FPY 1324
Dr. Aishah Badruzzaman
CHAPTER 3
ELECTRICITY &
DIRECT CURRENT
Prepared by Dr. Aishah (2020)
S2
LEARNING OUTCOME
After completing this chapter, you will be able to:
i.
ii.
iii.
iv.
v.
vi.
vii.
Define and use electric current
State and use Ohm’s Law
Define and use resistivity
Explain the factors that affect resistivity
Define e.m.f of a battery
Define power and electrical energy
Derive and determine the effective resistance in a
series and parallel circuit
viii. State and use Kirchhoff's laws
Prepared by Dr. Aishah (2020)
S3
ELECTRIC CURRENT, 𝑰
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S4
Current
An electric current is a flow of
microscopic particles called
electrons flowing through wires and
components.
Unit SI: Ampere(A)
Symbol: I
+
-
• The purpose of a battery is to produce a
potential difference, which can then make
charges move.
• When a continuous conducting path is
connected between the terminals of a battery,
we have an electric circuit.
• when such a circuit is formed, charge can
flow through the wires of the circuit.
•More precisely, the electric current in a wire is
defined as the net amount of charge that
passes through the wire’s full cross section at
any point per unit time.
Q
•The current, I is defined as: I 
t
∆Q= Amount of charge
∆t = time
Unit : Ampere (A)
ELECTRIC CURRENT, 𝑰
Definition: the rate of charge flow through
wire
𝐝𝑸
𝑰=
𝐝𝒕
No electric current:
• Potential difference is zero
• No electric field
• Electron in random motion
Prepared by Dr. Aishah (2020)
There is electric current:
• Potential difference exist
• Electric field is set up
• Charge flow from higher to
lower potential difference
S7
Current is flow of charge
A steady current of 2.5 A exists in a wire for 4.0 min.
(a) How much total charge passed by a given point in the circuit
during those 4.0 min?
(b) How many electrons would this be?
Solution:
Current is flow of charge
A steady current of 2.5 A exists in a wire for 4.0 min.
(a) How much total charge passed by a given point in the circuit
during those 4.0 min?
(b) How many electrons would this be?
Solution:
(a) The charge flow:
Q  It  (2.5C / s )( 240s)  600C
(b) The charge on one electron is 1.6 X 10-19 C, so 600 C
would consist of:
600C
21

