CHAPTER 3: ELECTRIC CURRENT & DIRECT-CURRENT CIRCUIT

CHAPTER 3

ELECTRIC CURRENT
&

DIRECT-CURRENT
CIRCUIT

KMM

TOPIC 3:
ELECTRIC CURRENT AND

DIRECT CURRENT

Student Learning Time (SLT)

Lecture = 2 hours

Tutorial = 12 hours

List of subtopics

3.1 Electrical current 3.5 Resistors in series and parallel
3.6 Kirchhoff’s Rules
3.2 Ohm’s law and resistivity 3.7 Electrical energy and power
3.8 Potential divider
3.3 Variation of resistance with 3.9 Potentiometer
temperature

3.4 Electromotive force (emf),
internal resistance and potential
difference

2

3.1 Electrical Current

Learning Outcomes:

a) Describe microscopic model of current.

*Emphasize on the flow of free electrons in a metal. Include concept of
drift velocity.

b) Define electric current, =



c) Use electric current, = , =



3

3.1 a) Describe microscopic model of current

• 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 electric field, E.
• The velocity of these free electrons is called drift velocity.

4



• The drift velocity vtdootfhtehdeirferecetioenlecotfrtohnes is the mean velocity of the
electrons parallel electric field when a
potential difference (battery) is applied.

5

Current I is the rate at which

+ + charge moves through an area, A
such as the cross-section of a wire.
+ Conventional current is defined to

++ move in the direction of the

electric field.

a) Positive charges move in the

direction of the electric field and in

the same direction as the

- - conventional current.

- b) Negative charges move in the
direction opposite to the electric
- - field.

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• The direction of current flow –
from the positive terminal to the
negative one – was decided
before it was realized that
electrons are negatively charged.

• Therefore, current flows around
a circuit in the direction a
positive charge would move;
electrons move the other way.

• However, this does not matter in
most circuits.

7

3.1 b) Define 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 steady current is defined as =


• The unit of current is Ampere (A).

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3.1 c) Use electric current =
=

Example 3.1.1
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?

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Solution:

a) I =



= = 1.5 5.0 = 7.5 Coulombs

b) = , =

=

= = (1.5)(1.0) = 9.38 × 1018 electrons
1.6×10−19

10

Example 3.1.2
A silver wire carries a current of 3.0 A. Determine
a. the number of electrons per second pass through the wire,
b. the amount of charge flows through a cross-sectional area of the

wire in 55 s.
(Given charge of electron, e = 1.60  1019 C)

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Solution : I  3.0 A

a. By applying the equation of average current, thus

IQ and Q  Ne
t
 3.0  N 1.601019 N  1.881019 electrons s1
I  Ne t t
t

b. Given t  55 s , thus the amount of charge flows is

Q  It

Q  3.055 Q  165 C

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3.2 Ohm’s Law and Resistivity

Learning Outcomes:

a) State and use Ohm’s Law.

b) Define and use resistivity, =


c) Sketch V-I graph

(Experiment 2:Ohm’s Law)

d) Verify Ohm’s Law

(Experiment 2:Ohm’s Law)

e) Determine effective resistance of resistors in series and parallel by
graphing method.

(Experiment 2:Ohm’s Law) 13

3.2 a) State and use Ohm’s Law

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.

∝
=


= → = Ohm’s Law

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• 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)

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The resistance of R offered
by a conductor depends on
following factors:
i. length, l
ii. cross-sectional area, A

of the conductor
iii. nature of the material
iv. temperature of the

conductor

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3.2 b) Define and use resistivity

• Resistivity, ρ is a measure of a material’s ability to oppose the flow of
electric current through the material.

• Resistivity, ρ is defined as the resistance of a sample of the material of
cross-sectional area 1 m2 and of length 1 m.

• It is a constant value depends on the types of materials.

• Its formula is given by =


where = resistance (Ω)
A = cross-sectional area of the conductor (m2)

= length (m)
• Its unit is Ω m.

