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					    Lecture 6 Current and Resistance Chp. 27

•   Cartoon -Invention of the battery and Voltaic Cell
•   Opening Demo - Lemon Battery
•   Warm-up problem
•   Physlet
•   Topics


•   Demos
     – Lemon Battery estimate internal resistance
     – Ohms Law demo on overhead projector
     – T dependence of resistance
     – Three 100 Watt light bulbs
•   Puzzles
     – Resistor network figure out equivalent resistance
Loop of copper wire

                        Nothing moving;
                        electrostatic equilibrium



                      E 0



                         Now battery forces charge
                         through the conductor. We
                         have a field in the wire.


                      E0
           What is Current?
            It is the amount of positive charge that moves past a certain point per
            unit time.
                                                                  Q Coulomb
                                                              I                 Amp
                                                                   t    sec ond
                                       I
                                 + +                         Copper wire with
                           +
                            +
                                 + +
                                                             voltage across it
                                + +


       A                        L

                          vt  L
   Drift
   velocity of          Q = charge per unit volume x volume
   charge               nq x Avt
                        Q = nqAv t
Density of electrons                   1.6 x 10-19 C
                                             Divide both sides by t.
                                                   Q
                                              I       nqAv
                                                   t
 Question         What causes charges to move in the wire?

                   How many charges are available to move?




Example            What is the drift velocity for 1 Amp of current flowing through
                   a 14 gauge copper wire of radius 0.815 mm?

                              I                        I = 1 Amp
 Drift velocity         vd 
                             nqA                       q = 1.6x10-19 C

                             No                      A = (.0815 cm)2
                         n    = 8.4x1022 atoms/cm3
                                                     = 8.9 grams/cm 3

                               1
                                                       No = 6x1023 atoms/mole
          vd 
                 8.4 1022 1.6 1019   (.0815) 2   M = 63.5 grams/mole

                                                       The higher the density
                                    5 m
                    v d  3.5 10          s
                                                       the smaller the drift
                                                       velocity
 Drift speed of electrons and current
               density

                                        Directions of current i is
                                        defined as the direction
                                        of positive charge.
                                        i  nAqvd
                                             i
                                        J
                                             A
                                        J  nqvd


(Note positive charge moves in
direction of E) electron flow is
opposite E.
Currents: Steady motion of charge and
       conservation of current
                      i = i1 + i2   (Kirchoff’s 2nd rule)




                                            Current is the same throughout all
                                            sections in the diagram below; it is
                                            continuous.
                                            Current density J does vary.
Question: How does the drift speed compare to the instantaneous speed?
     Instantaneous speed  106 m/s
     vd  3.5x10-11 • vinstant
     (This tiny ratio is why Ohm’s Law works so well for metals.)

     At this drift speed 3.5x10-5 m/s, it would take an electron 8 hours to go 1 meter.

Question: So why does the light come on immediately when you turn on the light switch?
      It’s like when the hose is full of water and you turn the faucet on, it immediately comes out the
      ends. The charge in the wire is like the water. A wave of electric field travels very rapidly
      down the wire, causing the free charges to begin drifting.

Example: Recall typical TV tube, CRT, or PC monitor. The electron beam has a speed 5x10 7 m/s. If
the current is I = 100 microamps, what is n?

          I                10 4 A                                    Take A = 1mm2
      n    
         qAv 1.6  10 19 C  10 6  5  10 7 m s                              = (10-3)2
                                                                                = 10-6 m2
        For CRT
        n = 1.2x1013 e/m3 = 1.2x107 e/cm3
                                                                     The lower the density the
        For Copper                                                   higher the speed.
        n = 8.5x1022 e/cm3
What is Resistance?
The collisions between the electrons and the atoms is the cause of resistance and a very
slow drift velocity of the electrons. The higher density, the more collisions.



                                                       field off

                                                               field on



                                           extra distance electron
                                           traveled


                           e-




 The dashed lines represent the straight line tracks of electrons in between collisions
 •Electric field is off.
 •Electric field is on. When the field is on, the electron traveled drifted further to B I.
Ohm’s Law
Want to emphasize here that as long as we have current (charge moving) due to an applied potential, the
electric field is no longer zero inside the conductor.
              I
                                          Potential difference
     •                  •
     A                  B                 VB - VA = E L                   Constant E
              L                          I = current  E L (Ohm’s Law)



 True for many materials – not all. Note that this is an experimental
 observation and is not a true law.                                                 Best conductors
 Constant of proportionality between V and I is known as the                        Silver – w/ sulpher
 resistance. The SI unit for resistance is called the ohm.
                                                                                    Copper – oxidizes
                                                                                    Gold – pretty inert
  V = RI              R = V/I                Ohm = volt/amp
                                                                                    Non-ohmic materials
                                                                                    Diodes
     Demo: Show Ohm’s Law
                                                                                    Superconductors
A test of whether or
not a material
satisfies Ohm’s Law




