resistance

iClicker Quiz





(1) I have completed at least 50% of the reading and study-

guide assignments associated with the lecture, as indicated on

the course schedule.



a) True b) False

In static equilibrium, E = 0 inside a conducting material.



Now consider non-equilibrium situations where E  0 and

V  0 in a conductor.



A non-zero E-field inside a conductor drives a non-zero

current density J, which requires a continous source of

flowing electric charge (e.g. a battery or power supply).



When the electric current in a conductor is constant in time,

we do not say that the charges present are in equilibrium,

but instead say that they are in a “steady state”.

I   J  dA

I = current or charge flux through a defined area

scalar quantity (coulombs/sec)



J = current density

vector quantity (coulombs/sec/m2)



Analogous to E-field flux: Φ 

 E  dA

Electric-fields vs flow-related quantities

carrier density flux density Flux (current)

 

 (current density)

  JA

J  v



Fluid flow kg/m3 kg/m2/sec kg/sec





Coulombs/m2/sec Coulombs/sec

Electricity flow Coulombs/m3

(Amps/m2) (Amps)



Joules/m2/sec Joule/sec

Energy flow Joules/m3

(Watts/m2) (Watts)



Electric field

Electric fields Fluxons/sec

Fluxons/m3 Fluxons/m2/sec

(Faraday’s analogy) (Volt·m)

(N/C = Volts/m)



Fluxon (made it up myself) = kgm3/Csec  The flow analogy requires imagination.

Rank the four regions in order of current magnitude.

Now define the positive direction to be to the right.



Rank the four regions in order of signed current.

 c  nc qc (carriers/m3)(charge/carrier) = charge/m3





J   c vd  nc qc vd (charge/m3)(m/sec) = charge/sec/area

Resistivity is determined entirely

by microscopic parameters

including the average thermal

collision rate (1/).







 qc  q c

v d  a    m E   m E



 c  c



 q c  n q c 

2

J  nc q c  E  E E

 m  mc

1 mc

 

 nc q c2

Ohm’s Law: linear response

E J Local



E (V / L)

I   J  A  JA  A A

 



 L

V  I     I R Global

 A

L

R

A

Analogous to Poiseuille’s Law of fluid flow (hyperphysics).

Ohmic non-Ohmic

Electric Current Flow Quantities

The resistor The flow

MICROSCOPIC

qc  carrier charge MICROSCOPIC

nc  carrier density vd  drift velocity

mc  carrier mass a  acceleration

  mean time between collisions



LOCAL LOCAL

 c  charge density J  currentdensity

  conductivity E  electric field

  resistivity

GLOBAL

GLOBAL V  potential difference

L  length I  current

A  area

R  resistance

Assume that the charge (qc), mass (m),

density of carriers (nc) and mean collision 1mc

time () in a wire are known.  

 nc qc2

Which of the following quantities can be

qc

determined? vd  E

mc

1) Resistivity ()

2) Resistance (R) J   c vd   E

3) Current (I)

V

4) Current density (J) E J

5) Drift velocity (vd) L

6) Electric field (E)  

I   J  dA  JA

If we are also given the shape of the wire (L and

A), which of these quantities can be determined? V  IR

L

If we are further given the potential difference (V) R

across the wire, which of these quantities can be A

determined?

V L

R 

I A



Quiz: If we double the current through a copper wire of length L and

area A, the resistance will be:

(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered





Quiz: If we double the length and the diameter of copper wire, its

resistance will be:

(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.





Quiz: If we stretch a copper wire to twice its original length at

constant volume, its resistance will be:

(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.

Work to move charge q through a potential difference V:



W  qV



If the average kinetic energy of the carriers doesn’t increase, then the

energy expended has been converted to heat via atomic collisions.

