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Slide 1
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CHAPTER
17 - CURRENT
AND RESISTANCE
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SECTION 17.1 ELECTRIC CURRENT
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SECTION
17.2 - CURRENT AND DRIFT SPEED<=
br>
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<=
/td>
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A net movement of charge occurs =
in a
medium due to the
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mot=
ion of charge
carriers (say electrons) The average
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electric curren=
t, I in the medium is defined as:=
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whe=
re Δq is
the amount of charge passing through a point in
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time interval, =
Δt.
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charge / time =3D C / S =3D Amp, A
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Δq can be =
found
from: Δq =3D
I Δt
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Slide 1
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/td>
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/td>
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/td>
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/td>
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E=
XAMPLE –
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A 0.5 A DC current flows in a light=
.
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) |
How=
much charge
passes through the light in 1 second ?
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) |
=
HOMEWORK – How many electrons wo=
uld
pass through the
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light in 1 second ?
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![](3D"v3_slide0014_image002.gif") |
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The=
Ampere was
defined first
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in the 1820’s b=
ased on the
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mag=
netic force
between two
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cur=
rent carrying
wires. The Coulomb
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was=
then defined
as an A s.
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(a) Δq =3D=
I Δt =3D
(0.5 C/s) 1 s
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(b) answer: no.=
of
electrons =3D 3.13 x 1018
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/td>
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bat=
teries charge
at full voltage is rated in A =
h
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(Amp hours)
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Slide 1
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SECTIONS 17.4
- RESISTANCE AND OHM’S =
LAW
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AND
RESISTAIVITY
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=
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A
conductor of length, L and cross sectional area, A has an electric curr=
ent
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density
J A / m2. An electric field, E exists along the
conductor axis. The
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tages at
either end of the conductor are V1 and V2<=
br>
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<=
/td>
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vd
– drift velocity of =
charge
carriers
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n –
number of charge carriers per unit volume
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e –=
charge per
charge carrier
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I -
electric current. I is the rate at which charge passes a point in
the circuit.
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1 C=
/S =3D 1
Ampere , 1 A
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ins=
tantaneously, I
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Slide 1
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| =
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The=
drift velocity is actually very small. La=
rge
currents result from
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the huge number=
of charge
carriers involved
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Slide 1
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NOT=
E: CURRENT, I, BY OUR DEFINITION IN CI=
RCUITS
IS
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CONVENTIONAL
CURRENT –A FLOW OF POSITIVE CHARGE
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(1)=
J
=3D I / A Amps / m2 where
J =3D J , the current density
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it
can be shown that:
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(2)=
J
=3D n e vd
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Emp=
irically, it
has been shown that for many substances:
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where: σ – conductivity of the substa=
nce;
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ρ =3D resistivity =
of the
substance
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(3)=
is the
vector form of Ohm’s Law
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![](3D"v3_slide0003_image01=) |
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Slide 1
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=
Ohm’s Law in
Terms of Voltage, ΔV
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![](3D"v3_slide0004_image011.gif") |
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E =3D ΔV / Δx =3D=
ΔV /
L
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![](3D"v3_slide0004_image013.gif") |
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Fro=
m (1) and
(2):
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I /
A =3D
σ ΔV / =
L or:
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![](3D"v3_slide0004_image014.gif") |
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Slide 1
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<=
/td>
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NOTE: R is the resistance of a particular
section of wire or any
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particular object. `=
1; is
the resistivity of the material of
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which the object is mad=
e. See
Table 17.1 p 577
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Units:
R - Ohms, Ω ρ
- Ohm m or Ω m I =3D
Volt / Ohm =3D Amp
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Let=
ting ΔV =3D
VR (4) is:<=
/font>
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(5)=
I
=3D VR / R where VR is the voltage
across R
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(5)=
is Ohm’s
Law in terms of V and R
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![](3D"v3_slide0005_image017.=) |
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Slide 1
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Calculate
the diameter of a 2 cm length of tungsten filament in a small
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lightbulb
if its resistance, R =3D 0.05 Ω.for the resistivity of tungsten us=
e
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=
61; =3D 5.6 x
10-8 Ω m.
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d =
=3D 2 r
=3D 1.68 x10-4 m
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/td>
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A =3D ρ
l / R
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π
r2 =3D ρ l / R=
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r2 =3D
5.6x10-8 Ω m
(2x10-2 m) / π
(0.05 Ω)
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/td>
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r2 =3D
7.13x10-9 m2
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r =
=3D 8.44x10-5 m
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Slide 1
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SECTION
17.7 - ELECTRICAL ENERGY AND
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POWER
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![](3D"v3_slide0007_image01=) |
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/td>
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/td>
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P -
power gained
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the charges as
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the =
element
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Slide 1
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dir=
ection of I
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<=
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The magnitude
of the power exchanged,
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P is=
:
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P =3D
I V
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I -
magnitude of current through the element
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V -
magnitude of voltage across the element
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(6) =
P =3D I V
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For a resistor of resistance, R with
voltage VR:
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P =3D
I VR and from Ohm’s Law:
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=3D VR / R then:
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<=
/td>
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Power
is always lost by charge
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carriers
in a resistor ( lost
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to =
thermal
energy)
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EXAMPLE: A light bulb uses 60 W of average p=
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Slide 1
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16;E =3D
(103 W) 1 hr
=3D 103 J/s<=
span
style=3D'mso-spacerun:yes'> (3.6x103 s)
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3.6x106 J =
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Slide 1
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/td>
| <=
/td>
| =
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=
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=
HOMEWORK –
|
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A 200 Ω resistor can safely
absorb 10 W of Power without damage.
|
The=
resistor is
connected to a regulatable DC power supply. At what
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Voltage will the resistor be damaged=
?