3
.
8

10
electrons
19
1.6 10 C / electron
Charged
-Electric charge is a
characteristic of some
subatomic particles.
-example: e =-1, p=+1
-Symbol: Q
electric charge: distribution in a thunderstorm
• The electric charge on a body may be positive or negative.
•Two positively charged bodies experience a mutual repulsive
force, as do two negatively charged bodies.
•A positively charged body and a negatively charged body
experience an attractive force.
• The SI unit of quantity of electric charge is the coulomb, which is
equivalent to about 6.25 × 1018 e (e is the charge on a single
electron or proton).
• Hence, the charge of an electron is approximately
−1.602×10−19 C.
• The quantity of electric charge can be directly measured with an
electrometer.
Voltage
•Voltage is electric
potential energy per unit
charge, measured in
joules per coulomb (=
volts).
•Symbol: V
Instrument to measure
voltage: voltmeter
• Voltage is commonly used as a short name for electrical potential
difference. Its corresponding SI unit is the volt.
• A common use of the term "voltage" is in specifying how many volts
are dropped across an electrical device (such as a resistor).
• The voltmeter symbol is seen in this example circuit diagram. A
voltmeter (V) and an ammeter (A) are shown measuring a voltage and
a current respectively, in a simple series circuit.
ELECTRIC CURRENT, 𝑰
Current = rate of charge
flow
𝐝𝑸
𝑰=
𝐝𝒕
𝑸
𝑰=
𝒕
Where:
𝐼 : Electric Current
𝑄: Electric Charge
𝑡 : Time taken
• SI Unit : Ampere (𝐴)
• Scalar quantity
• Can be measure using
ammeter
Prepared by Dr. Aishah (2020)
S14
MICROSCOPIC MODEL OF
CURRENT
𝐿
𝑒−
𝑒
−
𝑒
−
𝑒−
𝑒− −
𝑒
𝐴
𝑣𝑑
𝐸
• When there is electric force, electron will drift in a
direction opposing the electric field with drift velocity,
𝑣𝑑 across a cross-sectional area, 𝐴 in time 𝑡
• Drift velocity, 𝒗𝒅 is the average velocity of the free
moving electron in an electric field
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S15
Number of electron, 𝑁 = 𝑛𝐴𝐿
Total charge, 𝑄 = 𝑁𝑒
𝑄 = 𝑛𝐴𝐿𝑒
𝐿
Drift velocity, 𝑣𝑑 =
𝑡
𝑄
Where current, 𝐼 =
𝑡
𝑛𝐴𝐿𝑒
𝐼=
𝑡
𝐿
𝐼 = 𝑛𝐴𝑒
𝑡
𝐼 = 𝑛𝐴𝑒𝑣𝑑
Prepared by Dr. Aishah
(2020)
𝑰
𝒗𝒅 =
𝒏𝑨𝒆
Where:
𝑁 : Number electrons
𝑛 : Number of electrons per volume
𝑒 : charge of electron
1.6 × 10−19 C
𝑡 : Time taken
S16
Example
The current in a bulb is 2.0 𝐴.
What is the charge that flows through the bulb in 5
i.
minutes?
ii.
How many electrons flow through the bulb in 5
minutes?
iii. The diameter of the filament is 0.5 mm and the
number of free electrons in the filament , 𝑛 =
1.0 × 1029 m−3 . Calculate the drift velocity in the
filament.
Prepared by Dr. Aishah (2020)
S17
Solution
𝐼 =2𝐴
i. 𝑄 when 𝑡 = 5 min= 5 × 60 = 300 s
𝑄 = 𝐼𝑡 = 2 × 300 = 𝟔𝟎𝟎 𝐂
ii.
𝑁 when 𝑡 = 5 min= 5 × 60 = 300 s
𝑄 = 𝑁𝑒
𝑄
600
𝟐𝟏
𝑁= =
=
𝟑.
𝟕𝟓
×
𝟏𝟎
𝑒 1.6 × 10−19
iii. Drift velocity;
𝐼
2
𝑣𝑑 =
=
𝑛𝐴𝑒 (1.0 × 1029 )𝜋 0.25 × 10−3 2 (1.6 × 10−19 )
𝑣𝑑 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟒 𝐦𝐬 −𝟏
Prepared by Dr. Aishah (2020)
S18
Exercises
1.
2 × 1015 electrons flow through a cross-section of wire
in 2 s.
Determine
i. the amount of charge flows through the wire
ii. the current flow in the wire
3.2 × 10−4 C ; 1.6 × 10−4 A
2. A copper wire carries a current of 5 𝐴.
Determine
i. the number of electrons per second passing through
the wire
ii. the amount of charge flows through the wire in 45 s.
3.125 × 1019 s−1 ; 225 C
Prepared by Dr. Aishah (2020)
S19
OHM’S LAW &
RESISTIVITY
Prepared by Dr. Aishah (2020)
S20
OHM’S LAW
“The potential difference, 𝑉 across a conductor is
directly proportional to the current, 𝐼 flowing through it
at constant temperature”
𝑽
𝑽∝𝑰
𝑚=𝑅
𝑰
Prepared by Dr. Aishah (2020)
