ch 18 student_201516
TRANSCRIPT
18. Electric Current & Direct Current Circuits
SF026 2015/ 16
ANALOGY: Electric circuits
Voltage: A force
that pushes the
current through the
circuit (in this picture
it would be
equivalent to gravity)
Resistance:
Friction that
impedes flow of
current through
the circuit (rocks
in the river)
Current:
The actual
“substance”
that is
flowing
through the
wires of the
circuit
(electrons!)
18.1 Electrical Conduction
(a)Describe microscopic model of current
(b)Define and use electric current formulae,
dt
dQI
Microscopic Model of Current
In wire (metal) the charge carrier is free electrons
Without battery, V = 0 No
electric field, E
Free e undergoes
random motion
Microscopic Model of Current
When a potential difference is
applied across the metal, an
electric field, E is set up
This field exerts an electric force
on the freely moving electron
The freely moving electrons tend to drift with constant
average velocity (drift velocity, vd) along the metal in a direction opposite that of
the E
http://www.schoolphysics.co.uk/age16-
19/Electricity%20and%20magnetism/Current%20electricity/text/Electric_current/index.html
Electric Current, I
Consider a simple closed circuit consists of wires, a battery and a light bulb as shown
From the diagram,
Direction of electric field or electric current: (+)ve to (–)ve
Direction of electron flows: (–)ve to (+)ve
Electric current, I is defined as the rate of flow of chargeMathematically,
One Ampere is the flow of one C through an area in one second
t
QI
dt
dQI Instantaneous
current
Average
current
It is a base and scalar quantityS.I. unit: ampere (A).
1s C 1second 1
coulomb 1ampere 1
Example 18.1
(a) There is a current of 0.5
A in a flashlight bulb for
2 min. How much charge
passes through the bulb
during this time?
Solution:
(b) A silver wire carries a
current of 3.0 A.
Determine
(i) the number of electrons
per second pass
through the wire,
(ii) the amount of charge
flows through a cross-
sectional area of the
wire in 55 s.t
QI C60
119 s electrons 1088.1 t
N
C 165Q
18.2 Ohm’s Law & Resistivity
(a)State and use Ohm’s law
(b)Define and use resistivity,l
RAρ
Ohm’s Law
States that the potential
difference across a
conductor is
proportional to the
current flowing through
it if its temperature is
constant
where T is constant
Materials that obey Ohm’s law
are materials that have constant
resistance over a wide range of
voltage ohmic conductor
Materials that do not obey the
Ohm’s law non-ohmic
conductors
IV
IRV Then
R = resistance
(V)V
(A)I0
Gradient, m
= R
Ohmic conductor
(metal)
http://www.schoolphysics.co.uk/animations/Ohms_law/index.html
file:///C:/DOCUME~1/KMS/LOCALS~1/Temp/phet-ohms-law/ohms-law_en.html
Ohm’s Law – Non-ohmic conductor
V
I0
Semiconductor Carbon
V
I0
Electrolyte
V
I0
Resistance, R – electric property which is
impedes or limits current in an electrical circuit
is defined as a ratio of the potential difference across an electrical component to the current passing through it
Mathematically,
Scalar quantity, unit: ohm ( ) or V A1
I
VR
Resistance
Type of material
Length, lCross
sectional area, A
Temperature
In general, the
resistance of
a conductor
increases
with
temperature
V: potential difference (voltage),I: current
Resistivity &
Conductivity
Resistivity, (specific resistance)
is defined as the resistance of a unit cross-sectional area per unit length of the materialMathematically,
Scalar quantity Unit: ohm meter ( m) It is a measure of a material’s
ability to oppose the flow of an electric current
Resistivity depends on the type of the material & temperature
l
RAρ
ρσ
1
Conductivity,
is defined as the reciprocal of the resistivity of a material
Mathematically,
Scalar quantity,
unit: 1 m1
Material Resistivity, ( m)
Silver 1.59 108
Copper 1.68 108
Aluminum 2.82 108
Gold 2.44 108
Glass 10101014
A good electric conductors
have a very low resistivities,
good insulators have very
high resistivities
file:///C:/DOCUME~1/KMS/LOCALS~1/Temp/phet-resistance-in-a-wire/resistance-in-a-wire_en.html
Example 18.2
A constantan wire of
length 1.0m and cross
sectional area of 0.5 mm2
has a resistivity of 4.9 x 10–7 Ωm. Find the
resistance of the wire.
