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Transcript
106 PHYS - CH17 - Part1
CHAPTER 17
DIRECT CURENTS
106 Phys
Dr. Abdallah M. Azzeer
1
INTRODUCTION
¾Most applications of electricity and magnetism involve moving charges or
electrical currents in conductors.
¾Direct currents (dc) are produced when a conducting path exists between
the terminals of a battery or a dc generator.
¾These devices tend to maintain a constant potential difference between their
terminals and convert other kinds of energy, such as chemical or mechanical
energy, into electrical energy.
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106 PHYS - CH17 - Part1
¾An alternating current (ac) is produced by an ac generator, which has a
terminal potential difference that alternates in sign at some characteristic
frequency.
17.1 ELECTRIC CURRENT
The electric current in a wire is the rate at which charge moves in the wire
A
l
What causes the charges to move? ¨ applied electric field
If ΔQ ≡ net charge crosses the shaded cross-sectional area in time Δt,
The average current is
106 Phys
I =
ΔQ
Δt
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Dr. Abdallah M. Azzeer
The instantaneous current is
I = lim
t →0
ΔQ dQ
=
Δt
dt
The S.l. current unit is the ampere (A). 1 A =1 C/s
milliampere (mA); 1 mA = 10-3 A.
Example 17.1
I=0.5 A, t=1 hr
An electrochemical cell consists of two silver electrodes
placed in an aqueous solution of silver nitrate. A
constant 0.5-A current is passed through the cell for one
hour.
(a) Find the total charge transported through the cell in
coulombs and in multiples of the electronic charge.
(b) Each electron reaching the cell discharges one
positively charged silver ion, which is then deposited on
the negative electrode (cathode). What is the total mass
of the deposited silver? (The atomic mass of silver is
107.9 u.) (1 u = 1.66×10-27 kg)
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-
+
Ag+
AgNO3
Electrochemical cell
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106 PHYS - CH17 - Part1
(a)
Since the current is constant,
ΔQ = I Δt =(0.5)(1 hr) =(0.5)(3600 s)=1800 C
The number of silver ions transported through the cell and deposited in I hour.
N = ΔQ/e = 1.13×1022 ( where e is the electron charge)
(b) The mass of the deposited silver is the number of atoms N times the mass of an atom.
m = (1.13 x 1022)(107.9 )(1.66 x 10-27 ) = 2.02 X 10-3 kg
Conventionally the current in a conductor is assumed to be in the direction of motion of
positive charges
Metallic conductors ¨ the moving charges are electrons.
106 Phys
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Dr. Abdallah M. Azzeer
In metals, some of the electrons become detached, leaving behind positively charged
ions. The heavy ions form a regularly spaced crystalline lattice and vibrate about their
equilibrium positions with an energy and an amplitude that increases with the
temperature. The detached conduction electrons move randomly among the ions. In the
absence of an applied electric field, the average charge flow in any direction is zero.
When an electric field is applied, the electrons acquire an average drift velocity opposite
to the field, and there is a net current.
Relation of the current in a wire to the density of conduction electrons and their drift velocity v.
n = number of electrons per unit volume,
The total number of electrons in a volume V is n V.
A
volume V = Al .
v
Total # of electrons in the wire = nAl
Total # of charges in the wire = enAl
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106 PHYS - CH17 - Part1
The time needed for all of them to pass through the end of the segment shown is;
Δt = l / v
so the magnitude of the current is;
I =
ΔQ enAl
=
= enAv
l /v
Δt
Note that a positive charge flow in one direction is equivalent to a negative flow in the
opposite direction.
Example 17.2
A copper wire is often used to wire household electrical outlets. Its radius is 1 mm = 10-3 m. If
it carries a current of 10 A, what is the drift velocity of the electrons? (Metallic copper has one
conduction electron per atom, the atomic mass of copper is 64 u, and the density of copper is
8900 kg m-3.)
I
, A=π r 2
enA
n equals the number of atoms per unit volume.
The number of atoms per unit volume × the mass of one atom M = the mass of a
unit volume of copper, which is its density d.
v =
106 Phys
Dr. Abdallah M. Azzeer
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i.e d = nM , M = mass of one atom
n = d/M = 8.38×1028 m-3
∴ v = I / neA= 2.37×10-4 m/s
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106 PHYS - CH17 - Part1
17.2 RESISTANCE
The electrical resistance R of a conductor is the potential difference V between
its ends divided by the current
R=
V
I
The S.l. unit for resistance is the ohm (Ω )
Ohm is a volt per ampere.
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For many materials V ∝ I
R = constant
for a length of copper wire with its ends at an
adjustable potential difference V. The current
increases linearly with V and reverses
direction when V is reversed
Materials with a constant resistance are said
to obey Ohm's law and are called ohmic
conductors.
