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Файл:Steady electric current. Tutorial
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41
Using the expressions of Ohm's law for a closed circuit (2.19), we find
the external resistance R
r
I
ε
R −=)2
,
and obtain another expression for the efficiency
ε
Irε
η
−
=)3
.
Performing a substitution of numerical data.
Answer: η = 0.7 = 70 %.
Problem 2. Three galvanic elements with ε = 1.6 V and internal re-
sistance of 0.12 оhm are connected: a) in parallel; b) in series.
Find the currents if the external resistance is 0.5 оhm. At what value
of external resistance will the currents be the same and will not depend on
the method of connecting the elements?
Given
Solution
ε = 1.6 V
r = 0.12 Ω
R = 0.5 Ω
Drawing electrical circuits
I – ?
We find the currents based on expressions (2.35) and (2.36) for parallel and series connection of current sources, and also using Ohm's law (2.19)
for a closed circuit
1)
2)
ε
We substitute numerical data.
Answer: Ia = 3.0 A; Ib = 5.6 A.
42
By equating currents Ia and Ib, formulas (1) and (2) and solving the
resulting equation relative to R, we obtain the value of external resistance at
which currents Ia and Ib will be equal to each other.
Answer: R = r = 0.12 Ω.
Problem 3. Find the currents in a branched circuit, where the batteries
ε1 = 2 V, ε2 = 1 V, the resistor resistances R1 = 1 kΩ, R2 = 0.5 kΩ, R3 = 0.2 kΩ,
R4 = 0.2 kΩ. Consider the current sources ideal, that is, their internal resistances are zero.
Given
SI
Solution
ε
1
= 2 V
ε
2
= 1 V
R1 = 1 kΩ
R2 = 0.5 kΩ
R3 = 0.4 kΩ
R4 = 0.2 kΩ
1000 Ω
500 Ω
400 Ω
200 Ω
Drawing electrical circuits
I1 – ?
I2 – ?
I3 – ?
We designate the nodes, arbitrarily indicate the directions of the currents and indicate the positive direction of the circuit traversal.
We compose equations according to Kirchhoff's rules for node A and
circuits I and II.
I1 – I2 + I3= 0
1) –I1R1 – I2R
2
= ε1
I2R2 + I3(R3 + R4)= – ε
2
Using the Cramer method (determinant method) (2.28) we find the
currents I1, I2, I3. For example:
2)
=
=
43
To calculate the remaining I2 and I3 currents in the upper (substituted)
determinant, we replace the second and third columns with free terms, respectively. We obtain I2 = –1.57 mA, I3 = –0.36 mA.
The values of all currents turned out to be negative, which means that
the directions of the currents indicated in the diagram should be changed to
the opposite. Current I1 flows out of node A, current I2 flows into node A,
current I3 flows out of node A. The algebraic sum of the currents in the node
is zero.
Answer: I1 = 1.21 mA; I2 = 1.57 mA; I3 = 0.36 mA.
Problem 4. Copper and steel wires of the same length and cross-section are connected to a circuit: a) in parallel; b) in series.
Find the ratio of the heats released in the wires.
Given
Solution
l1 = l2
S1 = S2
ρ1=1.7∙10
−8
Ω∙m
ρ2=1∙10
−7
Ω∙m
Q1/Q2 – ?
If the conductors (wires) are connected in parallel (a), then the
voltаges on them will be equal; if in series (b), then the currents in them will
be the same. The heat released in conductors is determined by the Joule–
Lenz law (2.39) and for stationary conductors, taking into account Ohm's
law (2.1), is equal to
tUIt
R
U
RtIQ
2
2
===)1
.
The resistance of the wires is found using formula (2.2).
2)
.
2
1
2
1
1
2
2
1
ρ
ρ
Q
Q
:b;
ρ
ρ
Q
Q
:a ==)3
.
Substitute numerical data.
Answer: a) Q1/Q2 = 5,9; b) Q1/Q2 = 0,17.
44
Problem 5. Let us imagine a heating element in the form of two
coaxial cylindrical conductors, the space between which is filled with a conducting medium with a resistivity ρ. The radii of the inner and outer cylinders
are r1 and r2, respectively, the length is l. A constant current I is maintained
between the cylinders. Find the heat released in such a heater for time Δt.
Neglect the end effects.
Solution
Let us use the Joule‒Lenz law in differential form and determine the
heat released in a thin cylindrical layer of radius r and thickness dr, shown
in the right figure
dtdVjdQ
2
)1 = ρ
.