Example 18.3
Two wires P and Q with circular cross section are made of the same metal and have equal length. If the resistance of wire P is three times greater than that of wire Q, determine the ratio of their diameters.
Exercise 18.2
1. A wire 5.0 m long and 3.0 mm in diameter has a
resistance of 100 . A 15 V of potential difference is
applied across the wire. Determine
(a) the current in the wire,
(b) the resistivity of the wire,
(c) the rate at which heat is being produced in the wire.
(College Physics,6th edition, Wilson, Buffa & Lou, Q75,
p.589)ANS: 0.15 A; 1.414 104 m; 2.25 W
18.3 Variation of Resistance
With Temperature
(a)Explain the effect of temperature on
electrical resistance
(b)Use resistance,
00 1 TTRR
Effect of Temperature on Electrical
Resistance in Metals
When the temperature increases, the number of free electrons per unit volume in metal remains unchanged
Metal atoms in the crystal lattice vibrate with greater amplitude and cause the number of collisionsbetween the free electrons and metal atoms increase and slowing down the electron flow
Hence the resistance in the metal increases
Effect of Temperature on Electrical
Resistance in Metal
The resistance of a metal
can be represented by the
equation below
whereR = final resistance
Ro= initial resistance
= the temperature coefficient of resistivity
Material (C1)
Silver 4.10 103
Mercury 0.89 103
Iron 6.51 103
Aluminum 4.29 103
Copper 6.80 103
TαRR 10
α is defined as the fractional change in resistance per Celsius degree
temperature coefficients of resistivity for various materials
T
RR
0
Effect of Temperature on Electrical
Resistance in Metal
19
R
T0
0R
cT
Figure 18.8a : metal Figure 18.8b : semiconductor
R
T0
R
T0Figure 18.8c : superconductor
R
T0Figure 18.8d : carbon
Example 18.4
A copper wire has a resistance of 25 m at 20 C. When
the wire is carrying a current, heat produced by the current
causes the temperature of the wire to increase by 27 C.
(a) Calculate the change in the wire’s resistance
(b) If its original current was 10.0 mA and the potential
difference across wire remains constant, what is its final
current?
(copper = 6.80 103 C1)
Example 18.5
A platinum wire has a resistance of 0.5 Ω at 0°C.
It is placed in a water bath
where its resistance rises to a final value of 0.6 Ω.
What is the temperature
of the bath?
(platinum = 3.93 103
C1)
Exercise 18.3
1. A wire of unknown composition has a resistance of 35.0
when immersed in the water at 20.0 C. When the wire is
placed in the boiling water, its resistance rises to 47.6 .
Calculate the temperature on a hot day when the wire has a
resistance of 37.8 . (Physics,7th edition, Cutnell &
Johnson, Q15, p.639)
ANS: 37.78 C
2. A copper wire has a resistance of 25 m at 20 C. When the
wire is carrying a current, heat produced by the current
causes the temperature of the wire to increase by 27 C.
(a) Calculate the change in the wire’s resistance.
(b) If its original current was 10.0 mA and the potential
difference across wire remains constant, what is its final
current? (copper = 6.80 103 C1)ANS: (a) 4.59×10–3 Ω; (b) 8.45×10–3 A
18.4 Electromotive Force (emf),
Potential Difference & Resistance
(a)Define emf, of a battery
(b)Explain the relationship between emf of
a battery and potential difference across
the battery terminals
(c)Use terminal voltage,
IrεV
E.m.f, VS Potential Difference, V
Terminal voltage, V is the potential difference across the terminals of a battery when there is a current flowing through it
Electromotive force, e.m.f (ξ) of a battery is the maximum potential difference across its terminals when it is not connected to a circuits
Emf in electric
source, p.d
between two
points in a circuit
E.m.f, VS Potential Difference, V
Consider a circuit
consisting of a battery
(cell) that is connected
by wires to an external
resistor R as shown in
Figure
In reality, when a battery
is supplying current, its
terminal voltage is less than its e.m.f, ξ
This reduces of voltage
is due to energy
dissipation in the battery
In effect, the battery has
internal resistance (r)
E.m.f, VS Potential Difference, V
Mathematically
V = IR,
where = emf
V = terminal potential difference
I = current
R = external resistance
r = internal resistance
is the resistance due to chemicals inside the battery (cell)
It will constitutes part of the total resistance in a circuit
The emf of a battery is constant but the internal resistance of the battery increases with time as a result of chemical reaction
I R r
V Ir
Internal resistance, r
rε r εOR
Example 18.6
A battery has an emf of 9.0
V and an internal resistance
of 6.0 . Determine
(a) the potential difference
across its terminals
when it is supplying a
current of 0.50 A,
(b) the maximum current
which the battery could
supply.
