Transcript R 1
TOPIC 18 : ELECTRIC CURRENT AND DIRECT-CURRENT CIRCUITS
18.1 Electrical Conduction 18.2 Ohm’s law and Resistivity 18.3 Variation of resistance with temperature 18.4 Electromotive force (emf), internal resistance and potential difference 18.5 Electrical energy and power 18.6 Resistors in series and parallel 18.7 Kirchhoff’s Laws 18.8 Potential divider 18.9 Potentiometer and Wheatstone Bridge 1
SUBTOPIC : 18.1 Electrical Conduction LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) Describe microscopic model of current. b) Define and use electric current formulae,
I
dQ dt
2
18.1 Electrical Conduction
• • • • Conductors contain many
free electrons
and move randomly .
• If a continuous wire is connected to the terminal of a battery, the potential difference between the terminals of the battery sets up an electric field inside the wire and parallel to it, directed from the positive toward the negative terminal.
Thus free electrons are attracted into the positive terminal (are forced to drift in one direction) .
This direction is in the direction opposite to the field,
E
.
The
velocity of these free electrons is called
drift velocity.
3 battery
18.1 Electrical Conduction • The
drift velocity
v d
of the free electrons is
the mean velocity of the electrons parallel to the direction of the electric field when a potential difference is applied.
• Consider the circuit such as that in figure below, the battery creates an electric field within and parallel to the wire, directed from the positive toward the negative terminal.
wire battery 4
18.1 Electrical Conduction • Thus free electrons at one end of the wire are attracted into the positive terminal, and at the same time, electrons leave the negative terminal of the battery and enter the wire at the other end.
• There is a continuous flow of electrons through the wire that begins as soon as the wire is connected to both terminals.
• However, when the conventions of positive and negative charge were advised two centuries ago, it was assumed that positive charge flowed in a wire.
• For nearly all purposes, positive charge flowing in one direction is exactly equivalent to negative charge flowing in the opposite direction.
• Today we still use the historical convention of positive current when discussing the direction of a current. So when we speak of the current in a circuit, we mean the direction positive charge would flow.
5
5.1 Electrical Conduction
Electric Current, I
• Electric current is defined
as the amount of charge that passes through the wire’s full cross section at any point per unit time
( the rate of charge flow through a conductor).
•The average current , is defined as .
t
•The
instanstaneous current
is defined as
I
dQ dt
•The
steady current
is defined as
I
Q
• Unit of
I
is A (ampere).
t
6
5.1 Electrical Conduction
Example 18.1
A wire carries a current of 1.5 A. a) How much charge flows through a point in the wire in 5.0 s b) How many electrons cross a given area of the wire in 1.0 s ?
7
SUBTOPIC : 18.2 Ohm’s Law and Resistivity LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) State and use Ohm’s Law.
b) Define and use resistivity formulae,
RA l
8
18.2 Resistivity and Ohm’s Law
18.2 Ohm’s Law and Resistivity
Ohm’s Law
• Ohm’s law states that
the potential difference across a conductor, V is directly proportional to the current, I through it, if its physical conditions and the temperature are constant. V
I V
constant
I V
R
V
IR
Ohm's Law
I
9
V V
constant
I V
R I V
18.2 Resistivity and Ohm’s Law
Figure A
I
Figure B
I
• Ohmic conductors are conductors which obey Ohm’s law. Examples: pure metals. (
Figure A
) • Non-ohmic conductors do not obey Ohm’s law. Example: junction diode. (
Figure B
) 10
18.2 Resistivity and Ohm’s Law
Resistivity
• Resistivity is a measure of a material’s ability to oppose the flow of electric current through the material.
• Resistivity is defined as the resistance of a sample of the material of cross-sectional area 1 m 2 • It is a constant value.
RA
• Its formulae is given by .
l
where
l
= length of the conductor (m)
and of length 1m.
A
= area of cross-section of the conductor (m -2 )
• Its unit is Ωm.
• Its value depends on the material.
• All conductors have smaller resistivity.
• Insulators have larger resistivity. 11
5.2 Resistivity and Ohm’s Law
Example 18.2
A wire (length=2.0 m, diameter=1.0 mm) has a resistance of 0.45 Ω. What is the resistivity of the material used to make the wire ?
12
5.2 Resistivity and Ohm’s Law
Example 18.3
What voltage will be measured across a 1000 Ω resistor in a circuit if we determine that there is a current of 2.50 mA flowing through it ?