The current density is equal to the current divided by the lateral surface
area of the selected cylindrical layer
lr2π
I
S
I
j
==)2
.
Volume of the selected cylindrical layer
drlr2dV = π)3
.
4)
π
π‧‧‧‧
.
Carrying out the integration we arrive at the answer:
.
Problem 6. The tungsten filament of an electric bulb at 20 °C has a re-
sistance of 35.8 ohm. What will be the temperature of the bulb filament if after
it is connected to a 120 V network, a current of 0.33. A flows through the filament? The temperature coefficient of resistance of tungsten is 0.0046 deg−1.
Incandescent light bulb device.
Given
Solution
t1 = 20 °C
R1 = 35,8 Ω
U =120 V
I = 0,33 A
α = 0,0046 deg−1
Incandescent light bulb device: 1 – glass flask; 2 – tungsten filament; 3 – molybdenum holders for filament;
4 – electrical inputs; 5 – rod
for fixing molybdenum holders; 6 – light bulb base
t2 ‒ ?
Using Ohm's law, we find the resistance of the tungsten filament of the
I
U
R
2
=)1
)(
101
αt1RR +=)2
).(
202
αt1RR +=)3
.
1
1212
2
αR
tαRRR
t
+−
=)4
lamp in working condition
.
We write down expressions (2.48) for the resistance of a tungsten
filament in a cold and working state
,
We divide equations (2) and (3) by each other, eliminate the unknown
value R0, and find the temperature of the bulb in working condition
We perform substitutions of numerical data and obtain: t2 ≈ 2200 °C.
S e l f - c h e c k q u e s t i o n s
1. What electric current is called a steady current?
2. Formulate and write Ohm's law for a steady current in a metal con-
ductor.
is it measured?
units is it measured?
3. What is called the electrical resistance of a conductor? In what units
4. What is called the electrical conductance of a conductor? In what
45
5. What parameters determine the resistance of a homogeneous metal
conductor of constant cross-section?
6. What is called the specific electrical resistance or resistivity of the
substance? In what units is it measured?
7. What is called the specific electrical conductance or conductivity of
the substance? In what units is it measured?
8. Write Ohm's law in differential form.
9. Write Ohm's law for a non-uniform section of a circuit.
10. Write Ohm's law for a closed circuit.
11. What is a node, branch, contour in a branched chain?
12. Formulate Kirchhoff's first rule. What signs have currents flowing
into and out of a node?
13. Formulate the second Kirchhoff's rule. On what basis are the signs
of the products of currents and resistances and the signs of EMF established?
14. Write down and prove the formula for calculating the resistance of
series-connected resistors.
15. Write down and prove the formula for calculating the resistance of
parallel-connected resistors.
16. How to find the current in a closed circuit with series-connected
current sources?
17. How to find the current in a closed circuit with identical current
sources connected in parallel?
18. Write down and formulate the Joule–Lenz law in integral form.
19. What is called the unit thermal power of a current?
20. Write down and derive the Joule-Lenz law in differential form.
46
47
3 . C LA S SI C A L T H E O R Y O F M E T A L S C O N DU C TI V I T Y
The classical theory of electrical conductivity of metals appeared after the discovery of the electron by Thomson in 1997 in experiments with
cathode rays. The theory is based on the ideas of Riecke (1898), that neutral
atoms of a metal forming a solid are partially dissociated into positive ions
and electrons existing in the form of an “electron gas” (fig. 3.1). The theory
was proposed by Drude in 1900 and later improved by Lorentz, providing an
explanation for the laws of Ohm and Joule–Lenz.
Basic principles of the classical electron theory of metals:
1. Atoms in metals are dissociated into positive ions and electrons.
2. Ions form a crystalline ionic lattice of metals.
3. The set of free electrons in metals has the properties of a monatomic
ideal gas and is called an “electron gas”. In an electron gas, as in an ideal gas,
the mean free path is many times greater than the proper dimensions of the
electrons.
4. Current carriers in metals are free electrons of the “electron gas” –
conduction electrons. Atoms and ions of metals do not participate in charge
transfer and are not current carriers.
Fig. 3.1. According to the electron gas model, electrons move
between the stationary ions of the metal crystal lattice
The characteristics of the electron gas are the concentration n, the
mean free path
λ
~
, and the average velocity
V
~
of thermal motion. The concentration of the electron gas, assuming that one electron is split off from
each atom as a result of dissociation, will be equal to the concentration of
48
atoms in metals. This number is on average n = 1028−10
29 m−3
. The mean free
path is equal to the interatomic distance λ = 10
−10
m.