Example 18.7
A car battery has an emf of
12.0 V and an internal
resistance of 1.0 . The
external resistor of
resistance 5.0 is
connected in series with the
battery as shown.
Determine the reading of
the ammeter and voltmeter
if both meters are ideal.
Exercise 18.4
1. A battery of e.m.f 3.0 V and internal resistance 5.0 is connected to a switch by a wire of resistance 100 . The voltage across the battery is measured by a voltmeter. What is the voltmeter reading when the switch is
(a) off?(b) on? ANS: (a) 3V, (b) 2.86 V
2. An idealized voltmeter is connected across the terminals of a battery while the current is varied. Figure shows a graph of the voltmeter reading V as a function of the current I through the battery. Find
(a) the emf, ξ and(b) the internal resistance of the battery
ANS: (a) 9V, 4,5Ω
Exercise 18.4
3. A battery of emf 6.0 V is connected across a 10
resistor. If the potential difference across the resistor is
5.0 V,
(a) Determine
(i) the current in the circuit,
(ii) the internal resistance of the battery.
(b) When a 1.5 V dry cell is short-circuited, a current of 3.0
A flows through the cell. What is the internal resistance
of the cell?ANS: 0.50 A, 2.0 ; 0.50
Revision
ItQ
IV
Laws'Ohm
l
RAρ
ρσ
1
TRR 10
I Battery (cell)
A Brε
R
I R r
IRV
Use:
(i) power,
(ii) electrical energy,
IVP
IVtW
18.5 Electrical Energy & Power
Power, P
is defined as the energy liberated per unit time in the electrical device
The electrical power Psupplied to the electrical device is given by
When the electric current flows through wire or passive resistor, hence the potential difference across it is
then the electrical power can be written as
It is a scalar quantity and its unit is watts (W)
t
VIt
t
WP
IVP
IRV
RIP 2 OR
R
VP
2
Energy, E
Consider a circuit
consisting of a battery that
is connected by wires to
an electrical device (such
as a lamp, motor or battery
being charged) where the
potential different across
that electrical device is V
A current I flows from the terminal A to the terminal B, if it flows for time t, the charge Q which it carries from B to A is given by
Then the work done on this charge Qfrom B to A (equal to the electrical energy supplied) is
If the electrical device is passive resistor(device which convert all the electrical energy supplied into heat), the heat dissipated H is given by
Electrical device
A B
VI I
QVW
ItQ
VItEW
VItWH OR RtIH 2
Example 18.7
In figure below, a battery
has an emf of 12 V and an
internal resistance of 1.0 .
Determine
(a) the rate of energy
transferred to electrical
energy in the battery,
(b) the rate of heat
dissipated in the battery,
(c) the amount of heat loss
in the 5.0 resistor if the
current flows through it
for 20 minutes.
Exercise 18.5
An electric toy of resistance 2.50 is operated by a dry cell
of emf 1.50 V and an internal resistance 0.25 .
(a) What is the current does the toy drawn?
(b) If the cell delivers a steady current for 6.00 hours,
calculate the charge pass through the toy.
(c) Determine the energy was delivered to the toy.
