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httpslidepdfcomreaderfullemg08 119

1

Electric Currents and Resistance

Todayrsquos menu

bull The Electric Battery

bull Electric Current Defined

bull Resistance and Ohmrsquos Law

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2

The Electric Battery

bull Establishes electric current in a conductor bull Chemical energy stored in the battery istransformed to electric energy of the charge

carriersbull Contains two oppositely charged electrodes

(terminals)

bull Potential difference exists between the terminals

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4

Electric CurrentWhen charges of a like sign move a current is established Let

us define current more precisely Suppose the charges move

perpendicular to a surface of area A as shown

A

+

++

+

+

I

The current is the rate that charge flows through this surface

If ∆Q passes through A in time ∆t the the average current

during that period is

av

Q I

t

∆=

∆

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5

A Average current+

+

+

+av

Q I t

∆=∆+

Instantaneous current I

dQ I

dt

=Units Coulombssec or Ampere (A)

1C 1A1s

=

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7

Potential and Current

Georg Simon Ohm (1787-1854) in metal wires the current

flowing in the wire was proportional to the potential difference

between the two ends

I V prop ∆

bull Compare gravity

bull Water flowing down a hill -- the greater the change in

height (greater change in gravitational potential) the swifter

the water flows

bull For electrical current the greater the electrical potential

difference (or voltage) the greater the current

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8

Ohmrsquos Law

I V prop ∆Ohm Voltage and current were proportional

Proportionality constant is the resistance Ohmrsquos law

orV I V IR R∆= ∆ =

R is the resistance Different for different materialsand for different shapes of wire

1V1

1A= Ω

Units of R VoltsAmperes or Ohms (Ω)

Resistor Symbol

Conducting Wire (negligible resistance)

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9

Not all materials follow Ohmrsquos law

Those that do are called ohmic

Those that do not are called nonohmic

I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

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10

Resistivity

bull A wire of length l and cross-sectional area Abull Experiments show the resistance R is proportional to l

bull Experiments show the resistance R inversely proportional to A

l R

Aρ=

A

l

bull The constant of proportionality ρ is known as the resistivity

bull Different for different materials (copper aluminum iron etc)

bull Temperature dependent

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11

l R A

R or A l ρ ρ= = A

l

Resistivity has units of Ohmmeters ( minus

m)

The reciprocal of the resistivity is called the conductivity

1σ

ρ= Conductivity has units of (

minus

m)-1

bull Resistivity ρ not to be confused with mass density orcharge density

bull Conductivity σ not to be confused with surface charge

density

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12

Current and Potentiala b

I

bull Current flows from high potential to low potentialbull So the electrical potential at a is higher than at b

bull The current at a and b is however the same

P i l i R i L

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13

Potential in a Resistor Loop

-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

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14

Microscopic Description of Current

bull In a conductor charges (electrons) are always in motionbull They move with speeds of about 106 ms

bull If no electric field is applied the net electron velocity is zero

bull Then there is no net charge flowbull When electric field is applied to the conductor it causes the

electrons to drift in the opposite direction to the field

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15

Collisions in the conductor causes the electrons to reach a steady

average drift velocity vd

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16

bull Can relate the drift velocity to the

current

bull In time t the electrons move a

distance x = vd t

bull Wire has a cross-sectional area A In

time t electrons in the volume A

x = Avd t pass through the cross-section A

of the wire

bull Total charge that passes is Q = ( ofcharges)(charge per particle)

bull If n is the number of charges N per

unit volume (n = NV )

( )( ) ( )( ) d Q N e nV e n Av t e∆ = minus = minus = minus ∆

d ∆Q I - ne Av ∆t

= =And the current is

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17

d

∆Q I - ne Av

∆t

= =

Define a current density as the current per unit cross-sectional

area J = IA so

d J - n ev=

or in vector formd J - n ev=

r

r

The minus sign indicates that the direction of positive current isopposite the drift velocity of the electrons