13
SUBTOPIC : 18.3 Variation of Resistance with Temperature LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) Explain the effect of temperature on electrical resistance in metals and superconductors.
b) Define and use temperature coefficient of resistivity,
α
.
c) Apply resistance
R= R o [1 +α (T-T o )]
.
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18.3 Variation of Resistance with Temperature
Metal
• The
resistance
of a metal (conductor) depends on a) the
nature
( of the material, , resistivity) b) the
size
of the conductor, (
l
, the length and
A
, cross-sectional area) c) the
temperature
of the conductor.
• The resistance of metals increases with increasing temperature. (
T
↑,
R
↑) • As temperature increases, the ions of the conductor vibrate with greater amplitude.
15
18.3 Variation of Resistance with Temperature • More collisons occur between free electrons and ions.
• These electrons are slowed down thus increases the resistance.
• The resistance of a metal can be represented by the equation below
R=R o
[
1+α
∆
R=R o +α
∆
T
(∆
T
)] ,
R-R o
= ∆
R
where
R
= the resistance at temperature
T
,
R o
= the resistance at temperature
T o =
20 o C or 0 o C, = the temperature coefficient of resistance ( o C -1 ) 16
18.3 Variation of Resistance with Temperature • Temperature coefficient of resistance ,
α
is defined as the
fractional change in resistance per Celsius degree.
R
T R o
•
α
is a constant value and it is depends on the material. 17
18.3 Variation of Resistance with Temperature
Example 18.4
A platinum wire has a resistance of 0.50 Ω at 0 o C. It is placed in a water bath, where its resistance rises to a final value of 0.60 Ω . What is the temperature of the bath ? (
α
= 3.93 x 10 -3 o C -1 ) 18
18.3 Variation of resistance with temperature
Example 18.5
A narrow rod of pure iron has a resistance of 0.10 Ω at 20 o C. What is its resistance at 50 o C ? (
α
= 5.0 x 10 -3 o C -1 ) 19
18.3 Variation of Resistance with Temperature
Superconductor
• As the temperature decreases, the resistance at first decreases smoothly.
(
T
↓,
R
↓) • At a certain critical temperature T c (4.2 K for mercury) the resistance suddenly drops to zero.
R R
metal superconductor
T T
Graph of resistance
R
against temperature
T
20
SUBTOPIC : 18.4 Electromotive Force (emf), Internal Resistance and Potential Difference LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) Define emf , ε.
b) Explain the difference between emf of a battery and potential difference across the battery terminals.
c) Apply formulae
V= ε –Ir.
21
18.4 Electromotive Force (emf), Internal Resistance And Potential Difference
What is electromotive force,emf ( ε or ξ) ?
• The e.m.f of a source is the
work done per unit charge
. • The e.m.f of a source is the
p.d across the terminals of the source in open circuit
(no current is flowing,
I
= 0). • The e.m.f of a battery is the
potential difference across its terminal
when it is
not
connected to a circuit. • The e.m.f of a source is defined as the
electrical energy
that generated by a source so that the
charges can
flow from
one terminal to another terminal
of the source through any
resistor
.
• SI unit : Volt (V) 22
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
What is internal resistance?
• In a cell or battery, the negative ions are attracted by anode and the positive ions are attracted by the cathode. • The flow of these ions produces current. • However the collisions between the ions and the recombination of opposite ions reduce the flow of current. This resistance in the cell is called internal resistance,
r
.
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18.4 Electromotive Force (emf), Internal Resistance and Potential Difference • Suppose a battery of emf,
ε
and internal resistance
r
is connected to an external resistor,
R
.
• Total resistance in the circuit is (
R + r
).
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18.4 Electromotive Force (emf), Internal Resistance and Potential Difference • The
e.m.f
of this battery is given as
r
)
IR
ab V ab
V dc V ab = V b – V a =
terminal voltage (potential difference across the battery terminals)
R =
external resistance
r =
internal resistance 25
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference • In a circuit diagram, this symbol represents a
resistor
in a circuit that
dissipates
energy.
electrical • A straight line represents a conducting wire with
negligible resistance
.
26
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
terminal voltage (potential difference across terminals)
ab V ab
Ir
emf potential difference across internal resistance Notes:
a) b)
V ab
< ε
when the battery of emf external circuit with resistance
R
.
ε
V ab
> ε
when the battery of emf other battery.
ε
is connected to the is being charged by
c) V ab
= ε
when the battery of emf
ε
has no internal resistance (
r
=0) and connected to the external circuit with resistance
R.
.
27
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
Example 18.6
A battery with a terminal voltage of 11.5 V when delivering 0.50 A has an internal resistance of 0.10 Ω. What is its emf?
