 x 10-8Ω ⋅ m)(3.0 m)
                          R=       =                          ;        R =12.9 mΩ
                                A         4.00 x 10-6 m 2


27-19. Find the resistance of 40 m of tungsten (ρ = 5.5 x 10-8 Ω m) wire having a diameter of 0.8

       mm at 200C?

                                 π D 2 π (0.0008 m) 2
                            A=        =               ;        A = 5.03 x 10-7 m2
                                   4          4

                                ρ l (5.5 x 10-8Ω ⋅ m)(40.0 m)
                           R=      =                          ;         R = 4.37 Ω
                                A         5.03 x 10-7 m 2


27-20. A certain wire has a diameter of 3 mm and a length of 150 m. It has a resistance of

       3.00 Ω at 200C. What is the resistivity?         [ A = πD2/4= 7.07 x 10-7 m2. ]



                                                      122
Chapter 27. Current and Resistance                                                       Physics, 6th Edition


                      ρl      RA (3 Ω)(7.07 x 10-7 m 2 )
                 R=      ; ρ=    =                       ;              ρ = 1.41 x 10-8 Ωm
                      A        l        150 m

27-21. What is the resistance of 200 ft of iron (ρ = 9.5 x 10-8 Ω m) wire with a diameter of 0.002

        in. at 200C? (ρ = 9.5 x 10-8 Ω m). [ 200 ft = 61.0 m; 0.002 in. = 5.08 x 10-5 m ]

                              π D 2 π (5.08 x 10-5 m) 2
                       A=          =                    ;        A = 2.03 x 10-9 m2
                                4            4

                                  ρ l (9.5 x 10-8Ω ⋅ m)(61.0 m)
                         R=          =                          ;       R = 2860 Ω
                                  A         5.08 x 10-5 m 2


*27-22. A nichrome wire (ρ = 100 x 10-8 Ω m) has a length of 40 m at 200C What is the diameter

        if the total resistance is 5 Ω?

                       ρl              ρ l (100 x 10-8Ω ⋅ m)(40 m)
                  R=      ;       A=      =                        ;       A = 8 x 10-6 m2
                       A               R           5.00 Ω

                         π D2                 4A   2(8 x 10-6 m 2 )
                    A=        ;        D=        =                  ;      D = 2.26 mm
                           4                  π          π


*27-23. A 115-V source of emf is attached to a heating element which is a coil of nichrome wire

        (ρ = 100 x 10-8 Ω m ) of cross section 1.20 mm2 What must be the length of the wire if

        the resistive power loss is to be 800 W? [A = 1.20 mm2 = 1.20 x 10-6 m2 ]

                           V2               V 2 (115 V) 2
                      P=      ;        R=      =          = 16.5 Ω ;       R = 16.5 Ω
                           R                 P   800 W

                           ρl            RA (16.5 Ω)(1.20 x 10-6 m 2 )
                    R=        ;     l=     =                           ;     l = 19.8 m
                           A             ρ      100 x 10-8 Ω ⋅ m


Temperature Coefficient of Resistance




                                                      123
Chapter 27. Current and Resistance                                                Physics, 6th Edition


27-24. The resistance of a length of wire (α = 0.0065/C0) is 4.00 Ω at 200C. What is the

        resistance at 800C? [ ∆t = 800C – 200C = 60 C0 ]

       ∆R = α Ro ∆t = (0.0065 / C0 )(4 Ω)(60 C0 ) = 1.56 Ω ;   R = 4.00 Ω + 1.56 Ω = 5.56 Ω

27-25. If the resistance of a conductor is 100 Ω at 200C, and 116 Ω at 600C, what is its

        temperature coefficient of resistivity?    [ ∆t = 600C – 200C = 40 C0 ]

                                  ∆R    116 Ω - 100 Ω
                         α=           =                ;     α = 0.00400 /C0
                                 R0 ∆t (100 Ω)(40 C0 )


27-26. A length of copper (α = 0.0043/C0) wire has a resistance of 8 Ω at 200C. What is the

        resistance at 900C? At - 300C?

           ∆R = (0.0043 / C0 )(8 Ω)(70 C0 ) = 2.41 Ω ;     R = 8.00 Ω + 2.41 Ω = 10.41 Ω

       ∆R = (0.0043 / C0 )(8 Ω)(-300 C − 200 C) = −1.72 Ω ; R = 8.00 Ω - 1.72 Ω = 6.28 Ω


*27-27. The copper windings (α = 0.0043/C0) of a motor experience a 20 percent increase in

        resistance over their value at 200C. What is the operating temperature?

           ∆R             ∆R      0.2
              = 0.2; ∆t =    =             = 46.5 C0 ; t = 200C + 46.5 C0 = 66.5 0C
            R             R0α 0.0043 / C 0




*27-28. The resistivity of copper at 200C is 1.78 x 10-8 Ω m. What change in temperature will

        produce a 25 percent increase in resistivity?


Challenge Problems

27-29. A water turbine delivers 2000 kW to an electric generator which is 80 percent efficient

        and has an output terminal voltage of 1200 V. What current is delivered and what is the

        electrical resistance?          [ Pout = (0.80)(2000 kW) = 1600 kW ]


                                                  124
Chapter 27. Current and Resistance                                                    Physics, 6th Edition


                                              P 1600 x 103 W
                            P = VI ;     I=     =            ;         I = 1330 A
                                              V    1200 V

                                         V 1200 V
                                   R=     =       ;       R = 0.900 Ω
                                         I 1300 A

27-30. A 110-V radiant heater draws a current of 6.0 A. How much heat energy in joules is

       delivered in one hour?

                       E
                  P=     = VI ;        E = VIt = (110 V)(6 A)(3600 s);      E = 2.38 MJ
                       t


27-31. A power line has a total resistance of 4 kΩ. What is the power loss through the wire if the

       current is reduced to 6.0 mA?

                            P = I 2 R = (0.006 A) 2 (4000 Ω);      P =144 mW


27-32. A certain wire has a resistivity of 2 x 10-8 Ω m at 200C. If its length is 200 m and its cross

       section is 4 mm2, what will be its electrical resistance at 1000C. Assume that α = 0.005/C0

       for this material.    [ ∆t = 1000C – 200C = 80 C0 ]

                            ρ l (2 x 10-8Ω ⋅ m)(200 m)
                     R0 =      =                       ; R0 = 1.00 Ω at 200C
                            A         4 x 10-6 m 2

                 R = R0 + α R0 ∆t = 1.00 Ω + (0.005 / C0 )(1 Ω)(80 C0 );       R = 1.40 Ω


27-33. Determine the resistivity of a wire made of an unknown alloy if its diameter is 0.007 in.

       and 100 ft of the wire is found to have a resistance of 4.0 Ω.          [ D = 0.007 in. = 7 mil ]

                                                                ρl            RA
                            A = (7 mil)2 = 49 cmil;       R=       ;     ρ=
                                                                A              l

                                 RA (4 Ω)(49 cmil)
                            ∆=      =              ;        ρ = 1.96 Ω cmil/ft
                                  l     100 ft



                                                    125
Chapter 27. Current and Resistance                                                     Physics, 6th Edition


27-34. The resistivity of a certain wire is 1.72 x 10-8 Ω m at 200C. A 6-V battery is connected to

        a 20-m coil of this wire having a diameter of 0.8 mm. What is the current in the wire?