𝑽 = 𝑰𝑹
• The 𝑉 − 𝐼 graph shows the
linear relationship between 𝑉
and 𝐼
• The gradient of the graph
represent the resistance, 𝑅 of
the conductor
S21
RESISTANCE, R
• A property which opposes electric current
• Definition: the ratio of potential difference, 𝑉
across a conductor to the current, 𝐼 that flow
through it”
𝑽
𝑹=
𝑰
• Scalar quantity
• Unit: Ohm 𝛀 or 𝑽𝑨−𝟏
Prepared by Dr. Aishah (2020)
S22
Resistance:
Electrical resistance, R is a ratio of the
degree that an object opposes an electric
current through it, measured in Ohms(Ω).
V
R
I
V = voltage (volts)
I = current (ampere)
The electrical resistance of an object is a measure of its
opposition to the passage of a steady electric current.
Unit for resistance is ohm (1 Ω= 1 V/A)
Ohm’s Law
Ohm's Law states that in a simple electrical circuit, the
voltage equals the electrical current times the resistance.
𝑉 = 𝐼𝑅
Example :
Flashlight bulb resistance:
A small flashlight bulb draws 300mA
from its 1.5 V battery.
(a) What is the resistance of the
bulb?
(b) If the battery becomes weak and
the voltage drops to 1.2V, how
would the current change?
Example :
Flashlight bulb resistance:
A small flashlight bulb draws 300mA
from its 1.5 V battery.
(a) What is the resistance of the
bulb?
(b) If the battery becomes weak and
the voltage drops to 1.2V, how
would the current change?
Solution:
(a) The resistance is: R  V  1.5V  5.0
I
(b) The current is :
0.30 A
V 1.2V
I 
 0.24 A  240mA
R 5.0
Resistor
• Resistors are used to control the amount of
current.
•Resistors have resistances ranging from less than
an ohm to millions of ohms.
• The main types are “wire-wound” resistors which
consist of a coil of fine wire.
• “ composition” resistors which are usually made of
carbon and thin carbon or metal films.
•Symbol for resistor is:
Three composition
resistor.
RESISTIVITY, 𝝆
• Definition: the resistance of a unit cross-sectional area
per unit length of the conductor
𝑹𝑨
𝝆=
𝑳
• Scalar quantity
• Unit : Ohm meter (𝛀 𝐦)
• Measure the ability of a conductor to oppose the current
flow
• Depends on type of material and the temperature
• Good conductor → low resistivity
• Good insulator → high resistivity
Prepared by Dr. Aishah (2020)
S28
Example
A 2.05 m constantan wire of 2 Ω with a cross-section area
of 0.5 mm2 is connected to a power supply and 1000 C of
charge flow through the wire in 2 minutes
i. Determine the potential difference across the
constantan wire.
ii. Determine the resistivity of the constantan wire
Solution
i. 𝐼 = 𝑄 = 1000 = 8.33 A
𝑡 2 × 60
𝑉 = 𝐼𝑅 = 8.33 2 = 𝟏𝟔. 𝟔𝟔 𝐕
ii.
𝑅𝐴 2 × 0.5 1 × 10−3
𝜌=
=
𝑙
2.05
2
= 𝟒. 𝟗 × 𝟏𝟎−𝟕 𝜴 𝐦
Prepared by Dr. Aishah (2020)
S29
FACTORS AFFECT RESISTANCE
OF A CONDUCTOR
Type of material
R∝ρ
(resistivity)
Resistance depends upon resistivity which
is a proportionality constant and depends
on nature of substance and temperature.
Length
𝑅∝𝑙
The longer the length of conductor, the
larger the resistance
Cross-sectional
area
1
𝑅∝
𝐴
The smaller the cross-sectional area, the
larger the resistance
𝑅∝𝑇
The hotter the conductor, the larger the
resistivity. When a material gets hotter,
atoms will vibrate more. This makes it
difficult for electrons to move without
interaction with atoms.
Temperature
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S30
Series and parallel resistors
Series resistor:
To find their total equivalent resistance (Req):
Parallel Resistors:
To find their total equivalent resistance (Req):
For two resistors,
Combination of parallel and series
Example :
Resistance in Parallel
Given R1= 20Ω, R2= 15Ω,
R3 = 30Ω
(a) Find the equivalent resistance,
Req:
1
1
1
1
 