28
18.4 Electromotive Force (Emf), Internal Resistance and Potential Difference
Example 18.7
The battery in a circuit has an emf of 9.0 V. It is attached to a resistor and an ammeter that shows a current of 0.10 A. If a voltmeter across the battery’s terminals reads 8.9 V, what is its internal resistance ? 29
SUBTOPIC : 18.5 Electrical Energy and Power LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) Apply formula
P=IV
and electrical energy,
W=VIt
.
30
18.5 Electrical Energy and Power
• Electric energy is useful to us because it can be transformed into other forms of energy (thermal energy, light).
• According to the conversation of energy , all the energy delivered to the charge carriers by the battery must be lost in the circuit .
• That is, a charge carrier traversing the circuit must lose all the electrical potential energy it gained from the battery when that carrier returns to the negative terminal of the battery.
31
18.5 Electrical Energy and Power •The electrical (potential) energy,
W
is the energy gained by the charge
Q
from a voltage source (battery) having a terminal voltage
V
. •
W= QV
(the work done by the source on the charge)
• But
Q=It
, then
W= VIt
• Unit : Joule (J) • The rate of energy delivered to the external circuit by the battery is called the
electric power
given by,
P P
W t
IV
QV t
,
Q
@
P
It
•Unit : watt ( 1 W = 1J/s) 32
18.5 Electrical Energy and Power • The energy dissipated per second in an electric device (rate of energy dissipated) is given as
P
W t
VIt t
VI
for any device • A passive resistor is a resistor which converts all the
electrical energy into heat
. For example, a metal wire.
P
VI
but
V
IR P
or
P
V
2
R
only for resistor 33
18.5 Electrical Energy and Power
Example 18.8
Calculate the resistance of a 40 W automobile headlight designed for 12 V?
34
18.5 Electrical Energy and Power
Example 18.9
The current through a refrigerator of resistance 12 Ω is 13 A. What is the power consumed by the refrigerator?
35
18.5 Electrical Energy and Power
Example 18.10
An electric iron with a 15-ohm heating element operates at 120 V. How many joules of energy does the iron convert to heat in 1.0 h ?
36
SUBTOPIC : 18.6 Resistors in series and parallel LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) Deduce and calculate effective resistance of resistors in series and parallel.
37
18.6 Resistors in series and parallel
Resistors in Series
R 1 V 1 R 2 V 2 R 3 V 3 I I V
battery
, r
=0
• The properties of resistors in series are given below. o The same current
I
flows through each resistor where
I
I 1
I 2
I 3
38
18.6 Resistors in series and parallel o The sum of the voltages around a circuit loop (that is, the gains and losses with + and - ,respectively) is zero.
V V V
i i
V i
i
V
1
V
2 0
V
3
total potential difference
(Assuming that the connecting wires have no resistance) bur
V
1
IR
1
;
V 2
IR 2
;
V 3
IR 3
;
V
IR eq IR eq
IR 1
IR 2
IR 3 R eq
R 1
R 2
R 3
39 where
R eq
: equivalent (effective ) resistance
18.6 Resistors in series and parallel
I
Resistors in Parallel
I 3 I 2 I 1 R 3 V 3 R 2 V 2 R 1 V 1 V I
• The properties of resistors in parallel are given below. o There is the same potential difference,
V
across each resistor where
V
V 1
V 2
V 3
o Charge is conserved, therefore the total current
I
in the circuit is given by
I
I 1
I 2
I 3
40
18.6 Resistors in series and parallel but
I
1
V R
1
;
I 2
V R 2
;
I 3
V R 3
;
I
V R eq V R eq
V R 1
V R 2
V R 3 1 R eq
1 R 1
1 R 2
1 R 3
41
18.6 Resistors in series and parallel
Example 18.10
2
.
0
12
4
.
0
6
.
0 V
Calculate : a. the total resistance of the circuit.
b. the total current in the circuit.
c. the potential difference across 4.0 resistor.
42
Solution 18.10
2
.
0
18.6 Resistors in series and parallel
12
4
.
0
6
.
0 V
43
18.6 Resistors in series and parallel
Solution 18.10
b. Total current, c. The potential difference across
R 1
=2.0 is Therefore the potential difference across
R 3
is given by =4.0 44
18.6 Resistors in series and parallel
Example 18.11
y x
For the circuits shown below, calculate the equivalent resistance between points x and y.
1
.
0
2
.
0
1
.
0
2
.
0
3
.
0
x
(0.79 Ω)
8
.
0
16
.
0
16
.
0
9
.
0
18
.