                           π D 2 π (0.0008 m) 2                                   ρl
                      A=        =               = 5.03 x 10-7 m 2 ;         R=
                             4          4                                         A

                              ρ l (1.72 x 10-8Ω ⋅ m)(20 m)
27-34. (Cont.)           R=      =                         ;      R = 0.684 Ω
                              A        5.03 x 10-7 m 2

                                          V   6.00 V
                                     I=     =        ;   I = 8.77 A
                                          R 0.684 Ω


27-35. A certain resistor is used as a thermometer. Its resistance at 200C is 26.00 Ω, and its

        resistance at 400C is 26.20 Ω. What is the temperature coefficient of resistance for this

        material?

                         ∆R      (26.20 Ω - 26.00 Ω)
                    α=        =                         ;         α = 3.85 x 10-4/C0
                         R0 ∆t (26.00 Ω)(400 C - 200 C)


*27-36. What length of copper wire at 200C has the same resistance as 200 m of iron wire at

        200C? Assume the same cross section for each wire. [ Product RA doesn’t change. ]

                                ρl                                         ρ1l1
                           R=      ; RA = ρ l ;    ρ1R1 = ρ2l2;     l2 =
                                A                                          ρ2

                              ρ1l1 (9.5 x10−8 Ω ⋅ m)(200 m)
                       l2 =       =                         ;      l2 = 1100 m
                              ρ2       1.72 x 10-8Ω ⋅ m


*27-37. The power loss in a certain wire at 200C is 400 W. If α = 0.0036/C0, by what percentage

        will the power loss increase when the operating temperature is 680C?

                                      ∆R
                    ∆R = α R0 ∆t ;       = (0.0036 / C0 )(680 C - 200 C) = 0.173
                                       R

              Since P = I2R, the power loss increases by same percentage: 17.3 %


                                                  126
Chapter 27. Current and Resistance                                                 Physics, 6th Edition




Critical Thinking Problems

27-38. A 150-Ω resistor at 200C is rated at 2.0 W maximum power. What is the maximum

         voltage that can be applied across the resistor with exceeding the maximum allowable

         power? What is the current at this voltage?

                        V2
                     P=    ;     V = PR = (2.00 W)(150 Ω);            V = 17.3 V
                        R


27-39. The current in a home is alternating current, but the same formulas apply. Suppose a fan

        motor operating a home cooling system is rated at 10 A for a 120-V line. How much

        energy is required to operate the fan for a 24-h period? At a cost of 9 cents per kilowatt-

        hour, what is the cost of operating this fan continuously for 30 days?

          P = VI = (110 V)((10 A) = 1100 W;        E = Pt = (1100 W)(24 h) = 26.4 kW h

                      E = (26.4 kW h)(3600 s/h)(1000 W/kW); E = 95.0 MJ

                                   kW ⋅ h  $0.08 
                     Cost = 26.4                  (30 days);   Cost = $53.36
                                    day  kW ⋅ h 

*27-40. The power consumed in an electrical wire (α= 0.004/C0) is 40 W at 200C. If all other

         factors are held constant, what is the power consumption when (a) the length is doubled,

         (b) the diameter is doubled, (c) the resistivity is doubled, and (d) the absolute

         temperature is doubled? (Power loss is proportional to resistance R)

                                 ρl                  1
                            R=      ;   P ∝ l; P ∝     ;   P ∝ ∆R ∝ ∆T
                                 A                   A


                                                 127
Chapter 27. Current and Resistance                                                       Physics, 6th Edition


         (a) Double length and double power loss;        Loss = 2(40 W) = 80 W

         (b) doubling diameter gives 4A0 and one-fourth power loss:              Loss = ¼(40 W) = 10 W

         (c) Doubling resistivity doubles resistance, and also doubles power loss: Loss = 80 W

*27-40. (Cont.) (d) T = (200 + 2730) = 293 K; ∆T = 2T – T = T; ∆T = 293 K = 293 C0

              If absolute temperature doubles, the new resistance is given by:

                                          R
                    R = R0 (1 + α∆T );       = 1 + (0.004 / C0 )(293 C0 ) = 2.172;
                                          R0

                    P   R
                      =    = 2.172;       Loss = 2.172(40 W); Loss = 86.9 W
                    P0 R 0

           This of course presumes that resistivity remains linear, which is not likely.


*27-41. What must be the diameter of an aluminum wire if it is to have the same resistance as an

         equal length of copper wire of diameter 2.0 mm? What length of nichrome wire is

         needed to have the same resistance as 2 m of iron wire of the same cross section?

              ρl       R ρ                   ρc ρa            π D2            ρc         ρa
            R= ;         = = const.;           =   ;       A=      ;               2
                                                                                     =
              A        l  A                  Ac Aa              4           ( Dc )     ( Dc ) 2

                ρ a Dc2           ρc          1.72 x 10-8Ω ⋅ m
            D =
              2
              a         ; Da = Dc    = (2 mm)                  ;                 Da = 1.57 mm
                  ρc              ρa          2.80 x 10-8Ω ⋅ m

                            ρl                                                   ρi li
                       R=      ; RA = ρ l = const.;     ρ nln = ρi li ;   ln =
                            A                                                    ρn

                                 ρi li (1.72 x 10-8Ω ⋅ m)
                          ln =        =                   ;      ln = 1.72 cm
                                 ρn     (100 x 10-8Ω ⋅ m)


*27-42. An iron wire (α = 0.0065/C0) has a resistance of 6.00 Ω at 200C and a copper wire (α =

        0.0043/C0) has a resistance of 5.40 Ω at 200C. At what temperature will the two wires

        have the same resistance? [ Conditions: αiRoi∆ti - αcR0c∆tc = 6 Ω −5.4 Ω = -0.60.Ω. ]


                                                  128
Chapter 27. Current and Resistance                                                  Physics, 6th Edition


                −0.600 Ω                         −0.600 Ω
       ∆t =                    =                                             ;   ∆t = -38.0 C0
              α i Roi − α c Roc (0.0065 / C )(6.0 Ω) - (0.0043 / C0 )(5.4 Ω)
                                           0




                               ∆t = tf – 200 = -38.0 C0;    tf = -18.00C




                                                  129

Más contenido relacionado

La actualidad más candente

Problemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física II
Problemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física IIProblemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física II
Problemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física IILUIS POWELL
 
Problemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb Tippens
Problemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb   TippensProblemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb   Tippens
Problemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb TippensLUIS POWELL
 
Redes en escalera metodo 1
Redes en escalera metodo 1Redes en escalera metodo 1
Redes en escalera metodo 1Henry Alvarado
 
Solution manual for introduction to electric circuits
Solution manual for introduction to electric circuitsSolution manual for introduction to electric circuits
Solution manual for introduction to electric circuitschristian bastidas
 
[E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda]
[E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda][E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda]
[E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda]tensasparda
 
Vijay Balu Raskar E.T. Notes
Vijay Balu Raskar E.T. NotesVijay Balu Raskar E.T. Notes
Vijay Balu Raskar E.T. NotesVijay Raskar
 
1 potential & capacity 09
1 potential & capacity 091 potential & capacity 09
1 potential & capacity 09GODARAMANGERAM
 
Electronics hambley2nd
Electronics hambley2ndElectronics hambley2nd
Electronics hambley2ndfedericoblanco
 
Electrical machines solved objective
Electrical machines solved objectiveElectrical machines solved objective
Electrical machines solved objectivemaheshwareshwar
 
AST 406 Conductance & Resistance
AST 406 Conductance & ResistanceAST 406 Conductance & Resistance
AST 406 Conductance & ResistanceNeil MacIntosh
 
AST 406 Equivalent Resistance
AST 406 Equivalent ResistanceAST 406 Equivalent Resistance
AST 406 Equivalent ResistanceNeil MacIntosh
 
Eet3082 binod kumar sahu lecturer_14
Eet3082 binod kumar sahu lecturer_14Eet3082 binod kumar sahu lecturer_14
Eet3082 binod kumar sahu lecturer_14BinodKumarSahu5
 

