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= CHAPTER� 27= �� CURRENT� AND� RESISTANCE

=
SECTIONS 27.1 AND 27.2� ELECTRIC CURRENT= AND
�� OHM�S L= AW

=	=
A conductor of length, L and cross sectional area, A has an electric curr= ent
density J A / m². An electric field, E exists along the conductor axis. The
Vol= tages at either end of the conductor are V₁ and V₂<= br>

e

e

e

J<= /b>

E=

e

e

e

V1

V2

=	<= /td>
v_d � drift velocity of = charge carriers
n � number of charge carriers per unit volume
e �= charge per charge carrier

IAV� �=3D= � ΔQ / Δt�� C / s

1 C= /S =3D 1 Ampere , 1 A

ins= tantaneously,� I� =3D� dQ / dt

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=
NOT= E:� CURRENT, I, BY OUR DEFINITION IN CI= RCUITS IS
CONVENTIONAL CURRENT �A FLOW OF POSITIVE CHARGE

(1)= �� J� =3D� I / A�� Amps / m²� where� J� =3D�� J ,� the current density

=
it can be shown that:

(2)= �� J� =3D� n e v_d

Emp= irically, it has been shown that for many substances:

(3)= � J� =3D� σ E� =3D� E / ρ

=
where:� σ � conductivity of the substa= nce;

�� ρ =3D resistivity = of the substance

`= 1;� =3D� 1/σ

(3)= is the vector� form of Ohm�s Law

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= Ohm�s Law in Terms of Voltage, ΔV

E=

A

J<= /b>

V1

L

V2

[= 6;V� =3D� V₁ � V₂

and= since σ =3D 1 / ρ :

E� =3D�� E;�� J� =3D�� J

E� =3D ΔV / Δx� =3D= ΔV / L

=
Fro= m (1) and (2):
I / A� =3D� σ ΔV / = L or:

now = define:

then= :

(4)<= /font>

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=	<= /td>
NOTE:� R is the resistance of a particular section of wire or any

�� particular object. `= 1; is the resistivity of the material of

�� which the object is mad= e. See Table 27.1.

Units: R� - Ohms, Ω�� ρ� -� Ohm m or Ω m�� I =3D� Volt / Ohm =3D Amp

Let= ting �ΔV =3D� V_R�� (4) is:<= /font>

(5)= �� I� =3D� V_R / R�� where V_R is the voltage across R

(5)= is Ohm�s Law in terms of V and R

What= is this ?

v<= /b>

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= EXAMPLE� -� ( Ch 17 #13 Serway & Faughn)

=	=
Calculate the diameter of a 2 cm length of tungsten filament in a small
lightbulb if its resistance, R =3D 0.05 Ω.for the resistivity of tungsten us= e
= 61; =3D 5.6 x 10^-8 Ω m.

r

l

R� =3D� ρ l / A

d = =3D� 2 r� =3D� 1.69X10^{-4 m}

<= /td>
A� =3D� ρ l / R
π r²� =3D ρ l / R=

r2� =3D�� ρ l / π R

r2� =3D� 5.6x10^-8 Ω m� (2x10^-2 m) /� π (0.05 Ω)

<= /td>
r²� =3D� 7.13x10^-9 m²
r = =3D� 8.44X10^-5 m

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=
EXAMPLE � Sec. 27.2� Problem # 11

�� For tungsten, ρ =3D 5.6x10-8 Ω m

R = =3D ρ l / A =3D (5.6 x 10^-8 m) ( 1.5 m) / 0.6x10^-6 m^2<=
/sup>

R = =3D 0.14 Ω

I = =3D V / R� =3D 0.9 Volts / 0.14 Ω� =3D� 6.4 A

Wha= t Law is this ?

OHM�= S LAW

Is = Ohm�s Law truly a Law of Physics ?

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=
SECTION 27.6� -� ELECTRICAL ENERGY AND
�� POWER

I

<= /td>	<= /td>
P� -� power gained

by the charges as

they pass through

the = element

V1

V

V2

I

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I

P� =3D� I (V₂ � V₁)

V1

=
dir= ection of I �

from V₁�� V_= 2

V

V2

I

<= /td>	=	=	<= /td>
The magnitude of the power exchanged,

P is= :

�� P� =3D� I V

I� -� magnitude of current through the element

V� -� magnitude of voltage across the element

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= Units :�� P� =3D� I V

(C/s= ) (J/C)

=3D= � J/s =3D Watt, W

<= /td>
(6) = P =3D I V

�For a resistor of resistance, R with voltage V_R:

=
P� =3D� I V_R and from Ohm�s Law:

�I� =3D� V_R / R�� then:

V

R

I

=	<= /td>
Power is always lost by charge

carriers in a resistor ( lost

to = thermal energy)

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<= /td>		=		<= /td>
EXAMPLE:� A light bulb uses 60 W of average p= ower

when connected to a 120 Volt source.
(a)	What is the bulb�s resistance, R

(b)	What current, I flows in the bulb

I

(a)<= /font>

R

V

(b)<= /font>

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= A unit of energy, the kwh,� kilowatt-hour=

By d= efinition:

= 16;E� =3D� (10³ W) 1 hr� =3D� 10³ J/s<= span style=3D'mso-spacerun:yes'>� (3.6x10³ s)

= 16;E� =3D� 3.6x10⁶ J� = =3D� 1 kwh

[= 6;E in� J ?