Req R1 R2 R3
1
1
1
1



Req 20 15 30
1
3

Req 20
 Req  6.67
(b) If the voltage is 8V, how
much the current flow in the
circuit?
V
R
I
V
8V
I 

 1.2 A
Req 6.67
TRY THIS!!
1.
Given R1=25Ω, R2=30Ω and I = 10A.
Calculate the voltage?
(ans = 550 volts)
Find their total equivalent resistance (Req)
(a)
8.67 ohms
Find their total equivalent resistance (Req)
(b)
769 ohms
ELECTROMOTIVE
FORCE & INTERNAL
RESISTANCE
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S38
ELECTROMOTIVE FORCE (EMF), 𝜺
• Definition: Energy to drive current round the circuit
• Unit : Volt V or J C −1
• Basically, emf is the potential difference across the
terminal of the source in open circuit, when no
current flows from the source
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S39
• When no current flow
from the battery (open
circuit), the voltmeter
measures emf
Prepared by Dr. Aishah 2020
• When there is current
flow from the battery
(closed circuit), the
voltmeter measures
potential difference
across the terminal of
the source
S40
Internal Resistance, 𝑟
• As the electron flow through
the cell, the chemical energy
stored in the cell is converted
to electrical energy
• The chemical inside the cell
opposes the flow of electron
and produces resistance to
the current flow in the cell.
• This resistance is known as
internal resistance, 𝒓
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S41
Internal Resistance, 𝑟
• As the current, 𝐼 flow through
the internal resistance in a
closed circuit, some of the
electric potential energy is used
to overcome the internal
resistance causing the potential
difference, 𝑉
across the
terminals drops and becomes
less than emf, 𝜀
• The difference between the emf, 𝜀 and its terminal
potential, 𝑉 equal to the potential drop due to the
internal resistance, 𝑟
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S42
𝜺 − 𝑽 = 𝑰𝒓
𝐕 − 𝜺 = 𝑽 + 𝑰𝒓
𝐕 − 𝜺 = 𝑰𝑹 + 𝑰𝒓
𝐕 − 𝜺 = 𝑰(𝑹 + 𝒓)
Where;
𝜀
𝐼
𝑅
𝑟
𝑉
Prepared by Dr. Aishah 2020
: emf
: current through the circuit
: total external resistance
: internal resistance of the cell (battery)
: terminal potential difference (voltage)
S43
Example
a) The p.d across the terminals of a cell is 1.5 V when it is
not connected to a circuit. When the cell is connected to
a circuit, a current of 0.37 A is flowing the terminal, p.d.
falls to 1.1 V. What is the internal resistance of the cell?
b) The emf of the cell is always greater than its terminal
voltage. Why? Give reason.
Prepared by Dr. Aishah 2020
S44
Example
a) The p.d across the terminals of a cell is 1.5 V when it is
not connected to a circuit. When the cell is connected to
a circuit, a current of 0.37 A is flowing the terminal, p.d.
falls to 1.1 V. What is the internal resistance of the cell?
b) The emf of the cell is always greater than its terminal
voltage. Why? Give reason.
a)
𝑉 = 𝜀 − 𝐼𝑟
1.1 = 1.5 − 0.3𝑟
𝑟 = 1.08 Ω
b) Some of the electric potential energy is used to
overcome the internal resistance causing the potential
difference, 𝑉 across the terminals drops and becomes less
than emf, 𝜀
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S45
• Calculate the reading on the voltmeter when the
switch, S is
a) opened
b) closed
a) 2 V
b) 1.8 V
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S46
Kirchhoff’s Law
Kirchhoff’s first law :
“At any node (junction) in an electrical circuit, the
sum of currents flowing into that node is equal to
the sum of currents flowing out of that node.”
I in  I out
Kirchhoff’s second law :
“States that in any closed loop, the algebraic sum of emfs is equal to
the algebraic sum of the products of current and resistance.”
OR In any closed loop,
    IR
Sign convention
•For emf, : direction of
loop
ε
- +
direction of loop
ε
ε
For product of IR:
direction of loop
R
I
+ -
ε
direction of loop
 IR
R
I
 IR

For example, Consider a circuit as shown in
Figure below:
E
ε1
I1
I1
I2
ε2
D
Loop 3
I3
R3
C
I3
R1 I1
F
Loop 1
I1
R2 I 2
Loop 2
A
I3
ε3
I3
B
At junction A or D (applying the Kirchhoff’s first law) :
I
in