La actualidad más candente (18)

Anschp25
Anschp25Anschp25
Anschp25
 
Problemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física II
Problemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física IIProblemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física II
Problemas (17 Págs. - 45 Problemas) Resueltos del Laboratorio N° 2 De Física II
 
Problemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb Tippens
Problemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb   TippensProblemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb   Tippens
Problemas (13 Pág. - 37 Probl.) de Carga Eléctrica y Ley de Coulomb Tippens
 
Redes en escalera metodo 1
Redes en escalera metodo 1Redes en escalera metodo 1
Redes en escalera metodo 1
 
Solution manual for introduction to electric circuits
Solution manual for introduction to electric circuitsSolution manual for introduction to electric circuits
Solution manual for introduction to electric circuits
 
Chapter 01
Chapter 01Chapter 01
Chapter 01
 
Anschp39
Anschp39Anschp39
Anschp39
 
[E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda]
[E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda][E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda]
[E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda]
 
Vijay Balu Raskar E.T. Notes
Vijay Balu Raskar E.T. NotesVijay Balu Raskar E.T. Notes
Vijay Balu Raskar E.T. Notes
 
1 potential & capacity 09
1 potential & capacity 091 potential & capacity 09
1 potential & capacity 09
 
Electronics hambley2nd
Electronics hambley2ndElectronics hambley2nd
Electronics hambley2nd
 
Anschp23
Anschp23Anschp23
Anschp23
 
Electrical machines solved objective
Electrical machines solved objectiveElectrical machines solved objective
Electrical machines solved objective
 
AST 406 Conductance & Resistance
AST 406 Conductance & ResistanceAST 406 Conductance & Resistance
AST 406 Conductance & Resistance
 
Ac circuits
Ac circuitsAc circuits
Ac circuits
 
2 ohms law
2   ohms law2   ohms law
2 ohms law
 
AST 406 Equivalent Resistance
AST 406 Equivalent ResistanceAST 406 Equivalent Resistance
AST 406 Equivalent Resistance
 
Eet3082 binod kumar sahu lecturer_14
Eet3082 binod kumar sahu lecturer_14Eet3082 binod kumar sahu lecturer_14
Eet3082 binod kumar sahu lecturer_14
 

Destacado

Informe de práctica de física 1 eléctrización
Informe de práctica de física 1 eléctrizaciónInforme de práctica de física 1 eléctrización
Informe de práctica de física 1 eléctrizaciónMartín Vinces Alava
 
Informe de práctica de física 5 aplicaciónes del eléctromagnetismo
Informe de práctica de física 5 aplicaciónes del eléctromagnetismoInforme de práctica de física 5 aplicaciónes del eléctromagnetismo
Informe de práctica de física 5 aplicaciónes del eléctromagnetismoMartín Vinces Alava
 
Análisis de modelo de mercado, determinación de precios y rendimiento en su p...
Análisis de modelo de mercado, determinación de precios y rendimiento en su p...Análisis de modelo de mercado, determinación de precios y rendimiento en su p...
Análisis de modelo de mercado, determinación de precios y rendimiento en su p...Martín Vinces Alava
 
Sistema de distribución de energía
Sistema de distribución de energíaSistema de distribución de energía
Sistema de distribución de energíaMartín Vinces Alava
 
La psicología y sociología de la información
La psicología y sociología de la informaciónLa psicología y sociología de la información
La psicología y sociología de la informaciónMartín Vinces Alava
 
La psicología y sociología de la información
La psicología y sociología de la informaciónLa psicología y sociología de la información
La psicología y sociología de la informaciónMartín Vinces Alava
 
Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...
Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...
Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...Martín Vinces Alava
 
CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS
CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS
CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS Martín Vinces Alava
 
Informe de práctica de física 2 campo eléctrico
Informe de práctica de física 2 campo eléctricoInforme de práctica de física 2 campo eléctrico
Informe de práctica de física 2 campo eléctricoMartín Vinces Alava
 
Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.
Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.
Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.Martín Vinces Alava
 
Solucionario serway capitulo 15
Solucionario serway capitulo 15Solucionario serway capitulo 15
Solucionario serway capitulo 15franciscafloresg
 
Sistema de distribución de energíax
Sistema de distribución de energíaxSistema de distribución de energíax
Sistema de distribución de energíaxMartín Vinces Alava
 
Proyecto de etica profesional etica de un ingeniero industrial en si área de ...
Proyecto de etica profesional etica de un ingeniero industrial en si área de ...Proyecto de etica profesional etica de un ingeniero industrial en si área de ...
Proyecto de etica profesional etica de un ingeniero industrial en si área de ...Martín Vinces Alava
 
Informe de práctica de física 3 ley de ohm
Informe de práctica de física 3 ley de ohmInforme de práctica de física 3 ley de ohm
Informe de práctica de física 3 ley de ohmMartín Vinces Alava
 

Destacado (20)

Informe de práctica de física 1 eléctrización
Informe de práctica de física 1 eléctrizaciónInforme de práctica de física 1 eléctrización
Informe de práctica de física 1 eléctrización
 
Informe de práctica de física 5 aplicaciónes del eléctromagnetismo
Informe de práctica de física 5 aplicaciónes del eléctromagnetismoInforme de práctica de física 5 aplicaciónes del eléctromagnetismo
Informe de práctica de física 5 aplicaciónes del eléctromagnetismo
 
Análisis de modelo de mercado, determinación de precios y rendimiento en su p...
Análisis de modelo de mercado, determinación de precios y rendimiento en su p...Análisis de modelo de mercado, determinación de precios y rendimiento en su p...
Análisis de modelo de mercado, determinación de precios y rendimiento en su p...
 
PRESUPUESTO GENERAL DEL ESTADO 2016
PRESUPUESTO GENERAL DEL ESTADO 2016PRESUPUESTO GENERAL DEL ESTADO 2016
PRESUPUESTO GENERAL DEL ESTADO 2016
 
Sistema de distribución de energía
Sistema de distribución de energíaSistema de distribución de energía
Sistema de distribución de energía
 
La psicología y sociología de la información
La psicología y sociología de la informaciónLa psicología y sociología de la información
La psicología y sociología de la información
 
La psicología y sociología de la información
La psicología y sociología de la informaciónLa psicología y sociología de la información
La psicología y sociología de la información
 
Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...
Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...
Aplicación de las Ecuaciónes Diferenciales Ordinarias aplicadas en el vaciado...
 
Avena harinas
Avena harinasAvena harinas
Avena harinas
 
Analisis 3ero-f-proyecto
Analisis 3ero-f-proyectoAnalisis 3ero-f-proyecto
Analisis 3ero-f-proyecto
 
CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS
CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS
CENTRO DE DOCUMENTACIÓN ARCHIVOS Y MUSEOS
 
Informe de práctica de física 2 campo eléctrico
Informe de práctica de física 2 campo eléctricoInforme de práctica de física 2 campo eléctrico
Informe de práctica de física 2 campo eléctrico
 
Arroz
 Arroz Arroz
Arroz
 
Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.
Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.
Los problemas de liderazgo y su afectación en el funcionamiento de las empresas.
 
Solucionario serway capitulo 15
Solucionario serway capitulo 15Solucionario serway capitulo 15
Solucionario serway capitulo 15
 
Sistema de distribución de energíax
Sistema de distribución de energíaxSistema de distribución de energíax
Sistema de distribución de energíax
 
Proyecto de etica profesional etica de un ingeniero industrial en si área de ...
Proyecto de etica profesional etica de un ingeniero industrial en si área de ...Proyecto de etica profesional etica de un ingeniero industrial en si área de ...
Proyecto de etica profesional etica de un ingeniero industrial en si área de ...
 