                       V = IR                    Here the slope depends on
                       I = V/R                   the potential difference.
                                                 Ohm’s Law is violated.
                       Slope = 1/R = constant
                       Ohm’s Law is satisfied.
Resistance: What is it? Denote it by R

• Depends on shape, material, temperature.
• Most metals: R increases with increasing T
• Semi-conductors: R decreases with increasing T

Define a new constant which characterizes materials.
                                                                             L
Resistivity          R
                         A
                                                A
                                                                        R     
                         L                                 L                 A

Demo: Show temperature dependence of resistance
        For materials  = 10-8 to 1015 ohms-meters
Example: What is the resistance of a 14 gauge Cu wire? Find the resistance per
unit length.
R cu   1.7  10 8 m
                             8  10 3  m
L   A 3.14 (. 815  10 3 ) 2
              Build circuits with copper wire. We can neglect the resistance of
              the wire. For short wires 1-2 m, this is a good approximation.
 Example Temperature variation of resistivity.
   = 20 [ 1 +  (T-20) ]
      L
  R              can be positive or negative
      A

  Consider two examples of materials at T = 20oC.
                   (-m)            (k-1)            L              Area             R (20oC)
  Fe               10-7              .005              6x106 m        1mm2(10-6m2)     60,000 
  Si               640               -.075             1m             1 m2             640 
          Fe – conductor       -   a long 6x106 m wire.
          Si – insulator       -   a cube of Si 1 m on each side
Question: You might ask is there a temperature where a conductor and insulator are one
          and the same?
Condition: RFe = RSi at what temperature?                Set RFe = RSi and solve for T
                  L                               L
       Use   R           = 20 [ 1 +  (T-20) ] A              T – 20 = -196C   (pretty low temperature)
                  A
                                          6  10 6 m
       RFe = 10-7 -m [1 + .005 (T-20)]                          T = -176C
                                          10 6 m 2
                                         1m
       RSi = 640 -m [ 1 + .075 (T-20)]
                                        1m 2
Resistance at different Temperatures
Cu        .1194       .0152          conductor
Nb        .0235       .0209          impure
C         .0553       .069           semiconductor
T         = 300 K      = 77 K
Power dissipation resistors
                      I      Potential energy decrease


 U = Q (-V)
U Q
       (V )
t   t
 P = IV         (drop the minus sign)

 Rate of potential energy decreases equals rate of thermal energy increases in resistor.
 Called Joule heating
 • good for stove and electric oven
 • nuisance in a PC – need a fan to cool computer

 Also since V = IR,
 P = I2R or V2/R          All are equivalent.


 Example: How much power is dissipated when I = 2A flows through the Fe resistor of
             R = 10,000 .            P = I2R = 22x104  = 40,000 Watts
Batteries
A device that stores chemical energy and converts it to electrical energy.

Emf of a battery is the amount of increase of electrical potential of the charge when it
flows from negative to positive in the battery. (Emf stands for electromotive force.)

Carbon-zinc = Emf = 1.5V
Lead-acid in car = Emf = 2V per cell
(large areas of cells give lots of current)         Car battery has 6 cells or 12 volts.


 Power of a battery = P

  P = I          is the Emf
  Batteries are rated by their energy content. Normally they give an equivalent measure
  such as the charge content in mA-Hrs
                                                 milliamp-Hours

 Internal Resistance                             Charge = (coulomb/seconds) x seconds

 As the battery runs out of chemical energy the internal resistance increases.
 Terminal Voltage decreases quickly.               What is terminal
                                                   voltage?
 How do you visualize this?
What is the relationship between Emf, resistance, current, and terminal
voltage?
Circuit model looks like this:
                     I
                                        •
    r
                                  R            Terminal voltage = V
                                                         V = IR (decrease in PE)
                                        •

           = Ir + IR                 = I (r + R)
            - Ir = V = IR            I = /(r + R)



           The terminal voltage decrease =  - Ir as the internal resistance r increases or
           when I increases.
Example: This is called impedance matching. The question is what value of load resistor R
         do you want to maximize power transfer from the battery to the load.


        E = current from battery
  I
       rR
 P = I2R = power dissipated in load
                                            P
           2
          E
  P             R
       (r  R) 2
  dP
     0                                                      ?
                                                                                  R
  dR

  Solve for R

   R=r


   You get max. power when load resistor equals
   internal resistance of battery.
   (battery doesn’t last long)
Demo show lemon or apple batteries
          Estimate internal resistance by adjusting R until the terminal
          voltage is half of the open circuit voltage. r = R
          For lemons rlemons  3600 .