ANS: 0.55 A; 1.19 104 C; 16.3 kJ
18.6 Resistors in Series
& Parallel
Derive and determine effective resistance
of resistors in series and parallel
Resistors in
Series
From the definition of resistance, thus
Substituting for V1, V2 , V3
and V
where
Reff: effective) resistance
;22 IRV ;33 IRV ;11 IRV effIRV
321eff IRIRIRIR
321eff RRRR
Consider three resistors are
connected in series to the
battery
Characteristics:
Same current I flows through each resistor
Total potential difference, V (Assumption: the connecting wires have no resistance)
321 IIII
321 VVVV
Example
(a) What is the current in each resistor?
(b) What is the voltage across each resistor?
(c) What is the total resistance?
(d) What is the battery voltage?
ANS: 0.1A, 3V, 4V, 5V; 120; 12V
Resistors in
Parallel
Consider three resistors
are connected in parallel
to the battery
Characteristics:
Same potential difference, V across each resistor
Total current in the circuit
From the definition of resistance, thus
Substituting for I1, I2 , I3 and I321 VVVV
321 IIII
;2
2R
VI ;
3
3R
VI ;
1
1R
VI
effR
VI
321eff R
V
R
V
R
V
R
V
321eff
1111
RRRR
Example
(a) What is the total resistance of the circuit (watch
out for the bear trap)?
(b) What is the current through each resistor?
(c) What is the total current?
ANS: 12.8; 0.40A, 0.30A, 0.24A; 0.94A
Example 18.8
What is the equivalent
resistance of the resistors
in figure below?
R1= R2= R3= R4= 1 Ω
A
BR
1
R
2
R
3
R
4
A
B
R1
R2
R34
A
B
R1
R2
R3R4
5
3ER
A
B
R1R234
Example 18.9
Find the current in & voltage of the 10 Ωresistor shown below.
Series VS Parallel
When connected to the
same source, two light
bulbs in series draw
less power and glow
less brightly than when
they are in parallel
In the series case the same
current flows through both
bulbs.
If one of the bulbs burns out,
there will be no current at all in
the circuit, and neither bulb will
glow
In the parallel case the potential difference
across either bulb remains equal if one of
the bulbs burns out. The current through
the functional bulb remains equal and the
power delivered to that bulb remains the
same . This is another of the merits of a
parallel arrangement of light bulbs: If one
fails, the other bulbs are unaffected. This
principle is used in household wiring
systems
Example 18.10
For the circuit below, calculate
(a) the effective resistance of the circuit,
(b) the current passes through the 12 resistor,
(c) the potential difference across 4.0 resistor,
(d) the power delivered by the battery.
The internal resistance of the battery may be ignored.
Example 18.11
For the circuit above,
calculate the effective
resistance between the
points A and B.
Exercise 18.6
1. Determine the equivalent resistances of the resistors
below.
ANS: 0.80 ; 2.7 ; 8.0
0.2
0.2
0.2
0.2
0.6
01
0.6 0.4
18
16
0.8
0.9
16
0.6
20
Exercise 18.6
2. The circuit below includes a battery with a finite internal resistance, r = 0.50 .
(a) Determine the current flowing through the 7.1 and 3.2 resistors.
(b) How much current flows through the battery?
(c) What is the potential difference between the terminals of the battery? (Physics,3th edition, James S. Walker, Q39, p.728)
ANS: 1.1 A, 0.3 A; 1.4 A; 11.3 V
Exercise 18.6
3. Four identical resistors are connected to a battery as shown below. When the switch is open, the current through the battery is I0.
(a) When the switch is closed, will the current through the battery increase, decrease or stay the same? Explain.
(b) Calculate the current that flows through the battery when the switch is closed, Give your answer in terms of I0. (Physics,3th edition, James S. Walker, Q45, p.728)
ANS: U think
Exercise 18.6
4. Figure below shows the arrangement of five equal resistors in a circuit. Calculate
(a) the equivalent resistance between point xand y.
(b) the voltage across point b and c.
(c) the voltage across point c and y.
18.7 Kirchhoff’s Law
State and use
Kirchhoff’s law
1st Kirchhoff’s Law (KCL)
the sum of the currents entering any junctions in a
circuit must equal the sum of the currents leaving
that junction
outin II
1I
2I3I
321 III 123 III
3I 2I
1I
2nd Kirchhoff’s Law (KVL)
in any loop, the sum of emf is equal to the sum of the
products of current and resistance
For emf,
For product of IR
ε
ε
direction of loop
+-
ε-
ε+
direction of loop
IR
direction of loop
I
RIR
I
R
direction of loop
IR
Example 18.12
For the circuit
below, determine
the current and its
direction in the
circuit.