0
(8.0 Ω)
20
.
0
6
.
0
45
y
SUBTOPIC : 18.7 Kirchhoff’s Laws LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) State and use Kirchhoff’s Law’s.
46
18.7 Kirchhoff’s Laws
Kirchhoff’s first law (junction/current law)
• It states that the algebraic sum of the currents at any junction of a circuit is zero,
I
0 or
I in
I out
47
Kirchhoff’s second law (loop/voltage law)
18.7 Kirchhoff’s Laws.
• It states that
the algebraic sum of the voltages across all of the elements of any closed loop is zero.
or • It states that
in any closed loop, the algebraic sum of e.m.fs is equal to the algebraic sum of the products of current and resistance.
IR
Sign convention
-ε
-IR R I
+ ε
across battery
+IR
across resistor
48
18.7 Kirchhoff’s Laws
Example 18.12
Using Kirchhoff’s rules, find the current in each resistor.
1 20 V
R 2
= 20 Ω
R 1
= 10 Ω 10 V 49
Solution 18.12
R 2
= 20 Ω
Step
1. Draw current. (arbitrary) 2. Draw loop. (arbitrary) 3.
Apply Kirchhoff’s laws.
I
0 , 18.7 Kirchhoff’s Laws 1 20 V
I
R 1
= 10 Ω 10 V 50
18.7 Kirchhoff’s Laws
Example 18.13
Apply Kirchhoff’s rules to the circuit in figure below and find the current in each resistor.
1
R 1
= 3.0 Ω
R 4
= 3.0 Ω
R 2
= 3.0 Ω
R 3
= 3.0 Ω 51
18.7 Kirchhoff’s Laws
Solution 18.13
1
R 1
= 3.0 Ω
R 4
= 2.0 Ω
I
2 nd KL,
(
IR
)
R 3
= 5.0 Ω
R 2
= 4.0 Ω 52
18.7 Kirchhoff’s Laws
Example 18.14
Find the current in each resistor in the circuit shown below.
R 2
= 4.0 Ω
R 1
= 4.0 Ω 1 10 V
R 3
= 4.0 Ω
I 1
=3.75 A up,
I 2
= 1.25 A left,
I 3
= 1.25 A right 53
18.7 Kirchhoff’s Laws
Example 18.15
A cell of e.m.f. 4.0 V and internal resistance 1.0 Ω is connected in series with another cell of e.m.f. 2.5 internal resistance 0.5 Ω in a closed loop in such a way that the current in the loop is minimum. Draw a circuit diagram to show how the cells are connected and calculate the current. (
I
= 1.0 A) 54
18.7 Kirchhoff’s Laws
Example 18.16
Calculate the currents
I 1
,
I 2
resistance in each battery.
and
I 3
. Neglect the internal
I 1 R 1
1
I 2 ε 1
15 V R 2
0
.
5
ε 2
10 V I 3 R 3
0
.
1
ε 3
3
.
0 V I 1
17
.
69 A
;
I 2
14
.
62 A
;
I 3
3
.
07 A
55
18.7 Kirchhoff’s Laws
Example 18.17
Given
1 =8 V, R 2 =2
, R 3 =3
, R 1 =1
and
I=3 A
. Ignore the internal resistance in each battery.
I 1 ε 1
Calculate a. the currents
I 1
b. the e.m.f.
2
.
and
I 2 .
Ans. : 1 A, 4 A , 17 V
R 1 I 2 ε 2 R 2 I R 3
56
SUBTOPIC : 18.8 Potential divider LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) Explain the principle of a potential divider.
b) Apply equation of potential divider
V
1
R
1
R
1
R
2
V
.
57
18.8 Potential divider
• A potential divider is used to tap a fraction of the voltage supplied by a source of e.m.f.
• Two resistors are connected in series.
• The current flows in each resistor is the same ;
I
V R eq
,
R eq
R
1
R
2
I R 1 V R 2 I I
R 1 V
R 2
• Potential difference across
l
1 or
R
1 is
V 1 V 2
Potential divider circuit
V 1
IR 1 V
1
R
1
R
1
R
2
V
58
• 18.8 Potential divider.
Resistance
R
1 and
R
2 are replaced by a uniform homogeneous wire as shown in figure below.
V a I l V 1 1 c l 2 V 2 I b
R
l A
V
1
R
1
R
1
R
2
V
• Potential difference across
l
1 is
V 1
l 1 l 1
l 2
V
59
18.8 Potential divider
Example 18.18
Resistors of 3.0 Ω and 6.0 Ω are connected in series to a 12.0 V battery of negligible internal resistance. What are the potential difference across the (a) 3.0 Ω and (b) 6.0 Ω resistors ?