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• The material from which a wire is made is also affects its resistance:
current flows easily through conductors and only with difficulty
through insulators.

• The quantitative measure of a material’s opposition to the flow of
current is called resistivity.

• Resistivity depends only on the chemical composition of the material,
not its shape or size i.e. same material with different sizes will still
have the same resistivity.

• All conductors have smaller resistivity where insulators have larger
resistivity.

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20

Example 3.2.1
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?
Solution:
=

= (2.50 × 10−3)(1000)
= 2.50

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Example 3.2.2

A 2 m wire having diameter of 1 mm has a resistance of 0.45 Ω. What is
the resistivity of the material used to make the wire?

Solution:

Wire → Circular cross-section

= 2 = (1 × 10−3)2 = 7.85 × 10−7 m2
4 4

= = (0.45)(7.85 × 10−7) = 1.77 × 10−7Ω m
2

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Example 3.2.3

A wire of length 50 m carries a current of 10.0 A. The potential
difference across both end of the wire is 2.1 V. Determine
a. the electric field intensity.
b. the resistance of the wire.

Solution : l  50 m; I  10.0 A;V  2.1 V

a. The electric field intensity is

V  El

2.1  E50

E  0.042 V m1

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Example 3.2.3

Solution:

b. From the ohm’s law, therefore

V  IR
2.1  10.0R
R  0.21

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3.3 Variation of Resistance with Temperature

Learning Outcomes:
a) Explain the effect of temperature on electrical resistance in metals.
b) Use resistance, = 0 1 + − 0

* is at temperature 20 °C

25

3.3 a) Explain the effect of temperature on

electrical resistance in metals

Metal ion Free electron

• The resistance of metals increases with increasing temperature (T↑, R↑).

• As temperature increases, ions of the conductor vibrate with greater amplitude.

• More collisions occur between free electrons and ions.

• These electrons are slowed down thus resistance is increased. 26

3.3 b) Use resistance, = 0 1 + − 0

• Temperature coefficient of resistance, α is defined as the fractional
change in resistance per Celsius degree.

= ∆ ൗ 0
∆

• The resistance of a metal can be represented by the equation below:

= 0 1 + − 0 ; ∆ = − 0

= 0 + 0 ∆ ; ∆ = − 0

∆ = 0 ∆

• α is a constant value and it depends on the material (material

property). 27

Example 3.3.1

A platinum wire has a resistance of 0.50 Ω at 0 oC. 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 oC -1)

Solution:

∆ = ∆ = − 0 = 0.60 − 0.50 = 50.89 °
0 0 3.93 × 10−3(0.50)

∆ = − 0
= ∆ + 0

= 50.89 + 0

= 50.89 °

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Example 3.3.2
A narrow rod of pure iron has a resistance of 0.10 Ω at 20 oC. What is
its resistance at 50 oC? (α= 5.0 × 10−3 oC-1)
Solution:
Given: 0 = 0.10 Ω, 0 = 20 ° and = 50 °

= 0 1 + 5 . 0−× 100−3 50 − 20
= 0.10 1+

= 0.115 Ω

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Example 3.3.3

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.

Solution:

Given: = =2 , = =

= ; = 4 and = 3
= 3

= 3

4 4
2 = 3 2

= 3 OR = 1

3 30

Example 3.3.4

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?
(Given the temperature coefficient of resistivity for copper is 6.80 
103 C1)

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Solution : R0  25 103 ;T0  20 C; T  27 C

a. By using the equation for temperature variation of resistance, thus

R  R01 αT 

R  R0  R0αT and R  R0  R
R  R0αT

  R  25103 6.80103 27

R  4.59103 

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Solution : R0  25 103 ;T0  20 C; T  27 C

b. Given I0  10.0103 A

By using the equation for temperature variation of resistance,

thus R  R01 αT  where RV and R  V
I I0

V  V 1 αT 

I I0

  1 

 I
1 1 6.80103 27
10.0 103

I  8.45103 A

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3.4 Electromotive force (emf), internal
resistance and potential difference