Let power (P) be the rate at which energy is dissipated as heat.



dW dq

P V  VI 

dt dt

 V  V2

( IR) I  I R  V   

2



R R

2

V

PI R 2



R

V L V2

R  P  I 2R 

I A R

Quiz: If we double the resistance of an Ohmic wire of length L and

area A, the power dissipated in the wire will be:

(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.

(Not enough information! We could fix either V or I?)



Quiz: If we double the length of an Ohmic wire of area A and

resistivity , while holding the potential difference V across the wire

constant, the power dissipated in the wire will be:

(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.

Non-Ohmic behavior









Diode

Varistor









http://www.wikipedia.com

Cell Membrane http://www.onsemi.com

mc R

The resistivity of a metal    T

increases as it gets warmer. nc qc2 R

The resistivity of an insulator

or semiconductors decreases

as it gets warmer.









Practical definition of metal/insulator

slope of ρ vs. T > 0 → metal

slope of ρ vs. T < 0 → insulator

Superconductors ( = 0)

Apply 100 V to a 1 meter length of 12-gauge copper wire (r = 1 mm).



Cu :   1.7 10 8   m A   (0.001 ) 2  3.14 10 6 m 2

L (1.7 10 8 )(1.0)

R   5.4  10 3 

A (3.14  10 6 )

V (100 )

I   18 .5 10 3 A

R (5.4 10 3 )

P  I 2 R  (18 .5 10 3 ) 2 (5.4 10 3 )  1.85 10 6 W

I (18 .5  10 3 )

J   5.9  10 9 A/m2

A (3.14  10 6 )

E   J  (1.7 10 8 )(5.9 10 9 )  100 V/m

J J (5.9 109 )

vd     0.433 m/s

c nc qc (8.5 10 )(1.602 10 )

28 19





m vd (9.111031 )(0.433)

   24.6 1015 s

qc E (1.602 1019 )(100)

Apply 100 V to a 1 meter length of 12-gauge copper wire (r = 1 mm).



matomic  63.5 g / mol  m  8.96 g/cm3

cm  3.85 J/g/K H melt  13.14 kJ/mol H vap  300.5 kJ/mol

Tmelt  1356 Tboil  2840 K



m   m A L  (8.96 g/cm3 )(3.14 106 m3 )(1 m)  28.13 g

(28.13 g)

nmol  m / mmolar   0.443 mol

(63.5 g/mol)

Ew1  cm m T1  (3.85 J/g/K)(28.13 g)(1356  300 K )  11.4 kJ

Emelt  nmol H melt  (0.443 mol)(13.14 kJ/mol)  5.82 kJ

Ew2  cm m T2  (3.85 J/g/K)(28.13 g)(2840  1356 K )  161kJ

Evap  nmol H vap  (0.443 mol)(300.5 kJ/mol)  133 kJ

P

t w1   6.1ms tmelt  3.2 ms t w2  87 ms tvap  72 ms

Ew1

So what do you call a length of 12-gauge

copper wire connected across the terminals

of a good 120 V power source?





A fuse!

Cost of power used for home lighting

(50 bulbs, 25 Watts each, 50% use, $0.10 per kWhr)





 25 W 

P  50 bulbs    2250 W

 bulb 



P 25 W

I   0.21Amp/bulb

V 120 V



 $ 0.10  30  24 hr 

cost  (50 %)( 2.25 kW )    $81 / month

 kW  hr  month 

High-Voltage Transmission Line

(V = 120 V vs. V = 120 kV)



Power tran smitted :

P'trans  V ' I '  V I  Ptrans



Resistive power loss :

Ploss  I 2 R



2 2

P'loss I ' R ( P'trans / V ' )  V   1 

2 2

 2       106

 V '   1000 

2

Ploss I R ( Ptrans / V )

High-Voltage Transmission Line (aluminum)

L = 100 km, r = 0.5 cm, Ptrans = 10 MW, V = 120 kV)



Cable cost  (10 5 m)($30/m)  $3 10 6



L (2.82 10 8 )(10 5 )

R   36 

A  (0.005 ) 2





Ptrans (10  10 6 )

I   91 A

V (11 0  10 )

3







Ploss  I 2 R  (91) 2 (36 )  300 kW  30 $ /hr

Ploss (3 105 )

Vloss  IR  (91)(36)  3280 V   3%

Ptrans (110 )

7





Increasing A and cable cost by factor of 3 (adds 2×106 $) will

decrease R and Ploss by factor of 3 (saves 20 $/hr). The reduced

losses pay for the more expensive cable after 11.4 years.