1.15
.226
50.8 2 ,V 1.51
4 ,V 5.01Loop 1
I
I
II
Example 18.13
For the circuit below, determine
(a) the currents I1, I2 and I,
(b) the potential difference across the 6.7 resistor,
(c) the power dissipated from the 1.2 resistor.
8.9
9.3
V .09V 21
7.6
.21
I1I 2I
Example 18.14
Find the value of current, I , resistance R & the e.m.f, ξ
ξ
D
A
F E
B
IR
1
3
AI 11
AI 22
C
V12
Exercise 18.7
1. For a circuit below, given
1= 8V, R2= 2 , R3= 3
, R1 = 1 and I = 3 A.
Ignore the internal
resistance in each
battery. Calculate
(a) the currents I1 and I2.
(b) the emf, 2.
ANS: 1.0 A, 4.0 A; 17 V
Exercise 18.7
2. Determine the current in each resistor in the circuit below. (College Physics,6th edition, Wilson, Buffa & Lou, Q57, p.619)
ANS: 3.75 A; 1.25 A; 1.25 A
Exercise 18.7
3. Referring to the circuit above, calculate the current I1, I2and I3
ANS: I1=0.66A, I2 = I3 = 0.33A
18.8 Potential Divider
(a) Explain the principle of a potential divider.
(b) Apply equation of potential divider,
VRR
RV
21
11
Potential Divider
A potential divider produces an output voltage that is a fraction of the supply voltage V
This is done by connecting two resistors in series V
1V
1RI
2V
2RI 11 IRV V
RR
RV
21
11
VRR
RV
21
22
Since the current flowingthrough each resistor is the same,
Therefore, the potential difference (voltage) across R1 is given by
Similarly,
21eff RRR
effR
VI an
d
21 RR
VI
Potential Divider
Resistance R1 and R2 can
be replaced by a uniform
homogeneous wire
The potential difference
(voltage) across the wire
with length l1 is
Similarly,
V
I
2l1lI
BA C
2V1V
Vll
lV
21
11
Vll
lV
21
22
Example 18.15
For the circuit below,
(a) calculate the output
voltage
(b) If a voltmeter of
resistance 4000 is
connected across the
output, determine the
reading of the voltmeter.
18.8 Potentiometer &
Wheatstone Bridge
(a) Explain principles of potentiometer and
Wheatstone Bridge and their applications
(b) Use related equations for potentiometer,
and for Wheatstone Bridge,
x
3
2
1
R
R
R
R
2
1
2
1
l
l
ε
ε
Potentiometer
Consider a potentiometer circuit below
The potentiometer
is balanced when
the jockey is at
such a position on
wire AB that there is
no current through
the galvanometer
Thus
Galvanometer
reading = 0ACx VV
Potentiometer
When the potentiometer
in balanced, the unknown
voltage (potential
difference being
measured) is equal to
the voltage across AC
Application: Compare the emfs of two
cells or find unknown emf
o In this case, a potentiometer is set up as illustrated in Figure below, in which AB is a wire of uniform resistance and J is a sliding contact (jockey) onto the wire.
o An accumulator Xmaintains a steady current I through the wire AB
X
BAI
G
I
(2)
(1)
2εS
II
1ε
J
Application: Compare the emfs of two cells
or find unknown emf
Initially, switch S is
connected to the terminal (1)
and the jockey moved until
the emf ξ1 exactly balances
the potential difference (p.d.)
from the accumulator
(galvanometer reading is
zero) at point C. Hence
where
and
then
X
BAI
G
I
(2)
(1)
2εS
II
1ε
C
J
1l
1 ACV
ACAC IRV A
ρlR 1
AC
11
ρlI
A
(1)
Application: Compare the emfs of two cells
or find unknown emf
X
BAI
G
I
(2)
(1)
2εS
II
1ε
C
J
D1l2l
After that, the switch S is
connected to the terminal (2)
and the jockey moved until
the emf ξ2 balances the p.d.