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![](3D"v3_slide0013_image027.=) |
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Answer: Vmax =3D 44.7 Volts
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Slide 1
CHAPTER 17
- CURRENT AND RESISTANCE
=
;
SECTION 17.1&=
nbsp;
ELECTRIC CURRENT
SECTION 17.2&=
nbsp;
- CURRENT AND DRIFT S=
PEED
&nbs=
p;
A net movement of charge occurs in a medium due to the
=
motion of <=
u>charge
carriers (say electrons) The
average
electric current, I in the medium is defined as:
I =3D
Δq / Δt
where Δq is the amount of cha=
rge
passing through a point in
time interval, =
Δt.
SI un=
its of
I
![](3D"slide0001_image028.gif")
charge
/ time =3D C / S =3D Amp, A<=
/span>
Δq can be found from: Δq =3D
I Δt
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Slide 1
EXAM=
PLE
–
A 0.5 A DC current flows in a
light.
(a)How much charge passes through the light in 1 sec=
ond
?
(b) =
HOMEWORK – How many electrons would pass through th=
e
light in 1
second ?
![](3D"slide0014_image029.gif")
![](3D"slide0014_image030.gif")
![](3D"slide0014_image031.gif")
![](3D"slide0014_image032.gif")
![](3D"slide0014_image033.gif")
![](3D"slide0014_image034.gif")
![](3D"slide0014_image035.gif")
![](3D"slide0014_image036.gif")
(a) Δq =3D I [=
6;t =3D (0.5 C/s) 1 s
(b) answer: no. of electrons =3D 3.13 x 10=
18=
batteries charge at full voltage is rated in A h
(Amp hours)
The Ampere was defined first
in the 1820’s based on the
magnetic force between two
current carrying wires. The Coulomb
was then defined as an A s.
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Slide 1
SECTIONS 17.4 -
RESISTANCE AND OHM’S LAW
=
; &n=
bsp;
AND
RESISTAIVITY
=
A conductor of leng=
th, L
and cross sectional area, A has an electric current
density J A / m2. An electric field, EVoltages at either =
end of
the conductor are V1 and V2
![](3D"slide0002_image037.gif")
E
![](3D"slide0002_image038.gif")
J
e
e
e
e
e
e
=
vd – <=
/b>drift velocity of c=
harge
carriers
n – number of
charge carriers per unit volume
e – charge per
charge carrier
I -
electric current. I is the rate at which charge passes a point in t=
he
circuit.
IAV <=
/span> =3D
&=
#916;Q
/ Δt C / s<=
/span>
instantaneously, I
=3D dQ / dt
1 C/S =3D 1 Ampere =
, 1 A
V1
V2
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Slide 1
![](3D"slide0015_image039.gif")
E
J
e
e
e
e
e
e
V1
V2
e
e
e
![](3D"slide0015_image040.gif")
![](3D"slide0015_image041.gif")
![](3D"slide0015_image042.gif")
![](3D"slide0015_image043.gif")
![](3D"slide0015_image044.gif")
![](3D"slide0015_image045.gif")
e
vd
vt
The “drift velocity”, vd res=
ults
from the application of the electric field,
E and
represents an average drift<=
/span> of the + charge carriers in the direction
of E. It is this vd tha=
t causes
the electric current density, J. In addition
to v=
d, the charge carriers have “thermal
velocity”,vt tha=
t is
random in direction
and is very large in magnitude (at room temperatu=
re)
compared to vd
The thermal random motion of the charge carriers =
leads
to collisions between
the carriers and the conductor’s atoms. It =
is
these collisions that account
for the resistance of the conductor.
The drift
velocity is actually very small. Large currents result from
the huge number of charge carriers involved
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Slide 1
=
NOTE: CURRENT, I, BY OUR DEFINITION IN
CIRCUITS IS
CONVENTIONAL CUR=
RENT –A FLOW OF
POSITIVE CHARGE
(1)
J =3D I / A Amps / m=
2 where J
=3D J , the current density
![](3D"slide0003_image046.gif")
![](3D"slide0003_image047.gif")
it can be shown
that:
<=
/span>
(2)
J =3D n e vd
Empirically, it has=
been
shown that for many substances:
=
where: σ – conductivity of the substance;
<=
/div>
=
;
ρ =3D resistivity of the substance&=
#13;
(3) is the vector form of Ohm’s Law
=
If both equations a=
re
valid, it should be possible to derive equation
(3) from equation
(2) by relating the drift
velocity, vd to the
electric field and =
some
other factors using classical electricity and
classical mechanics=
. This
proved not to be true
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Slide 1
Ohm’s Law =
in
Terms of Voltage, ΔV
=
![](3D"slide0004_image=)
A
J
E
V1
V2
L
![](3D"slide0004_image049.gif")
![](3D"slide0004_image050.gif")
![](3D"slide0004_image051.gif")
![](3D"slide0004_image052.gif")
![](3D"slide0004_image053.gif")
=
From (1) and (2):&#=
13;
I / A =3D
&=
#963;
ΔV / L or:
![](3D"slide0004_ima=)
and since σ =3D 1 / `=
1; :
now define:<=
/div>
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Slide 1
=
NOTE: R is the resistance of a particul=
ar
section of wire or any
=
;
particular object. ρ is the resistivity of the material of
=
;
which the object is made. See Table 17.1 p 577
<=
/span>
Units: R - Ohms, Ω ρ -
Ohm m or Ω m I
=3D Volt / Ohm =3D Amp
Letting ΔV =3D VR (4) is:
(5) =
;
I =3D VR / R where VR is the volta=
ge
across R
(5) is Ohm’s =
Law in
terms of V and R
What is this ?
![](3D"slide0005_image062.gif")
v
![3D"Collapse](3D"v3_outline_collapse.gif")