Effective resistance: two resistors in series



R1 R2











R1 and R2 experience the same

current but different voltages.

Largest R has largest V.







V V1  V2 I R1  I R2 Req is larger than

Req     R1  R2 either R1 or R2.

I I I

Effective resistance: two resistors in parallel

R1







R2









R1 and R2 experience the same

voltage but different currents.

Smallest R has largest I.





1 Req is smaller

V V V 1 1 

Req       than smallest

I I1  I 2 V / R1  V / R2  R1 R2 

  of R1 and R2.

Reduction of a resistive network





R R

R R

R R



R R

R R











2R



R/2

 R



2R

Reduction of a resistive network



R R b R R b R

a a











R



R R R

a b

R







R R/3 R











(7/3)R

Reduction of a resistive network







Apply 42 V between a and c.



What is I between a and c?

I=3A



What is Vbc?

Vbc = 6 V



What is I2?

I2 = 2 A

The switch is initially open. When the switch is closed, the current

measured by the ammeter will:

(1) increase (2) decrease (3) stay the same (4) fall to zero.

Compare the brightness of

the four identical bulbs in

this circuit.



R R

R





V









D is in parallel with a zero-resistance wire. The current will

take the zero-resistance path and bypass D altogether.



A and B are in series. So they will burn equally bright.

Together, they see the full battery voltage.

C experiences the full battery voltage, or twice the voltage

experienced by A or B. So C is four times as bright.

If R1 is removed, R2 will glow

(1) more brightly.

(2) less brightly.

(3) same brightness as before.









If R1 is removed, R2 will glow

(1) more brightly.

(2) less brightly.

(3) same brightness as before.





Household devices are wired to run in parallel!

Strings of 50 Christmas lights in

series. Assume ~100 V source

and 25 W power consumption.









http://www.ciphersbyritter.com/RADELECT/LITES/XMSLITES.HTM







What is the resistance of a single bulb?

(1) 2 (2) 4 (3) 8 (4) 10 

Resistance of an object with arbitrary shape









 length  dz 

dR  d     dz

 area  A(z )  (b  a )

2 2



End-to-end:

 L L

R   dR 

 (b 2  a 2 ) 0

dz 

 (b 2  a 2 )





 length  dr dr

dR  d     

 area  A(r ) 2 r L

Inside-out:

 b  b

R   dR 

2 L a

dr  ln  

2 L  a 

Resistance of an object with arbitrary shape

dz

a

b





L



dL dz dz

R   dR        2

A A( z )  r ( z)

L dz   L  L du

     z 0 2

0  [a  (b  a ) z / L]   b  a 

2

u



  L  1 1   L

     

  b  a  a b   ab

Batteries as circuit elements

• A battery or other constant-voltage device (e.g. power suppy) is

usually the energy source in a direct-current (DC) circuit.

• The positive terminal of the battery has higher potential than the

negative terminal.

• The electromotive force (emf, ) of a battery is the open-circuit

voltage between its terminals when no current is flowing.

• Ideally, the battery has no internal resistance of its own.

• We often idealize the wires in a circuit to have zero resistance.





+



Real batteries



• Chemical energy: an electrolyte solution allows negative

ions to flow toward and react with the anode (), while

positive ions flow toward and react with the cathode (+).

• Charge build-up prevents the reaction from proceeding

unless an external circuit allows electrons accumulating at

the anode to return to the cathode.