Arroz
ArrozArroz
Arroz
 
Centeno
CentenoCenteno
Centeno
 
Informe de práctica de física 3 ley de ohm
Informe de práctica de física 3 ley de ohmInforme de práctica de física 3 ley de ohm
Informe de práctica de física 3 ley de ohm
 

Similar a PROBLEMAS RESUELTOS (42) DE LABORATORIO N° 1 DE FÍSICA II - TIPPENS

Problemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física II
Problemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física IIProblemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física II
Problemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física IILUIS POWELL
 
Ch 21 Alternating Current
Ch 21 Alternating CurrentCh 21 Alternating Current
Ch 21 Alternating CurrentRohit Mohd
 
presentation_chapter_2_ac_networks_1516086872_20707.ppt
presentation_chapter_2_ac_networks_1516086872_20707.pptpresentation_chapter_2_ac_networks_1516086872_20707.ppt
presentation_chapter_2_ac_networks_1516086872_20707.pptraghuRAGHU56
 
What is electronics
What is electronicsWhat is electronics
What is electronicsMrinal Pal
 
Ac wave forms theroy
Ac wave forms theroyAc wave forms theroy
Ac wave forms theroyReece Hancock
 
Rangkain arus bolak balik kelas xii
Rangkain arus bolak balik kelas xiiRangkain arus bolak balik kelas xii
Rangkain arus bolak balik kelas xiiemri3
 
UNIT-III complex reactive three phase.ppt
UNIT-III complex reactive three phase.pptUNIT-III complex reactive three phase.ppt
UNIT-III complex reactive three phase.pptAbinaya Saraswathy T
 
Current Electricity (NA)
Current Electricity (NA)Current Electricity (NA)
Current Electricity (NA)guest5e66ab3
 
Chapter20 powerpoint 090223020055-phpapp01
Chapter20 powerpoint 090223020055-phpapp01Chapter20 powerpoint 090223020055-phpapp01
Chapter20 powerpoint 090223020055-phpapp01Cleophas Rwemera
 
Initially the circuit of Figure 1 is --driven-- by a DC source V of va.docx
Initially the circuit of Figure 1 is --driven-- by a DC source V of va.docxInitially the circuit of Figure 1 is --driven-- by a DC source V of va.docx
Initially the circuit of Figure 1 is --driven-- by a DC source V of va.docxdiegor62
 
Electricity, without notes
Electricity, without notesElectricity, without notes
Electricity, without notesMrPolko
 

Similar a PROBLEMAS RESUELTOS (42) DE LABORATORIO N° 1 DE FÍSICA II - TIPPENS (20)

Problemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física II
Problemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física IIProblemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física II
Problemas (10 Págs. - 42 Probl.) del Laboratorio N° 1 de Física II
 
AC Circuit Theory
AC Circuit TheoryAC Circuit Theory
AC Circuit Theory
 
Chapter32A.ppt
Chapter32A.pptChapter32A.ppt
Chapter32A.ppt
 
Theory of AC and DC Meter Testing
Theory of AC and DC Meter TestingTheory of AC and DC Meter Testing
Theory of AC and DC Meter Testing
 
Theory of AC and DC Meter Testing_Advanced.pptx
Theory of AC and DC Meter Testing_Advanced.pptxTheory of AC and DC Meter Testing_Advanced.pptx
Theory of AC and DC Meter Testing_Advanced.pptx
 
Theory of AC and DC Meter Testing
Theory of AC and DC Meter TestingTheory of AC and DC Meter Testing
Theory of AC and DC Meter Testing
 
Ch 21 Alternating Current
Ch 21 Alternating CurrentCh 21 Alternating Current
Ch 21 Alternating Current
 
presentation_chapter_2_ac_networks_1516086872_20707.ppt
presentation_chapter_2_ac_networks_1516086872_20707.pptpresentation_chapter_2_ac_networks_1516086872_20707.ppt
presentation_chapter_2_ac_networks_1516086872_20707.ppt
 
Theory of AC and DC Meter Testing (Polyphase)
Theory of AC and DC Meter Testing (Polyphase)Theory of AC and DC Meter Testing (Polyphase)
Theory of AC and DC Meter Testing (Polyphase)
 
What is electronics
What is electronicsWhat is electronics
What is electronics
 
Theory of AC and DC Meter Testing (Single Phase)
Theory of AC and DC Meter Testing (Single Phase)Theory of AC and DC Meter Testing (Single Phase)
Theory of AC and DC Meter Testing (Single Phase)
 
Ac wave forms theroy
Ac wave forms theroyAc wave forms theroy
Ac wave forms theroy
 
Rangkain arus bolak balik kelas xii
Rangkain arus bolak balik kelas xiiRangkain arus bolak balik kelas xii
Rangkain arus bolak balik kelas xii
 
UNIT-III complex reactive three phase.ppt
UNIT-III complex reactive three phase.pptUNIT-III complex reactive three phase.ppt
UNIT-III complex reactive three phase.ppt
 
Electric power
Electric powerElectric power
Electric power
 
Current Electricity (NA)
Current Electricity (NA)Current Electricity (NA)
Current Electricity (NA)
 
Chapter20 powerpoint 090223020055-phpapp01
Chapter20 powerpoint 090223020055-phpapp01Chapter20 powerpoint 090223020055-phpapp01
Chapter20 powerpoint 090223020055-phpapp01
 
Initially the circuit of Figure 1 is --driven-- by a DC source V of va.docx
Initially the circuit of Figure 1 is --driven-- by a DC source V of va.docxInitially the circuit of Figure 1 is --driven-- by a DC source V of va.docx
Initially the circuit of Figure 1 is --driven-- by a DC source V of va.docx
 
Ohmslawweb
OhmslawwebOhmslawweb
Ohmslawweb
 
Electricity, without notes
Electricity, without notesElectricity, without notes
Electricity, without notes
 

Más de LUIS POWELL

CAPÍTULO I (22) DE LABORATORIO DE FÍSICA II
CAPÍTULO I (22) DE LABORATORIO DE FÍSICA IICAPÍTULO I (22) DE LABORATORIO DE FÍSICA II
CAPÍTULO I (22) DE LABORATORIO DE FÍSICA IILUIS POWELL
 
CATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA II
CATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA IICATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA II
CATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA IILUIS POWELL
 
CAPÍTULO II DE LABORATORIO (28) DE FÍSICA II
CAPÍTULO II DE LABORATORIO (28) DE FÍSICA IICAPÍTULO II DE LABORATORIO (28) DE FÍSICA II
CAPÍTULO II DE LABORATORIO (28) DE FÍSICA IILUIS POWELL
 
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARS
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARSPROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARS
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARSLUIS POWELL
 
Problemas (67) del Capítulo III de física II Ley de Gauss
Problemas (67) del Capítulo III de física II   Ley de GaussProblemas (67) del Capítulo III de física II   Ley de Gauss
Problemas (67) del Capítulo III de física II Ley de GaussLUIS POWELL
 
Capítulo III (68) de física II Ley de Gauss - definitivo
Capítulo III (68) de física II   Ley de Gauss - definitivoCapítulo III (68) de física II   Ley de Gauss - definitivo
Capítulo III (68) de física II Ley de Gauss - definitivoLUIS POWELL
 
Problemas (16 Págs. - 42 Probl.) del Capítulo II de Física II
Problemas (16 Págs. - 42 Probl.) del Capítulo II de Física IIProblemas (16 Págs. - 42 Probl.) del Capítulo II de Física II
Problemas (16 Págs. - 42 Probl.) del Capítulo II de Física IILUIS POWELL
 
Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...
Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...
Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...LUIS POWELL
 
Capítulo II (31) de Física II - Campo Eléctrico - Definitivo
Capítulo II (31) de Física II - Campo Eléctrico - DefinitivoCapítulo II (31) de Física II - Campo Eléctrico - Definitivo
Capítulo II (31) de Física II - Campo Eléctrico - DefinitivoLUIS POWELL
 
PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1
PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1
PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1LUIS POWELL
 
Problemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico - Sears
Problemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico -  SearsProblemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico -  Sears
Problemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico - SearsLUIS POWELL
 
Programa (6) Sintético de Física II
Programa (6) Sintético de Física IIPrograma (6) Sintético de Física II
Programa (6) Sintético de Física IILUIS POWELL
 
Tecnología Educativa (26) y Educación Virtual
Tecnología Educativa (26) y Educación VirtualTecnología Educativa (26) y Educación Virtual
Tecnología Educativa (26) y Educación VirtualLUIS POWELL
 
Pasos (17) de la Investigación Científica
Pasos (17) de la Investigación CientíficaPasos (17) de la Investigación Científica
Pasos (17) de la Investigación CientíficaLUIS POWELL
 
Presentación (12) sobre El Desapego - Versión Alternativa
Presentación (12) sobre El Desapego - Versión AlternativaPresentación (12) sobre El Desapego - Versión Alternativa
Presentación (12) sobre El Desapego - Versión AlternativaLUIS POWELL
 

Más de LUIS POWELL (15)

CAPÍTULO I (22) DE LABORATORIO DE FÍSICA II
CAPÍTULO I (22) DE LABORATORIO DE FÍSICA IICAPÍTULO I (22) DE LABORATORIO DE FÍSICA II
CAPÍTULO I (22) DE LABORATORIO DE FÍSICA II
 
CATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA II
CATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA IICATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA II
CATÁLOGO DE (23) VIDEOS SOBRE LOS CAPÍTULOS I, II Y II DE FÍSICA II
 
CAPÍTULO II DE LABORATORIO (28) DE FÍSICA II
CAPÍTULO II DE LABORATORIO (28) DE FÍSICA IICAPÍTULO II DE LABORATORIO (28) DE FÍSICA II
CAPÍTULO II DE LABORATORIO (28) DE FÍSICA II
 
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARS
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARSPROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARS
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARS
 
Problemas (67) del Capítulo III de física II Ley de Gauss
Problemas (67) del Capítulo III de física II   Ley de GaussProblemas (67) del Capítulo III de física II   Ley de Gauss
Problemas (67) del Capítulo III de física II Ley de Gauss
 
Capítulo III (68) de física II Ley de Gauss - definitivo
Capítulo III (68) de física II   Ley de Gauss - definitivoCapítulo III (68) de física II   Ley de Gauss - definitivo
Capítulo III (68) de física II Ley de Gauss - definitivo
 
Problemas (16 Págs. - 42 Probl.) del Capítulo II de Física II
Problemas (16 Págs. - 42 Probl.) del Capítulo II de Física IIProblemas (16 Págs. - 42 Probl.) del Capítulo II de Física II
Problemas (16 Págs. - 42 Probl.) del Capítulo II de Física II
 
Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...
Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...
Capítulo I (24) de Física II - La Carga Eléctrica y La Ley de Coulomb - Defin...
 
Capítulo II (31) de Física II - Campo Eléctrico - Definitivo
Capítulo II (31) de Física II - Campo Eléctrico - DefinitivoCapítulo II (31) de Física II - Campo Eléctrico - Definitivo
Capítulo II (31) de Física II - Campo Eléctrico - Definitivo
 
PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1
PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1
PRESENTACIÓN (40) DE FÍSICA II - SEMANA 1
 
Problemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico - Sears
Problemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico -  SearsProblemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico -  Sears
Problemas (43 Pág. - 107 Probl.) de Carga Eléctrica y Campo Eléctrico - Sears
 
Programa (6) Sintético de Física II
Programa (6) Sintético de Física IIPrograma (6) Sintético de Física II
Programa (6) Sintético de Física II
 
Tecnología Educativa (26) y Educación Virtual
Tecnología Educativa (26) y Educación VirtualTecnología Educativa (26) y Educación Virtual
Tecnología Educativa (26) y Educación Virtual
 
Pasos (17) de la Investigación Científica
Pasos (17) de la Investigación CientíficaPasos (17) de la Investigación Científica
Pasos (17) de la Investigación Científica
 
Presentación (12) sobre El Desapego - Versión Alternativa
Presentación (12) sobre El Desapego - Versión AlternativaPresentación (12) sobre El Desapego - Versión Alternativa
Presentación (12) sobre El Desapego - Versión Alternativa
 

Último

Practical Research 1: Lesson 8 Writing the Thesis Statement.pptx
Practical Research 1: Lesson 8 Writing the Thesis Statement.pptxPractical Research 1: Lesson 8 Writing the Thesis Statement.pptx
Practical Research 1: Lesson 8 Writing the Thesis Statement.pptxKatherine Villaluna
 
AUDIENCE THEORY -- FANDOM -- JENKINS.pptx
AUDIENCE THEORY -- FANDOM -- JENKINS.pptxAUDIENCE THEORY -- FANDOM -- JENKINS.pptx
AUDIENCE THEORY -- FANDOM -- JENKINS.pptxiammrhaywood
 
M-2- General Reactions of amino acids.pptx
M-2- General Reactions of amino acids.pptxM-2- General Reactions of amino acids.pptx
M-2- General Reactions of amino acids.pptxDr. Santhosh Kumar. N
 
In - Vivo and In - Vitro Correlation.pptx
In - Vivo and In - Vitro Correlation.pptxIn - Vivo and In - Vitro Correlation.pptx
In - Vivo and In - Vitro Correlation.pptxAditiChauhan701637
 
What is the Future of QuickBooks DeskTop?
What is the Future of QuickBooks DeskTop?What is the Future of QuickBooks DeskTop?
What is the Future of QuickBooks DeskTop?TechSoup
 
General views of Histopathology and step
General views of Histopathology and stepGeneral views of Histopathology and step
General views of Histopathology and stepobaje godwin sunday
 
2024.03.23 What do successful readers do - Sandy Millin for PARK.pptx
2024.03.23 What do successful readers do - Sandy Millin for PARK.pptx2024.03.23 What do successful readers do - Sandy Millin for PARK.pptx
2024.03.23 What do successful readers do - Sandy Millin for PARK.pptxSandy Millin
 
P4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdf
P4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdfP4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdf
P4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdfYu Kanazawa / Osaka University
 
The basics of sentences session 10pptx.pptx
The basics of sentences session 10pptx.pptxThe basics of sentences session 10pptx.pptx
The basics of sentences session 10pptx.pptxheathfieldcps1
 
Prescribed medication order and communication skills.pptx
Prescribed medication order and communication skills.pptxPrescribed medication order and communication skills.pptx
Prescribed medication order and communication skills.pptxraviapr7
 
3.21.24 The Origins of Black Power.pptx
3.21.24  The Origins of Black Power.pptx3.21.24  The Origins of Black Power.pptx
3.21.24 The Origins of Black Power.pptxmary850239
 
Presentation on the Basics of Writing. Writing a Paragraph
Presentation on the Basics of Writing. Writing a ParagraphPresentation on the Basics of Writing. Writing a Paragraph
Presentation on the Basics of Writing. Writing a ParagraphNetziValdelomar1
 