4.0 V, 8.0 V
60
SUBTOPIC : 18.9 Potentiometer and Wheatstone Bridge LEARNING OUTCOMES :
At the end of this lesson, the students should be able to : a) Explain the principle of potentiometer and Wheatstone Bridge and their applications.
b) Use related equations. 61
18.9 Potentiometer and Wheatstone Bridge
Potentiometer
• A potentiometer is mainly used to measure potentiometer.
• It consists of a uniform wire.
• Basically a potentiometer circuit consists of a uniform wire AB of length 100.0cm, connected in series to a driver cell with emf
V
of negligible internal resistance.
V
(Driver cell -accumulator)
I I A
G +
V
-
x
(Unknown Voltage)
C I
Jockey
I B
62
18.9 Potentiometer and Wheatstone Bridge • The potentiometer is
balanced
when the jockey (sliding contact) is at such a position on wire AB that there is
no current through the galvanometer
. Thus •
Galvanometer reading = 0
When the potentiometer in balanced, the unknown voltage (potential difference being measured) is equal to the voltage across AC.
V x
V AC
• Potentiometer can be used to : i) Measure an unknown e.m.f. of a cell. ii) Compare the e.m.f.s of two cells.
iii) Measure the internal resistance of a cell.
63
18.9 Potentiometer and Wheatstone Bridge
If the galvanometer shows deflection in one direction only
,
it may be due to :
• The connections of the terminals of the cells are wrong. The positive terminal of the cell must be connected to the positive terminal of another cell.
• The emf of the unknown cell is more then the emf of the cell connected across the wire of the potentiometer, AB.
• The connections are not tight and the current does not flow in certain part of the circuit.
64
A I
18.9 Potentiometer and Wheatstone Bridge
i) Measure an unknown e.m.f. of a cell.
I V
(Driver cell -accumulator)
C I I B
G Jockey +
(Unknown cell)
4 and 5 into 1,
V AC
V R AB
l l AC AB R AB
l AC l AB V
•When the potentiometer is balanced,
I G
= 0 Balance length =
l AC
V AC = ε V AC = IR AC …
1
R AC
l AC A
…2 ,
R AB
l AB A
2 , 3
R AC
l AC l AB R AB
...4
…3
I
V R AB
...5
65
A I
18.9 Potentiometer and Wheatstone Bridge
ii) Compare the e.m.f.s of two cells.
V
ε l 1 l 2 I ε 1 2 C
J
D
(1) S G (2)
I I B
• When the potentiometer is balanced,
I G
= 0 • Balance length,
l AC = l 1
for ε 1 and
l CD = l 2
for
ε 2
1
l l AB V l AC l AB V
...1 and 2
l CD l AB V
...2
l CD
1
l AC
2
l l
1 2 1 66
• • • 18.9 Potentiometer and Wheatstone Bridge
Wheatstone Bridge
It is used to measure the unknown resistance of the resistor.
Figure below shows the Wheatstone bridge circuit consists of a cell of e.m.f. (accumulator), a galvanometer , known resistances (
R 1
,
R 2
and
R 3
) and unknown resistance
R x
.
I I 1 A I 2 R 1 R 3 C
G
I 1
0 R 2 D I 2 R X I
I B
Then ,
Potential at C = Potential at D
Therefore
V AC
Since
V
I 1 R 1
IR V I 2 R 3 AD
and thus and
V I 1 R 2 BC
V BD I 2 R X I 1 R 1 I 1 R 2
I 2 R I 2 R X 3 R X
R 2 R 1
R 3
67
ε
18.9 Potentiometer and Wheatstone Bridge
I I 1 A I 2 R 1 R 3 C
G
I 1
0 R 2 D I 2 R X I B R X
R 2 R 1
R 3
R X R
3
R
2
R
1
Thick copper strip
I 1 R X
(Unknown resistance)
I A
Wire of uniform resistance
0
G
R I 1
Jockey
I 2 ε J
Accumulator (resistance box)
B I
R X R
l l
2 1 68
18.9 Potentiometer and Wheatstone Bridge
Example 18.19
An unknown length of platinum wire 0.920 mm in diameter is placed as the unknown resistance in a Wheatstone bridge as shown in figure below.
Resistors R 38.0 1 and R 2 and 46.0 have resistance of respectively. Balance is achieved when the switch closed and R 3 is 3.48 . Find the length of the platinum wire if its resistivity is 10.6 x 10 -8 m. 69