I
out
I1  I 2  I 3
(1)
For the closed loop (either clockwise or anticlockwise),
apply the Kirchhoff’s second law.
LOOP 1 :
LOOP 2 :
LOOP 3 :
    IR
    IR
    IR
ε1  ε 2  I 2 R 2  I 1 R1
(2)
ε 2  ε 3  I 2 R 2  I 3 R3
(3)
ε1  ε 3  I 3 R3  I 1 R1
(4)
By solving equation (1) and any two equations from the
closed loop, hence each current in the circuit can be
determined.
resistance of a wire can be expressed as

R
L
A
depends on the material used. (unit: Ωm)
constant of proportionality
measured in ohms, Ω
measured in metres, m
measured in square metres, m²
Example :speaker wires
Suppose you want to connect your stereo to
remote speakers.
Given ρcopper= 1.68 X 10-8 Ωm.
(a) If each wire must be 20m long, what
diameter copper wire should you use to
keep the resistance less than 0.10 Ω per
wire?
(b) If the current to each speaker is 4.0 A,
what is the potential difference, or
voltage drop across each wire?
Solutions :
(a)
L (1.68 108 .m)( 20m)
A  
 3.4 106 m2
R
(0.10)
Then, A= π r2.
r
A

 1.04 103 m  1.04mm
The diameter, d = 2r. So d = 2.1 mm
(b) The voltage drop across each wire is
V = IR = (4.0 A)(0.10Ω) = 0.40 V
There are 2 different ways that electricity is produced
DIRECT CURRENT (DC)
Definition
Unidirectional flow of electric
charge→direction remains
constant
ALTERNATING CURRENT (AC)
Electric charge whose
direction reverses cyclically
Direct current
Symbol
Alternating current
Current
Graph
I vs t
Examples
Battery, CD player
time
Power Plant
• Direct current may be obtained from an alternating current supply
by use of a current-switching arrangement called a rectifier, which
allow current to flow only in one direction.
Rectifier
• Because of the advantage of alternating current over direct
current in transforming and transmission, electric power
distribution today is nearly all alternating current.
For an alternating current, the voltage is:
V  V0 sin t where
  2f
V0 = peak voltage
The current:
I  I 0 sin t
I0 = peak current
I0 = V0/R
The power transformed in a resistance R :
P  I R  I R sin t
2
2
0
2
The rms (root-mean-square) value of the current or voltage:
rms current I rms 
I0
I 
 0.707 I 0
2
2
rms voltage V  V 2  V0  0.707V
rms
0
2
The average power:
P  I rmsVrms
1 2
2
P  I O R  I rms
R
2
2
1 V02 Vrms
P

2 R
R
Example : Hair dryer
(a) Calculate the resistance and the peak
current in a 1000-W hair dryer
connected to a 120V line.
(b) What happens if it is connected to a
240V line?
Solutions:
(a) The rms current
I rms
Then,
P 1000W


 8.33 A
Vrms
120V
I 0  2I rms  11.8 A
The resistance is:
Vrms 120V
R

 14.4
I rms 8.33 A
The resistance could equally well
be calculated using peak values:
V0 170V
R

 14.4
I 0 11.8 A
(b) When connected to a 240 V line, more current would flow
and the resistance would change with the increased
temperature. The average power would be:
2
Vrms
(240V ) 2
P

 4000W
R
(14.4)
This is four times the dryer’s
power rating and would
undoubtedly melt the heating
element or the wire coils of the
motor.
Electric devices
 References 
• Poh Liong Yong, (2018),Physics for Matriculation
Semester 2, Oxford Fajar, Fifth edition updated.
• Zainal Abidin Sulaiman et al.(2019), Physics for
Matriculation Semester 2, SAP Publication, 1st
Edition
• H. E. Ting (2020), Diagrams Physics for
Matriculation Semester 2, SAP Publication, 1st
Edition
• N. Sabirin M., Zainal A. S. (2019). Comprehensive
College Physics Upgraded. SAP Publication (M)
Sdn. Bhd.
Kata-kata Pujangga (bebas)