• Dead when the reactants are used up. Rechargeable if the

anode/cathode reactions are reversible.

• Internal resistance tends to increase with age, use and

multiple recharge cycles.

Internal resistance in non-ideal batteries

Terminal voltage:

V  Vab    I r  



Current:

 V 

I  

Rr R R



Two AA batteries yield a combined  = 3 V. You observe a

terminal voltage of 2.7 V while delivering 300 mA of current to

an ultra-bright flashlight.



What is the load resistance (R) ? (1) 1 (2) 3 (3) 9 (4) 10 



What is the internal resistance of a single AA battery?

(1) 0.5 (2) 1 (3) 2 (4) 3

Typical Alkaline (Zn/MnO2)

1.5 to 1.6 V open circuit, 1.1 to 1.3 V closed circuit.

163 W-hr/kg (590 J/g) – 400 kJ total for AA

0.034 (122.4 C) to 15 A-hr (54 kC) of charge depending on size.

85% Capacity after 4 years of non-use.



Lithium Ion

3.2 V open circuit, 2.5 to 3.0 V closed circuit.

230 W-hr/kg (828 J/g) – 460 kJ total for AA

0.160 (576 C) to 1.4 Amp-hrs (5.04 kC) of charge

95% capacity after 5 years of non-use



NiCd

1.2 V open circuit

50 W-hr/kg (180 J/g) – 140 kJ total for AA

70% capacity after one month of non use (500-5000 cycles)



Zn-Air bus battery:

200 W-hr/kg (720 J/g)

320 kW-hr (1.15 GJ) of energy

Multi-loop circuits









Branch: An independent current path experiences only one current

at a given moment. It may be a simple wire or may also contain one

or more circuit elements connected in series.

Junction: A point where three or more circuit branches meet.



Loop: A current path that begins and ends at the same circuit point,

traversing one or more circuit branches, but without ever passing the

same point twice.

Multi-loop circuits









The circuit above has: 3 branches, 2 junctions, 3 loops.



To solve for 3 unknown branch currents, we need 3 equations.



To get these equations, use all but one (2  1 = 1) junction, and as

many independent loops as needed ( 3  1 = 2).

Kirchoff’s current rule:  I n  0

Current rule: The total current flowing into a junction is zero.

Arrows define positive branch-current directions. A current later

determined to flow opposite its arrow is “negative”.



















 I  I1  I 2  0  I  I1  I 2  0  I  I1  I 2  0

Kirchoff’s Voltage Rule:  Vn  0

Voltage rule: The voltage changes around a loop sum to zero.

Arrows define positive branch-current directions.



V = + for a battery crossing from – to + terminal.



Use V = –I R when crossing a resistor in the positive direction.

Use V = +I R when crossing a resistor in the negative direction.







Alpine loop elevation

   I1R1  0



Single-loop circuit example



 20  I (2000)  30  I (1000)

 I (1500)  25  I (500)

 15  I (5000)  0





I I  15 / 5000  3 mA





VA  Vground  20  I (2000)  30  I (1000)

 10  I (3000)  10  (0.003)(3000)

 19 V

Multi-loop circuit example





Bottom loop:

I2 = I1

 5  I 2 (1)  I 3 (2)  0

Substitute I2 in junction Eq:

I3 = 2I1

 I 2  I1  I 3  0

Substitute I2 and I3 in top loop:

I1 = 5/3  5  I1 (1)  I 2 (1)  5  0



Solve for currents:

I1 = 5/3, I2 = 5/3, I3 = 10/3

Multi-loop circuit example

 6 I1  10  4 I 2  14  0

I 2  1.5I1  6



 2 I 3  10  6 I1  0

I 3  3I1  5



 I1  I 2  I 3  0

I1  (1.5I1  6)  (3I1  5)

 5.5I1  11  0



I1  2 I 2  3 I 3  1