How to Solve Singleton Error in the Odoo 17
How to Solve Singleton Error in the  Odoo 17How to Solve Singleton Error in the  Odoo 17
How to Solve Singleton Error in the Odoo 17Celine George
 
CAULIFLOWER BREEDING 1 Parmar pptx
CAULIFLOWER BREEDING 1 Parmar pptxCAULIFLOWER BREEDING 1 Parmar pptx
CAULIFLOWER BREEDING 1 Parmar pptxSaurabhParmar42
 
How to Manage Cross-Selling in Odoo 17 Sales
How to Manage Cross-Selling in Odoo 17 SalesHow to Manage Cross-Selling in Odoo 17 Sales
How to Manage Cross-Selling in Odoo 17 SalesCeline George
 
Drug Information Services- DIC and Sources.
Drug Information Services- DIC and Sources.Drug Information Services- DIC and Sources.
Drug Information Services- DIC and Sources.raviapr7
 
Patient Counselling. Definition of patient counseling; steps involved in pati...
Patient Counselling. Definition of patient counseling; steps involved in pati...Patient Counselling. Definition of patient counseling; steps involved in pati...
Patient Counselling. Definition of patient counseling; steps involved in pati...raviapr7
 
Education and training program in the hospital APR.pptx
Education and training program in the hospital APR.pptxEducation and training program in the hospital APR.pptx
Education and training program in the hospital APR.pptxraviapr7
 

Último (20)

Practical Research 1: Lesson 8 Writing the Thesis Statement.pptx
Practical Research 1: Lesson 8 Writing the Thesis Statement.pptxPractical Research 1: Lesson 8 Writing the Thesis Statement.pptx
Practical Research 1: Lesson 8 Writing the Thesis Statement.pptx
 
AUDIENCE THEORY -- FANDOM -- JENKINS.pptx
AUDIENCE THEORY -- FANDOM -- JENKINS.pptxAUDIENCE THEORY -- FANDOM -- JENKINS.pptx
AUDIENCE THEORY -- FANDOM -- JENKINS.pptx
 
M-2- General Reactions of amino acids.pptx
M-2- General Reactions of amino acids.pptxM-2- General Reactions of amino acids.pptx
M-2- General Reactions of amino acids.pptx
 
Finals of Kant get Marx 2.0 : a general politics quiz
Finals of Kant get Marx 2.0 : a general politics quizFinals of Kant get Marx 2.0 : a general politics quiz
Finals of Kant get Marx 2.0 : a general politics quiz
 
Personal Resilience in Project Management 2 - TV Edit 1a.pdf
Personal Resilience in Project Management 2 - TV Edit 1a.pdfPersonal Resilience in Project Management 2 - TV Edit 1a.pdf
Personal Resilience in Project Management 2 - TV Edit 1a.pdf
 
In - Vivo and In - Vitro Correlation.pptx
In - Vivo and In - Vitro Correlation.pptxIn - Vivo and In - Vitro Correlation.pptx
In - Vivo and In - Vitro Correlation.pptx
 
What is the Future of QuickBooks DeskTop?
What is the Future of QuickBooks DeskTop?What is the Future of QuickBooks DeskTop?
What is the Future of QuickBooks DeskTop?
 
General views of Histopathology and step
General views of Histopathology and stepGeneral views of Histopathology and step
General views of Histopathology and step
 
2024.03.23 What do successful readers do - Sandy Millin for PARK.pptx
2024.03.23 What do successful readers do - Sandy Millin for PARK.pptx2024.03.23 What do successful readers do - Sandy Millin for PARK.pptx
2024.03.23 What do successful readers do - Sandy Millin for PARK.pptx
 
P4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdf
P4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdfP4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdf
P4C x ELT = P4ELT: Its Theoretical Background (Kanazawa, 2024 March).pdf
 
The basics of sentences session 10pptx.pptx
The basics of sentences session 10pptx.pptxThe basics of sentences session 10pptx.pptx
The basics of sentences session 10pptx.pptx
 
Prescribed medication order and communication skills.pptx
Prescribed medication order and communication skills.pptxPrescribed medication order and communication skills.pptx
Prescribed medication order and communication skills.pptx
 
3.21.24 The Origins of Black Power.pptx
3.21.24  The Origins of Black Power.pptx3.21.24  The Origins of Black Power.pptx
3.21.24 The Origins of Black Power.pptx
 
Presentation on the Basics of Writing. Writing a Paragraph
Presentation on the Basics of Writing. Writing a ParagraphPresentation on the Basics of Writing. Writing a Paragraph
Presentation on the Basics of Writing. Writing a Paragraph
 
How to Solve Singleton Error in the Odoo 17
How to Solve Singleton Error in the  Odoo 17How to Solve Singleton Error in the  Odoo 17
How to Solve Singleton Error in the Odoo 17
 
CAULIFLOWER BREEDING 1 Parmar pptx
CAULIFLOWER BREEDING 1 Parmar pptxCAULIFLOWER BREEDING 1 Parmar pptx
CAULIFLOWER BREEDING 1 Parmar pptx
 
How to Manage Cross-Selling in Odoo 17 Sales
How to Manage Cross-Selling in Odoo 17 SalesHow to Manage Cross-Selling in Odoo 17 Sales
How to Manage Cross-Selling in Odoo 17 Sales
 
Drug Information Services- DIC and Sources.
Drug Information Services- DIC and Sources.Drug Information Services- DIC and Sources.
Drug Information Services- DIC and Sources.
 
Patient Counselling. Definition of patient counseling; steps involved in pati...
Patient Counselling. Definition of patient counseling; steps involved in pati...Patient Counselling. Definition of patient counseling; steps involved in pati...
Patient Counselling. Definition of patient counseling; steps involved in pati...
 
Education and training program in the hospital APR.pptx
Education and training program in the hospital APR.pptxEducation and training program in the hospital APR.pptx
Education and training program in the hospital APR.pptx
 