           - Ir = V
open circuit            terminal
voltage                 voltage


                       
    Note: When V           , r=R
                        2
Combination of resistors
   Resistors in series

                                       V = R1I + R2I = (R1 + R2)I
                                       Requiv = R1 + R2




  Resistors in parallel
                                        Voltages are the same, currents add.
                                         I = I1 + I2
                                         V/R = V/R1 + V/R2
                                                        1/R = 1/R1 + 1/R2

                                         Requiv = R1R2 /(R1 + R2)



   Demo: 3 lightbulb resistor puzzle
Equivalence of two versions of Ohm’s Law

 E = J             V = RI


 LE = L J

  V = L J

 R =  (L/A)

 L = AR

 V = ARJ = RJA

                 I
     V = RI
                                Warm-up Set 6
1. HRW6 27.TB.08. [119812] Current is a measure of:
  amount of charge that moves past a point per unit time
  force that moves a charge past a point
  energy used to move a charge past a point
  speed with which a charge moves past a point
  resistance to the movement of a charge past a point

2. HRW6 27.TB.14. [119818] In a conductor carrying a current we expect the electron drift speed to be:
   about the same as the average electron speed
   much less than the average electron speed
   less than the electron speed at high temperature and greater than the electron speed at low temperature
   less than the electron speed at low temperature and greater than the electron speed at high temperature
   much greater than the average electron speed

3. HRW6 27.TB.49. [119853] You buy a "75 W" light bulb. The label means that:
   the bulb is expected to "burn out" after you use up its 75 watts
   none of these
   no matter how you use the bulb, the power will be 75 W
   the bulb was filled with 75 W at the factory
   the actual power dissipated will be much higher than 75 W since most of the power appears as heat
What is the electric field in a sphere of uniform
distribution of positive charge. (nucleus of protons)
                                         Q
                                   
                                        4 3
                                          R
      R                                 3
                                                qenc
                          E
                                       EdA 
                                                0
                      r
                                            4 r 3
                                            
                                   E 4 r 
                                         2      3
                                                 0
                          
                                       r     Q
                                    E              r
                                     30 4 0 R 3




                              
    Cq            V
                        qS
               V        0A
    V  E0S             0A
S                 C
    E0                  S
         0
     qA

            q
    E0 
           0A


           E0
    E             Cq
                        V
         E 0S           q
    V             C
                       V0
           V0
    V             C  C0
           
        Find the capacitance of a ordinary piece of coaxial cable (TV cable)

                                                2k                                 metal braid
      For a long wire we found that        Er              outer insulator         with - q
                                                 r                                                 signal
      where r is radial to the wire.                                                    •          wire
                                           • r
                    a                  a                    a                                      radius a
                                           dr
       Va  Vb    E. ds  2k             2k ln r             radius b
                                                                                                   with + q
                                                                                             Insulator
                                           r
                   b            b
                                                            b                                (dielectric )
       E. ds  Edscos180  Eds  Edr

      ds = - dr because path of integration is radially inward                              a = 0.5 mm
                             a                                            C 2 0
         Va Vb  2k ln                                                   b
                                                                                            b = 2.0 mm
                             b                                            L ln a            2
                  or
                       b                          Q b                     C 6 10 11 6 10 11
            V  2k ln                 V        ln                                 
                       a                  2 0 L a                       L   ln 4      1.38
                                                      
          Va is higher than Vb                  Q2 0 L
                                       C  QV 
                                                                          C                   0 (air)
                                                                             43 PF m
            QL                                 Qln b                    L
                                                   a
                           
                  1                              2 0 L                  C
                                                                             86 PF m          =2
            k                             C
                 4 0  air                      ln b
                                                     a
                                                                          L
                         
Capacitance of two concentric spherical shells
    -q            Integration path
                                                  a         a
                      E
        +q                       Va  Vb    E. ds    Edr
                                                  b         b
    a
                                 E. ds  Edscos180  Eds  Edr
                                     ds = - dr
              b
                                          a           a              a
                                                                         dr
                       Va  Vb    Edr    kq/r dr  kq 
                                                     2

                                          b           b              b   r2
                                      a
                                1     1 1        b  a)
                          V  kq  kq(  )  kq(
                                rb    a b          ab

                                                ab              ab
                          C  q /V                     4 0
                                              k(b  a)         ba
             



             
Model of coaxial cable for calculation of capacitance


                                      Outer metal braid




                                     Signal wire

				
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