= 16;E� =3D� 6.22x10^8�J

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<= /td>	<= /td>	=	=
= HOMEWORK �

�� 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
�Voltage will the resistor be damaged= ?

I

R

V

Answer:� V_max =3D� 44.7 Volts

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CHAPTER 27 CURRENT AND RESISTANCE

SECTIONS 27.1 AND 27.2 ELECTRIC CURRENT AND

= ; &n= bsp; OHM’S LAW<= /div>

A conductor of length, L and cross sectional area,= A has an electric current

density J A / = m2<= /span>. An electric field, E exists along the conductor axis. The <= /div>

Voltages at either end of the conductor are V1<= /span> and = V2<= /span>

E

J

e

e

e

e=

e

e

= vd – drift velocity of charge carriers

n – number of charge carriers per unit volum= e

e – charge per charge carrier

I -<= span style=3D'mso-spacerun:yes'> electric current. I is the rate at which charge passes a point in= the circuit.

IAV <= /span> =3D<= span style=3D'mso-spacerun:yes'> ΔQ / Δt C / s

instantaneously,&= nbsp; I =3D dQ / dt

1 C/S =3D 1 Ampere , 1 A

V1

V2

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= NOTE: CURRENT, I, BY OUR DEFINITION IN CIRCUITS IS

CONVENTIONAL CURRENT –A FLOW OF POSITIVE CHARGE

(1) J =3D I / A Amps / m= 2<= /span> whe= re J =3D J , the current density

it can be shown that:

(2) J =3D n e vd

Empirically, it has been shown that for many substances:

(3) J =3D σ E =3D E<= /b> / ρ

= where: σ – conductivity of the substance;

        = ;      ρ =3D resistivity of the substance

ρ =3D 1/σ

(3) is the vector form of Ohm= 217;s Law

= If both equations are valid, it should be possible to derive equation
(3) from equation (2) by relatin= g 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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Ohm’s Law in Terms of Voltage, ΔV

A

J

E

V1

V2

L

ΔV =3D V1 – V2

E =3D    E;       J =3D    J

E =3D ΔV / Δx =3D ΔV / L

= From (1) and (2):

I / A =3D σ &#= 916;V / L or:

and since σ =3D 1 / ρ :

now define:

then:

(4)

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= 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 27.1.

=

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 voltage across R

(5) is Ohm’s Law in terms of V and R

What is this ?

v

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EXAMPLE - ( Ch 17 #13 Serway = & Faughn)

Calculate the diameter of a 2 cm length of tungsten filament in a small

lightbulb if its resistance, R =3D 0.05 Ω.for the resistivity of tungsten use

ρ =3D 5.6 x 10-8 &#= 937; m.<= span lang=3DEL style=3D'mso-bidi-font-family:"Times New Roman";font-size:83%; mso-special-format:lastCR;display:none'>

l

r

R =3D ρ l / A

= A =3D ρ l / R

π r2 =3D ρ= l / R <= /span>

r2<= /span> =3D   ρ l / π R

r2 =3D 5.6x10-8 &#= 937; m (2x10-2 m) / &#= 960; (0.05 Ω)

r2 =3D 7.13x10-9 m2

r =3D 8.44X10-5= m

d =3D 2 r =3D 1.69X10-4 m

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EXAMPLE – Sec. 27.2 Problem # 11

    For tungsten, ρ =3D 5.6x10= -8 &#= 937; m

R =3D ρ l / A =3D = (5.6 x 10-8 m) ( 1.5 m) / 0.6x10-6 m2&#= 13;

R =3D 0.14 Ω

I =3D V / R = =3D 0.9 Volts / 0.14 Ω =3D 6.4 A

What Law is this ?

OHM’S LAW

Is Ohm’s Law truly a Law of Physics ?=

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= SECTION 27.6 - ELECTRICAL ENERGY AND
        = ;            &n= bsp;            POWER=

= Conventional current, I passes through a circuit element with a voltage

difference, V =3D V2 – V1<= /span>

V2

V1

V
=

e

l

e

m

e

n

t

I

I

= P - power gained

by the charges as

they pass through

the element

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V2

V1

V
=

e

l

e

m

e

n

t

I

I

P = =3D I (V2 R= 11; V1<= /span>)

= direction of I –

from V1        = ;    V2 =

If V2<= /span> > V1 := I toward a voltage

        = ;             rise, P > 0

        = ;             charges gain energy

        = ;         (conventional current)

If V2 < V1 : I toward a voltage

        = ;            drop, P < 0

        = ;             charges lose energy

        = ;          (conventional current)

The magnitude of the power exchanged,

P is:

        = ;            &n= bsp; P =3D I V

I - magnitude of current through the element

V - magnitude of voltage across the element

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Units :   P =3D I V <= /u>

(C/s) (J/C)

=3D J/s =3D Watt, W

= (6) P =3D I V

For a resistor of resistance, R wi= th voltage VR:

V

R

I

P =3D I VR and from Ohm&#= 8217;s Law:

I= =3D VR / R   then:

P in (7) and (8) represent the= 3;

rate at which charge carriers

lose energy on passing through R

Power is always lost by charge

carriers in a resistor ( lost

to thermal energy)

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quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation

EXAMPLE: A light bulb uses 60 W of average power

when connected to a 120 Volt source.= 3;

(a)What is the bulb’s resistan= ce, R

(b)What current, I flows in the bulb=

R

V

I

(a)

(b)