PROBLEMAS RESUELTOS (42) DE LABORATORIO N° 1 DE FÍSICA II - TIPPENS

  • 1. Chapter 27. Current and Resistance Physics, 6th Edition Chapter 27. Current and Resistance Electric Current and Ohm’s Law 27-1. How many electrons pass a point every second in a wire carrying a current of 20 A? How much time is needed to transport 40 C of charge past this point? C 1e  Q = It = (20 C/s)(1 s); Q = 20  -19 ; Q = 1.25 x 1020 electrons/s s  1.6 x 10 C  Q Q 40 C I= ; t= = = 2.00 s ; t = 2.00 s t I 20 A 27-2. If 600 C of charge pass a given point in 3 s, what is the electric current in amperes? Q 600 C I= = ; I = 20 A t 3s 27-3. Find the current in amperes when 690 C of charge pass a given point in 2 min. Q 690 C I= = ; I = 5.75 A t 120 s 27-4. If a current of 24 A exists for 50 s, how many coulombs of charge have passed through the wire? Q = It = (24 A)(50 s); Q = 1200 C 27-5. What is the potential drop across a 4-Ω resistor with a current of 8 A passing through it? V = IR = (8 A)(4 Ω); V = 32.0 V 27-6. Find the resistance of a rheostat if the drop in potential is 48 V and the current is 4 A. V 48 V R= = ; R = 12.0 Ω I 4A 119
  • 2. Chapter 27. Current and Resistance Physics, 6th Edition 27-7. Determine the current through a 5-Ω resistor that has a 40-V drop in potential across it? V 40 V I= = ; I = 8.00 A R 5Ω 27-8. A 2-A fuse is placed in a circuit with a battery having a terminal voltage of 12 V. What is the minimum resistance for a circuit containing this fuse? V 12 V R= = ; R = 6.00 Ω I 2A 7-9. What emf is required to pass 60 mA through a resistance of 20 kΩ? If this same emf is applied to a resistance of 300 Ω, what will be the new current? E = IR = 60 x 10-3 A)(20 x 103 Ω); E = 1200 V E 1200 V I= = ; I = 4.00 A R 300 Ω Electric Power and Heat Loss 27-10. A soldering iron draws 0.75 A at 120 V. How much energy will it use in 15 min? P = IV = (0.75 A)(120 V); P = 90.0 W; t = 15 min = 900 s Work P= ; Energy = Pt = (90 W)(900 s) ; E = 81,000 J t 27-11. An electric lamp has an 80-Ω filament connected to a 110-V direct-current line. What is the current through the filament? What is the power loss in watts? V 110 V I= = ; I = 1.38 A R 80 Ω 120
  • 3. Chapter 27. Current and Resistance Physics, 6th Edition V 2 (110 V) 2 P= = ; P = 151 W R 80 Ω 27-12. Assume that the cost of energy in a home is 8 cents per kilowatt-hour. A family goes on a 2-week vacation leaving a single 80-W light bulb burning. What is the cost? E = Pt = (80 W)(2 wk)(7 day/wk)(24 h/day)(3600 s/h) = 26.9 kW h E = (26.9 kW h)(0.08 c/kw h) = $2.15 (Rates vary) 27-13. A 120-V, direct-current generator delivers 2.4 kW to an electric furnace. What current is supplied? What is the resistance? P 2400W V 120 V I= = ; I = 20 A; R= = ; R = 6.00 Ω V 120 V I 20 A 27-14. A resistor develops heat at the rate of 250 W when the potential difference across its ends is 120 V. What is its resistance? V2 V 2 (120 V) 2 P= ; R= = R = 57.6 Ω R P 250 W; 27-15. A 120-V motor draws a current of 4.0 A. How many joules of electrical energy is used in one hour? How many kilowatt-hours? P = VI = (120 V)(4.0 A) = 480 W E P= ; E = Pt = (480 W)(3600 s); E = 1.73 MJ t  1 kW ⋅ h  E = 1.73 x 106 J  6  E = 0.480 kW h  3.60 x 10 J  27-16. A household hair dryer is rated at 2000 W and is designed to operate on a 120-V outlet. What is the resistance of the device? 121
  • 4. Chapter 27. Current and Resistance Physics, 6th Edition V2 V 2 (120 V) 2 P= ; R= = ; R = 7.20 Ω R P 2000 W Resistivity 27-17. What length of copper wire 1/16 in. in diameter is required to construct a 20-Ω resistor at 200C? What length of nichrome wire is needed? Copper: ρ = 1.78 x 10-8 Ω m ; nichrome: ρ = 100 x 10-8 Ω m 1 16 ft = 0.0625 in. = 62.5 mil; A = (62.5 mil) 2 = 3906 cmil ρl RA (20 Ω)(3906 cmil) R= ; l= = ; l = 7510 ft A ρ 10.4 Ω ⋅ cmil/ft ρl RA (20 Ω)(3906 cmil) R= ; l= = ; l = 130 ft A ρ 600 Ω ⋅ cmil/ft 27-18. A 3.0-m length of copper wire (ρ = 1.78 x 10-8 Ω m) at 200C has a cross section of 4 mm2.What is the electrical resistance of this wire? [ A = 4 mm2 = 4 x 10-6 m2 ] ρ l (1.72 x 10-8Ω ⋅ m)(3.0 m) R= = ; R =12.9 mΩ A 4.00 x 10-6 m 2 27-19. Find the resistance of 40 m of tungsten (ρ = 5.5 x 10-8 Ω m) wire having a diameter of 0.8 mm at 200C? π D 2 π (0.0008 m) 2 A= = ; A = 5.03 x 10-7 m2 4 4 ρ l (5.5 x 10-8Ω ⋅ m)(40.0 m) R= = ; R = 4.37 Ω A 5.03 x 10-7 m 2 27-20. A certain wire has a diameter of 3 mm and a length of 150 m. It has a resistance of 3.00 Ω at 200C. What is the resistivity? [ A = πD2/4= 7.07 x 10-7 m2. ] 122
  • 5. Chapter 27. Current and Resistance Physics, 6th Edition ρl RA (3 Ω)(7.07 x 10-7 m 2 ) R= ; ρ= = ; ρ = 1.41 x 10-8 Ωm A l 150 m 27-21. What is the resistance of 200 ft of iron (ρ = 9.5 x 10-8 Ω m) wire with a diameter of 0.002 in. at 200C? (ρ = 9.5 x 10-8 Ω m). [ 200 ft = 61.0 m; 0.002 in. = 5.08 x 10-5 m ] π D 2 π (5.08 x 10-5 m) 2 A= = ; A = 2.03 x 10-9 m2 4 4 ρ l (9.5 x 10-8Ω ⋅ m)(61.0 m) R= = ; R = 2860 Ω A 5.08 x 10-5 m 2 *27-22. A nichrome wire (ρ = 100 x 10-8 Ω m) has a length of 40 m at 200C What is the diameter if the total resistance is 5 Ω? ρl ρ l (100 x 10-8Ω ⋅ m)(40 m) R= ; A= = ; A = 8 x 10-6 m2 A R 5.00 Ω π D2 4A 2(8 x 10-6 m 2 ) A= ; D= = ; D = 2.26 mm 4 π π *27-23. A 115-V source of emf is attached to a heating element which is a coil of nichrome wire (ρ = 100 x 10-8 Ω m ) of cross section 1.20 mm2 What must be the length of the wire if the resistive power loss is to be 800 W? [A = 1.20 mm2 = 1.20 x 10-6 m2 ] V2 V 2 (115 V) 2 P= ; R= = = 16.5 Ω ; R = 16.5 Ω R P 800 W ρl RA (16.5 Ω)(1.20 x 10-6 m 2 ) R= ; l= = ; l = 19.8 m A ρ 100 x 10-8 Ω ⋅ m Temperature Coefficient of Resistance 123