Some conductors R varies with the
magnitude or direction of the applied
potential difference ¨ nonohmic conductors
e.g vacuum tubes and transistors
Example 17.3 (READ)
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106 PHYS - CH17 - Part1
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106 Phys
Resistivity
The resistance of a conductor depends on; its
(i)
geometry (length and cross sectional area)
(ii)
properties of the material (resistivity)
Put two identical wires side by
side, the current doubles and
hence the resistance is halved
¨
A
R ∝ 1/A
l
If we halve the length l, the
potential change and hence the
resistance are also halved
¨
R∝l
R=
ρl
A
ρ≡ Resistivity, depends only on the properties of the material.
The S.I. unit for resistivity is the ohm meter (Ω m).
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106 PHYS - CH17 - Part1
1
ρ
≡ σ ≡ conductivity
The S.I. unit of conductivity is the ohm-1m-1 .
Dr. Abdallah M. Azzeer
106 Phys
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Example 17.4
Find the room temperature resistance of a copper wire 100 m long with a radius of 1
mm.
According to Table 17.1, the resistivity of copper at room temperature (20° C) is
1.72 × 10-8 Ω m.
R=
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ρl
A
=
ρl
= 0.547 Ω
πr 2
Dr. Abdallah M. Azzeer
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106 PHYS - CH17 - Part1
MORE example:
(a) Calculate the resistance per unit length of a nichrome wire with radius 0.321 mm.
A = π r2 = π (0.321× 10−3 m )2 = 3.24 × 10−7 m 2
R = (ρ)
Ω
l
R ρ 1.5 × 10−6 Ω⋅ m
⇒
= =
= 4.6
A
l A 3.24 × 10−7 m 2
m
(b) If a potential difference of 10 V is applied across a 1 meter length of this wire, find
the current in the wire.
Ω ⎞
⎛
⋅ ( 1 m ) = 4.60 Ω
R = ⎜ 4.60
m ⎟⎠
⎝
V
10.0 V
=
= 2.20 A
I =
R
4.60 Ω
106 Phys
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Dr. Abdallah M. Azzeer
17.3 ENERGY SOURCES IN CIRCUITS
The source that maintains the current in a closed circuit is called a source of
electromotive force (emf)
Any devices that increase the potential energy of charges circulating in circuits
Water flow as analogy
are sources of emf
for electric current
The emf may originate from a chemical reaction as
in a battery or from mechanical motion such as in a
generator.
The EMF E of a battery or generator is defined as
the work done per unit charge by the nonelectrical
forces.
S.l. unit of E is a J/C or volt, which is also the unit
of the electric potential.
106 Phys
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106 PHYS - CH17 - Part1
The Electric Battery
Volta discovered that electricity could be created if
dissimilar metals were connected by a conductive
solution called an electrolyte.
This is a simple electric cell.
A battery transforms chemical energy into
electrical energy. Chemical reactions within the
cell create a potential difference between the
terminals by slowly dissolving them. This potential
difference can be maintained even if a current is
kept flowing, until one or the other terminal is
completely dissolved.
Several cells connected together make a battery,
although now we refer to a single cell as a battery
as well.
106 Phys
Dr. Abdallah M. Azzeer
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Electric Current
By convention, current is defined as flowing from + to -. Electrons actually flow
in the opposite direction, but not all currents consist of electrons.
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106 PHYS - CH17 - Part1
Potential in a resistor loop
V
c
I
b
V0
V0
R
a
d
a
c
b
d
a
As (positive) charge moves from c to d in the resistor it loses electrical potential
by an amount ΔV because Vb = Vc and Va = Vd.
What goes up, must come down!
voltage rises and drops
loses
must be equal around
12 J/C
any closed loop.
ε = 12 J/C
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Dr. Abdallah M. Azzeer
106 Phys
From a to b: V increases by E(ΔV = E).
From b to c: V is constant (ΔV = 0); R=0.
From c to d: V decreases by IR (ΔV = -IR).
From d to a: V is constant (ΔV = 0).
E
When a charge goes around a closed path
and returns to the starting point, its
potential energy must return to its original
value, since the conservative electrical
forces do no net work.
ΣΔV = 0
¨ E - IR =0
E = IR
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106 PHYS - CH17 - Part1
Circuits (Conceptual)
106 Phys
Dr. Abdallah M. Azzeer
21
Example 17.5
AA dry
dry cell
cell with
with an
an EMF
EMF of
of 66 VV isis connected
connected to
to aa light
light bulb
bulb with
with resistance
resistance 44 ohms.
ohms. Find
Find
the
the current.
current.
READ
Voltage sources
Batteries (chemical reactions)
Solar cells (conversion of solar energy to voltage output)
Generators (conversion of mechanical energy to voltage
output)
Real batteries and generators usually have
various dissipative effects associated with
them that can be thought of as internal
resistances
This resistance behaves as though it were
in series with the emf.
106 Phys
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106 PHYS - CH17 - Part1
Real emf’s often have energy losses during operation so that the ideal voltage E
is reduced. This energy loss can be modeled by an internal resistance r
represented by the equivalent circuit:
In this case the current
is then determined by
E = (r + R)I
I=
E
r+R
106 Phys
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Dr. Abdallah M. Azzeer
Example 17.6
A battery with an EMF of 6 V and an internal resistance r = 2 Ω is connected to an
R = 4 Ω light bulb (Fig. 17.7). Find
(a)the current;
(b)the terminal potential difference of the battery.
b
E= 6 V, r =2 Ω , R = 4 Ω
(a)I = ? , (b) ΔV = ?