E l 1

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18

Example 1

bull 12-gauge copper wire in a typical residential building has a cross-

sectional area of 331 x 10-6 m2bull It carries a current of 10 A

bull What is the drift speed of the electrons The conduction electron

density for copper is 849 x 1028 electronsm3

d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus

minus minus= = timestimes times times

R Oh rsquo L

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19

Return to Ohmrsquos Law

l

A

I

E

V aV b

For wire have ∆V = El and I = JA Using Ohmrsquos Law ∆V = IR can

use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =

Ohmrsquos Law in terms of Current density and Electric Field

8112019 emg08


2

The Electric Battery

bull Establishes electric current in a conductor bull Chemical energy stored in the battery istransformed to electric energy of the charge

carriersbull Contains two oppositely charged electrodes

(terminals)

bull Potential difference exists between the terminals

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8112019 emg08


4




A

+

++

+

+

I




av

Q I

t

∆=

∆

8112019 emg08


5

A Average current+

+

+

+av

Q I t

∆=∆+


dQ I

dt


1C 1A1s

=

8112019 emg08


8112019 emg08


7





I V prop ∆




the water flows



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8

Ohmrsquos Law





1V1

1A= Ω


Resistor Symbol


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9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


8112019 emg08


4




A

+

++

+

+

I




av

Q I

t

∆=

∆

8112019 emg08


5

A Average current+

+

+

+av

Q I t

∆=∆+


dQ I

dt


1C 1A1s

=

8112019 emg08


8112019 emg08


7





I V prop ∆




the water flows



8112019 emg08


8

Ohmrsquos Law





1V1

1A= Ω


Resistor Symbol


8112019 emg08


9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


4




A

+

++

+

+

I




av

Q I

t

∆=

∆

8112019 emg08


5

A Average current+

+

+

+av

Q I t

∆=∆+


dQ I

dt


1C 1A1s

=

8112019 emg08


8112019 emg08


7





I V prop ∆




the water flows



8112019 emg08


8

Ohmrsquos Law





1V1

1A= Ω


Resistor Symbol


8112019 emg08


9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


5

A Average current+

+

+

+av

Q I t

∆=∆+


dQ I

dt


1C 1A1s

=

8112019 emg08


8112019 emg08


7





I V prop ∆




the water flows



8112019 emg08


8

Ohmrsquos Law





1V1

1A= Ω


Resistor Symbol


8112019 emg08


9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


8112019 emg08


7





I V prop ∆




the water flows



8112019 emg08


8

Ohmrsquos Law





1V1

1A= Ω


Resistor Symbol


8112019 emg08


9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


7





I V prop ∆




the water flows



8112019 emg08


8

Ohmrsquos Law





1V1

1A= Ω


Resistor Symbol


8112019 emg08


9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


8

Ohmrsquos Law





1V1

1A= Ω


Resistor Symbol


8112019 emg08


9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


9




I I

∆V

Nonohmic

V IR∆ =

∆V

Ohmic

8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


10

Resistivity



l R

Aρ=

A

l




8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


11

l R A

R or A l ρ ρ= = A

l


m)


1σ


minus

m)-1



density

8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


12


I



P i l i R i L

8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


13


-+∆V 0

R b c

da

I

a b c dd

V

∆V 0

i i i i f

8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


14






8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


15



8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


16


current


distance x = vd t




of the wire







8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


17

d

∆Q I - ne Av

∆t

= =


area J = IA so

d J - n ev=


r

r


E l 1

8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


18

Example 1





d J - n ev=

d

J I v

ne neA= =

4

28 3 19 6 2

10Cs222 10 ms

(849 10 m )(160 10 C)(331 10 m )d v minus


R Oh rsquo L

8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


8112019 emg08


19


l

A

I

E

V aV b


use earlier results

( )

V I R

l El JA J l

A

ρ ρ

∆ = = =

1 J E E σ

ρ

= =r r r

or in vector form1

J E E σ

ρ

= =


emg08

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