from the accumulator at
point D
Hence
where
and
then
2 ADV
ADAD IRV A
ρlR 2
AD
22
ρlI
A
(2)
Application: Compare the emfs of two cells
or find unknown emf
By dividing eq. (1) and eq. (2) then
Since
Equation above can be written as
1 1
2 2
l
l
1
1
22
ρlI
A
ρlI
A
1ρlR R l
A
1 1
2 2
R
R
Example 18.16
Consider a potentiometer
with a standard battery with
an e.m.f. of 1.0186 V is
used in the circuit. When the resistance is 36 Ω, the
galvanometer reads zero. If
the standard battery is
replaced by an unknown
e.m.f. the galvanometer
reads zero when the
resistance is adjusted to 48 Ω. What is the value of the
unknown e.m.f. ?
Exercise 18.8(a)
1. In figure below, PQ is a uniform
wire of length 1.0 m and
resistance 10.0 . ξ1 is an
accumulator of emf 2.0 V and
negligible internal resistance.
R1 is a 15 resistor and R2 is a
5.0 resistor when S1 and S2
open, galvanometer G is
balanced when QT is 62.5 cm.
When both S1 and S2 are
closed, the balance length is
10.0 cm. Calculate
(a) the emf of cell ξ2
(b) the internal resistance of cell ξ2.
Exercise 18.8(a)
2. Cells A and B and centre-zero
galvanometer G are connected to a
uniform wire OS using jockeys X and Y as
below. The length of the uniform wire OS is
1.00 m and its resistance is 12 . When
OY is 75.0 cm, the galvanometer does not
show any deflection when OX= 50.0 cm. If
Y touches the end S of the wire, OX = 62.5
cm when the galvanometer is balanced.
The emf of the cell B is 1.0 V. Calculate
(a) the potential difference across OY when
OY = 75.0 cm,
(b) the potential difference across OY when Y
touches S and the galvanometer is
balanced,
(c) the internal resistance of the cell A,
(d) the emf of cell A.
Wheatstone Bridge
It is used to measured the
unknown resistance of the
resistor
Figure below shows the Wheatstone bridge circuit consists of a cell of emf , a galvanometer , know resistances (R1, R2 and R3) and unknown resistance Rx.
The Wheatstone bridge is said to be balanced when no current flows through the galvanometer
Wheatstone Bridge
Hence
then
Therefore
Since thus
Dividing gives
1CBAC III 2DBAD III and
Potential at C = Potential at D
ADAC VV BDBC VV and
IRV
3211 RIRI and x221 RIRI
x2
32
21
11
RI
RI
RI
RI
3
1
2x R
R
RR
Example 18.17
A wheatstone bridge is used to make a precise
measurement of the resistance of a wire connector. If R = 1 kΩ & the bridge is balanced by adjusting P such that P =
2.5 Q, what is the value of X ?
Wheatstone Bridge:
Application (meter bridge)
The application of the Wheatstone bridge is Metre Bridge
The metre bridge is balanced when the jockey J is at such a
position on wire AB that there is no current through the
galvanometer. Thus the current I1 flows through the
resistance Rx and R but current I2 flows in the wire AB.
0
Accumulator
Jockey
Thick copper strip
(Unknown resistance)(resistance box)
Wire of uniform resistance
xR
A
ε
G
B
R
J
2l1lI I
1I
2I
1I
Wheatstone Bridge:
Application (meter bridge)
Let Vx : p.d. across Rx and V : p.d. across R,
At balance condition,
By applying Ohm’s law, thus
Dividing gives
AJx VV JBVV and
AJ2x1 RIRI JB21 RIRI and
A
ρlR 1
AJ JB2
AJ2
1
x1
RI
RI
RI
RI where and
A
ρlR 2
JB
A
ρl
A
ρl
R
R
2
1
x
Rl
lR
2
1x
Exercise 18.8(b)
The circuit shown in figure is
known as a Wheatstone
bridge. Determine the value
of the resistor R such that
the current through the 85.0
resistor is zero.
(Physics,3th edition, James
S. Walker, Q93, p.731)
ANS: 7.50
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