The average velocity of thermal motion of electrons can be calculated
using the conclusions and formulas of the molecular kinetic theory of gases.
The arithmetic average velocity of translational motion of ideal gas molecules is equal to
πm
8kT
V =
~
, (3.1)
where k is the Boltzmann constant, T is the absolute temperature, and m is
the mass of a gas molecules.
Replacing the mass of a gas molecule with the mass of an electron in
formula (3.1), we obtain that at room temperature (~300 K) the average velocity of thermal motion of electrons in an electron gas will be:
. (3.2)
For comparison, let us estimate the average velocity
U
~
of the ordered
movement of electrons, which they acquire under the influence of an electric
field. To do this, we will relate the average velocity
U
~
of the ordered move-
ment of electrons and the current density
j
in a conductor. Let us consider
an elementary small element of a conductor (fig. 3.2), whose cross-sectional
area is equal to dS, length
dtUdl
~
=
,
and the element is oriented in the direction of the current density vector
j
.
Fig. 3.2. Elementary small element of the conductor
The length of a conductor element is equal to the distance traveled by
electrons under the influence of an electric field during time dt. During this
time, a number of electrons dN will pass through the cross section dS, equal
to the product of the electron gas density n and the volume of the conductor
element dV, and the charge dq will be transferred
49
dtUdSnedVnedNedq
~
===
. (3.3)
The current in this element of the conductor will be equal to
UdSne
dt
dq
dI
~
==
, (3.4)
and finally, we obtain an expression connecting the current density with the
velocity of ordered movement of the electron gas
.Une
dS
dI
j
~
==
(3.5)
Let's perform calculations, taking copper as an example as an excellent
conductor of electricity. Maximum permissible current density for copper
j
max
= 107 A/m2, current carrier concentration n = 1029 m−3. Substituting these
values into formula (3.5), we obtain:
. (3.6)
Thus, the average velocity of ordered motion of electrons is significantly less
than the average velocity of thermal motion.
The ionic crystal lattice presents obstacles to the flow of current and
is in thermal equilibrium with the electron gas. Ions with a positive charge
are held in the nodes of the metal crystal lattice by electrical interactions with
the electron gas, just as the electron gas is held by the lattice ions. Such a mutual bond (III Newton's law) is called a metallic bond.
The electrical resistance of metals and the release of Joule heat during
the flow of current are caused by collisions of free electrons with ions in the
nodes of the crystal lattice. Let us consider these processes in more detail,
and obtain Ohm's law and the Joule–Lenz law within the framework of classical electron theory.
3.1. O h m ' s l a w in t h e c l a s s i c a l e l e c t r o n
t h e o r y o f m e t a l s
When colliding with lattice ions, electrons of the electron gas con-
stantly lose and regain the velocity of ordered motion, moving between collisions under the influence of an electric field
E
with acceleration
50
m
Ee
m
F
a ==
. (3.7)
The average velocity of ordered motion of electrons during uniformly
accelerated motion between two successive collisions is equal to
2
UU
U
max0
+
=
~
, (3.8)
where U0 = 0;
τaU
max
~
=
;
τ
~
is the s the mean free path time of electrons.
The mean free path time
τ
~
is determined by the free path length
λ
~
and
the average velocity
V
~
of thermal motion of electrons in the electron gas
V
τ
~
~
~
λ
=
. (3.9)
Therefore, the average speed of ordered motion is equal to
V2m
λEe
U
~
~
~
=
. (3.10)
Substituting (3.10) into (3.5), we obtain an expression for the current
density
E
V2m
λne
V2m
λeE
nej
2
==
~
~
~
~
. (3.11)
Comparing (3.11) with Ohm's law in differential form (2.9), we obtain
an expression for the specific electrical conductivity
V2m
λne
2
~
~
=σ
. (3.12)
The classical electron theory of metals gives proportionality of current
density and field strength and gives an expression for conductivity.
3.2. J o u l e – L e n z l a w i n t h e c l a ss i c a l e l e c t r o n
t h e o r y o f m e t a l s
When colliding with lattice ions, electrons lose the velocity of ordered
motion they have gained U
max
and transfer the kinetic energy acquired be-
tween collisions to the lattice.
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