Learning Outcomes:
a) Define emf , ε and internal resistance, r of a battery.
b) State factors that influence internal resistance.
c) Explain the relationship between emf of a battery and potential

difference across the battery terminals.
d) Use terminal voltage, = −

34

3.4 a) Define emf , ε and internal resistance, r
of a battery

• The emf of a battery is defined as the electrical energy that is
generated by a battery so that the charges can flow from one
terminal to another terminal of the battery through any resistor.

or
• The emf of a battery is the work done per unit charge

or
• The emf of a battery is the potential difference across the terminals

of the source in open circuit (open circuit = circuit is not complete,
therefore no current is flowing, I = 0).
• SI Unit: Volt (V)

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36

Emf and internal resistance
• The source that maintains the current in a closed circuit is called a

source of emf.
• Any devices that increase the potential energy of charges circulating

in circuits are sources of emf.
• A real battery has some internal resistance, r.
• Therefore, the terminal voltage is not necessarily equal to the emf.

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Battery

r

Resistor 38

• The terminal voltage, = − = −
• For the entire circuit,

= +
= +
= +

where = terminal voltage
Ɛ = emf of the battery

r = internal resistance

R = external resistance

3.4 b) State factors that influence internal
resistance

a) The surface area of electrodes - the larger the surface area of
electrodes, the less is the internal resistance.

b) The distance between the electrodes – the greater the distance
between the electrodes, the greater is the internal resistance.

c) The nature and concentration of the electrolyte – the less ionic the
electrolyte or the higher the concentration of electrolyte, the
greater is the internal resistance.

d) The temperature of the electrolyte – the higher the temperature of
the electrolyte, the less is the internal resistance.

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3.4 c) Explain the relationship between emf of a battery
and potential difference across the battery terminals

Notes:
• Vab < ε when the battery of emf, ε with internal resistance r is

connected to the external circuit with resistance R.
• Vab > ε when the battery of emf, ε is being charged by other battery.
• Vab = ε when the battery of emf, ε has no internal resistance (r =0)

and connected to the external circuit with resistance R.
40

Example 3.4.1
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?
Solution:
= +

= 0.50 0.10 + 11.5
= 11.55

41

Example 3.4.2

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?

Solution:

= +

= − = 9.0 − 8.9 = 1 Ω
0.10

42

Example 3.4.3

Refer to the circuit shown in figure above, R is an ohmic resistor of
resistance 10.0 Ω. Initially the switch is open. At time T the switch is
closed. The graph shows the voltmeter reading V before and after time
T. Determine the internal resistance of the battery and the current in
the circuit when the switch is closed.

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Solution:

Before time T, the voltmeter records the emf, Ɛ of the battery Ɛ = 6.0 V.
After time T, the voltmeter records the terminal p.d. across R

= = 5.0
5.0
= = 10.0 = 0.50

= +

= − = 6.0−5.0 = 2.0 Ω

0.50

44

Example 3.4.4

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.

Solution : ε  9.0 V; r  6.0  45

a. Given I  0.50 A

By applying the eεxpVressIrion for emf, thus

9.0  V  0.506.0

V  6.0 V

Example 3.4.4

Solution :

b. The current is maximum when the total external resistance, R =0,

therefore ε  IR  r

9.0  Imax 0  6.0

Imax  1.5 A

46

Example 3.4.5

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 in below.

V

rε

R

A

Determine the reading of the ammeter and voltmeter if both meters
are ideal.

47

Example 3.4.5

Solution : ε  12.0 V;r  1.0 ; R  5.0 

By applying the equation of e.m.f., the current in the circuit is

ε  IR  r

12.0  I5.0 1.0

I  2.0 A

Therefore the reading of the ammeter is 2.0 A

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Example 3.4.5

Solution : ε  12.0 V;r  1.0 ; R  5.0 

The voltmeter measures the potential difference across the terminals
of the battery equal to the potential difference across the total
external resistor, thus its reading is

V  IR

V  2.05.0

V 10 V

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