  • 6. Chapter 27. Current and Resistance Physics, 6th Edition 27-24. The resistance of a length of wire (α = 0.0065/C0) is 4.00 Ω at 200C. What is the resistance at 800C? [ ∆t = 800C – 200C = 60 C0 ] ∆R = α Ro ∆t = (0.0065 / C0 )(4 Ω)(60 C0 ) = 1.56 Ω ; R = 4.00 Ω + 1.56 Ω = 5.56 Ω 27-25. If the resistance of a conductor is 100 Ω at 200C, and 116 Ω at 600C, what is its temperature coefficient of resistivity? [ ∆t = 600C – 200C = 40 C0 ] ∆R 116 Ω - 100 Ω α= = ; α = 0.00400 /C0 R0 ∆t (100 Ω)(40 C0 ) 27-26. A length of copper (α = 0.0043/C0) wire has a resistance of 8 Ω at 200C. What is the resistance at 900C? At - 300C? ∆R = (0.0043 / C0 )(8 Ω)(70 C0 ) = 2.41 Ω ; R = 8.00 Ω + 2.41 Ω = 10.41 Ω ∆R = (0.0043 / C0 )(8 Ω)(-300 C − 200 C) = −1.72 Ω ; R = 8.00 Ω - 1.72 Ω = 6.28 Ω *27-27. The copper windings (α = 0.0043/C0) of a motor experience a 20 percent increase in resistance over their value at 200C. What is the operating temperature? ∆R ∆R 0.2 = 0.2; ∆t = = = 46.5 C0 ; t = 200C + 46.5 C0 = 66.5 0C R R0α 0.0043 / C 0 *27-28. The resistivity of copper at 200C is 1.78 x 10-8 Ω m. What change in temperature will produce a 25 percent increase in resistivity? Challenge Problems 27-29. A water turbine delivers 2000 kW to an electric generator which is 80 percent efficient and has an output terminal voltage of 1200 V. What current is delivered and what is the electrical resistance? [ Pout = (0.80)(2000 kW) = 1600 kW ] 124
  • 7. Chapter 27. Current and Resistance Physics, 6th Edition P 1600 x 103 W P = VI ; I= = ; I = 1330 A V 1200 V V 1200 V R= = ; R = 0.900 Ω I 1300 A 27-30. A 110-V radiant heater draws a current of 6.0 A. How much heat energy in joules is delivered in one hour? E P= = VI ; E = VIt = (110 V)(6 A)(3600 s); E = 2.38 MJ t 27-31. A power line has a total resistance of 4 kΩ. What is the power loss through the wire if the current is reduced to 6.0 mA? P = I 2 R = (0.006 A) 2 (4000 Ω); P =144 mW 27-32. A certain wire has a resistivity of 2 x 10-8 Ω m at 200C. If its length is 200 m and its cross section is 4 mm2, what will be its electrical resistance at 1000C. Assume that α = 0.005/C0 for this material. [ ∆t = 1000C – 200C = 80 C0 ] ρ l (2 x 10-8Ω ⋅ m)(200 m) R0 = = ; R0 = 1.00 Ω at 200C A 4 x 10-6 m 2 R = R0 + α R0 ∆t = 1.00 Ω + (0.005 / C0 )(1 Ω)(80 C0 ); R = 1.40 Ω 27-33. Determine the resistivity of a wire made of an unknown alloy if its diameter is 0.007 in. and 100 ft of the wire is found to have a resistance of 4.0 Ω. [ D = 0.007 in. = 7 mil ] ρl RA A = (7 mil)2 = 49 cmil; R= ; ρ= A l RA (4 Ω)(49 cmil) ∆= = ; ρ = 1.96 Ω cmil/ft l 100 ft 125
  • 8. Chapter 27. Current and Resistance Physics, 6th Edition 27-34. The resistivity of a certain wire is 1.72 x 10-8 Ω m at 200C. A 6-V battery is connected to a 20-m coil of this wire having a diameter of 0.8 mm. What is the current in the wire? π D 2 π (0.0008 m) 2 ρl A= = = 5.03 x 10-7 m 2 ; R= 4 4 A ρ l (1.72 x 10-8Ω ⋅ m)(20 m) 27-34. (Cont.) R= = ; R = 0.684 Ω A 5.03 x 10-7 m 2 V 6.00 V I= = ; I = 8.77 A R 0.684 Ω 27-35. A certain resistor is used as a thermometer. Its resistance at 200C is 26.00 Ω, and its resistance at 400C is 26.20 Ω. What is the temperature coefficient of resistance for this material? ∆R (26.20 Ω - 26.00 Ω) α= = ; α = 3.85 x 10-4/C0 R0 ∆t (26.00 Ω)(400 C - 200 C) *27-36. What length of copper wire at 200C has the same resistance as 200 m of iron wire at 200C? Assume the same cross section for each wire. [ Product RA doesn’t change. ] ρl ρ1l1 R= ; RA = ρ l ; ρ1R1 = ρ2l2; l2 = A ρ2 ρ1l1 (9.5 x10−8 Ω ⋅ m)(200 m) l2 = = ; l2 = 1100 m ρ2 1.72 x 10-8Ω ⋅ m *27-37. The power loss in a certain wire at 200C is 400 W. If α = 0.0036/C0, by what percentage will the power loss increase when the operating temperature is 680C? ∆R ∆R = α R0 ∆t ; = (0.0036 / C0 )(680 C - 200 C) = 0.173 R Since P = I2R, the power loss increases by same percentage: 17.3 % 126
  • 9. Chapter 27. Current and Resistance Physics, 6th Edition Critical Thinking Problems 27-38. A 150-Ω resistor at 200C is rated at 2.0 W maximum power. What is the maximum voltage that can be applied across the resistor with exceeding the maximum allowable power? What is the current at this voltage? V2 P= ; V = PR = (2.00 W)(150 Ω); V = 17.3 V R 27-39. The current in a home is alternating current, but the same formulas apply. Suppose a fan motor operating a home cooling system is rated at 10 A for a 120-V line. How much energy is required to operate the fan for a 24-h period? At a cost of 9 cents per kilowatt- hour, what is the cost of operating this fan continuously for 30 days? P = VI = (110 V)((10 A) = 1100 W; E = Pt = (1100 W)(24 h) = 26.4 kW h E = (26.4 kW h)(3600 s/h)(1000 W/kW); E = 95.0 MJ kW ⋅ h  $0.08  Cost = 26.4   (30 days); Cost = $53.36 day  kW ⋅ h  *27-40. The power consumed in an electrical wire (α= 0.004/C0) is 40 W at 200C. If all other factors are held constant, what is the power consumption when (a) the length is doubled, (b) the diameter is doubled, (c) the resistivity is doubled, and (d) the absolute temperature is doubled? (Power loss is proportional to resistance R) ρl 1 R= ; P ∝ l; P ∝ ; P ∝ ∆R ∝ ∆T A A 127
  • 10. Chapter 27. Current and Resistance Physics, 6th Edition (a) Double length and double power loss; Loss = 2(40 W) = 80 W (b) doubling diameter gives 4A0 and one-fourth power loss: Loss = ¼(40 W) = 10 W (c) Doubling resistivity doubles resistance, and also doubles power loss: Loss = 80 W *27-40. (Cont.) (d) T = (200 + 2730) = 293 K; ∆T = 2T – T = T; ∆T = 293 K = 293 C0 If absolute temperature doubles, the new resistance is given by: R R = R0 (1 + α∆T ); = 1 + (0.004 / C0 )(293 C0 ) = 2.172; R0 P R = = 2.172; Loss = 2.172(40 W); Loss = 86.9 W P0 R 0 This of course presumes that resistivity remains linear, which is not likely. *27-41. What must be the diameter of an aluminum wire if it is to have the same resistance as an equal length of copper wire of diameter 2.0 mm? What length of nichrome wire is needed to have the same resistance as 2 m of iron wire of the same cross section? ρl R ρ ρc ρa π D2 ρc ρa R= ; = = const.; = ; A= ; 2 = A l A Ac Aa 4 ( Dc ) ( Dc ) 2 ρ a Dc2 ρc 1.72 x 10-8Ω ⋅ m D = 2 a ; Da = Dc = (2 mm) ; Da = 1.57 mm ρc ρa 2.80 x 10-8Ω ⋅ m ρl ρi li R= ; RA = ρ l = const.; ρ nln = ρi li ; ln = A ρn ρi li (1.72 x 10-8Ω ⋅ m) ln = = ; ln = 1.72 cm ρn (100 x 10-8Ω ⋅ m) *27-42. An iron wire (α = 0.0065/C0) has a resistance of 6.00 Ω at 200C and a copper wire (α = 0.0043/C0) has a resistance of 5.40 Ω at 200C. At what temperature will the two wires have the same resistance? [ Conditions: αiRoi∆ti - αcR0c∆tc = 6 Ω −5.4 Ω = -0.60.Ω. ] 128
  • 11. Chapter 27. Current and Resistance Physics, 6th Edition −0.600 Ω −0.600 Ω ∆t = = ; ∆t = -38.0 C0 α i Roi − α c Roc (0.0065 / C )(6.0 Ω) - (0.0043 / C0 )(5.4 Ω) 0 ∆t = tf – 200 = -38.0 C0; tf = -18.00C 129