E
+
r
a
R
E = IR+Ir ¨ I = E/(R+r) = 1 A
Vab = E – Ir = 4 V
Vab = IR = 4 V
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106 PHYS - CH17 - Part1
Example 17.7
Two
Two batteries
batteries are
are connected
connected to
to aa resistor
resistor as
as shown
shown in
in Fig.
Fig. 17.8.
17.8. Find
Find (a)
(a) the
the
current;
current; (b)
(b) the
the terminal
terminal potential
potential difference
difference of
of each
each battery.
battery.
(a) Since the batteries are
connected so that their polarities
are opposite, the direction of the
current is determined by the larger
EMF and is counterclockwise in
Fig. 17.8.
c
E1=6 V
+ r1=2 Ω
d
R=3 Ω
The potential increases by E2, and
in the other EMF the potential
decreases by E1.
I
b
E2=18 V
+ r2=1 Ω
a
E
E22 –– E
E11 == I(r
I(r11+r
+r22+R)
+R) ¨
¨ II == E
E22 –– E
E11 // (r
(r11+r
+r22+R)
+R) =2
=2 A
A
106 Phys
Dr. Abdallah M. Azzeer
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(b)
Vab = E2 - Ir2 = 18 V - (2 A)(1 Ω) = 16 V
Vdc = E1 + Ir1 = 6 V + (2 A)(2 Ω) = 10 V
Note that the terminal potential difference Vdc is greater than the EMF in this
case. In calculating the potential difference between points c and d, we are
proceeding opposite to the current. Thus we see a potential rise in the
resistance as weII as in the EMF. The terminal potential difference exceeds the
EMF only when the current is driven "backward" through the battery by another
larger EMF.
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106 PHYS - CH17 - Part1
1 7.5 SERIES AND PARALLEL RESISTORS;
KIRCHHOFF'S RULES
There are ways in which resistors can be connected so that the circuits
formed can be reduced to a single equivalent resistor
Two rules, called Kirchhoff’s rules, can be used instead
Statement of Kirchhoff’
Kirchhoff’s Rules
• Junction Rule
The sum of the currents entering any junction must
equal the sum of the currents leaving that junction
¾ A statement of Conservation of Charge
• Loop Rule
The sum of the potential differences across all the
elements around any closed circuit loop must be zero
¾ A statement of Conservation of Energy
27
Dr. Abdallah M. Azzeer
106 Phys
Mathematical Statement of Kirchhoff’s Rules
• Junction Rule: I
3
Σ Iin = Σ Iout
I4
Iout
I2
I1 - I2 - I3 - I4 + I5 = 0
I5
Iin
I1
I1 = I2 + I3
• Loop Rule:
∑ ΔV
=0
closed
loop
i.e the sum of resistance times the current for every
part in the circuit = the sum of emf’s in the circuit
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106 PHYS - CH17 - Part1
More about the Loop Rule
••
Traveling
Traveling around
around the
the loop
loop from
from aa to
to bb
••
In
In (a),
(a), the
the resistor
resistor is
is traversed
traversed in
in the
the
direction
direction of
of the
the current,
current, the
the potential
potential
across
across the
the resistor
resistor is
is –– IR
IR
••
In
In (b),
(b), the
the resistor
resistor is
is traversed
traversed in
in the
the
direction
opposite
of
the
current,
the
direction opposite of the current, the
potential
potential across
across the
the resistor
resistor is
is is
is ++ IR
IR
••
In
In (c),
(c), the
the source
source of
of (electromotive
(electromotive
force)
force) emf
emf is
is traversed
traversed in
in the
the direction
direction
of
of the
the emf
emf (from
(from –– to
to +),
+), and
and the
the
change
change in
in the
the electric
electric potential
potential is
is +ε
+ε
••
In
In (d),
(d), the
the source
source of
of emf
emf is
is traversed
traversed in
in
the
the direction
direction opposite
opposite of
of the
the emf
emf (from
(from
++ to
to -),
-), and
and the
the change
change in
in the
the electric
electric
potential
potential is
is -ε
-ε
Dr. Abdallah M. Azzeer
106 Phys
29
More about the Loop Rule
ΔV = ΔVab = Vb −Va
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106 PHYS - CH17 - Part1
Loop Rule, final
E1 +
I2
R2
I1
R1
Σ E = Σ IR
R3
R5
I5
E2
+
¾When
¾When apply
apply LOOP
LOOP RULE,
RULE, assume
assume specific
specific
direction
for
the
current
,
as
shown
direction for the current , as shown
¾Potential
¾Potential difference
difference between
between connection
connection
is
is POSITIVE
POSITIVE in
in this
this DIRECTION
DIRECTION and
and
NEGATIVE
NEGATIVE in
in the
the opposite
opposite .. Also
Also the
the same
same
for
for all
all EMF.
EMF.
I3
I4
R4
I1R1+I2R2+I3R3-I4R4-I5R5= E1 – E2
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