LECTURE 6: CURRENT ELECTRICITY

ELECTRICITY AND MAGNETISM I

ISAAC KWESI ACQUAH

 CURRENT ELECTRICITY
Within a battery, a chemical reaction occurs that transfers
electrons from one terminal (leaving it positively
charged) to another terminal (leaving it negatively
charged).

The symbol ( ) represent a cell in circuit

drawings.

Because of the +ve and –ve charges on the battery

terminals, an electric potential difference (Pd) exists
between them. The maximum Pd is called the
electromotive force (emf) of the battery, for which the
symbol E is used.

In a typical car battery, the chemical reaction maintains

the potential of the +ve terminal at a maximum of 12 volts
(12 joules/coulomb) higher than the potential of the –ve
terminal, so the emf is E = 12 V.
 CURRENT ELECTRICITY
Thus one 1C of charge emerging from the battery
and entering a circuit has at most 12 J of energy.

In a circuit, the battery creates an electric field

within and parallel to the wire, directed from the +ve
towards the –ve terminal.

The electric field exerts a force on the free electrons

in the wire, and they respond by moving. This flow of
charge is known as an electric current.
The electric current I is defined as the amount of
charge per unit time that crosses the imaginary
surface that is perpendicular to the motion of the
charges.
 CURRENT ELECTRICITY

Each elctron is assumed to move with same constant

velocity v, and in time dt each advances by vdt. The
electrons have a volume = Avdt
Let n be free electrons per unit volume, then the total
number of electrons in the volume = nAvdt, if e
represent a charge on electron then dq = nevAdt
The rate at which charge is transported across a
section of the wire, or dq/dt is called the current I in
the wire.
Since the units for charge and time are the coulomb
(C) and the second (s), the SI unit for current is a
coulomb per second (C/s).
 CURRENT ELECTRICITY
One coulomb per second is referred to as an
ampere (A).

If the charges move around a circuit in the same

direction at all times, the current is said to be direct
current (dc), which is the kind produced by batteries.

In contrast, the current is said to be alternating

current (ac) when the charges move first one way
and then the opposite way, changing direction from
moment to moment, for example generators and
microphones produce ac.
 CURRENT ELECTRICITY
Electrons leave the negative terminal of the battery,
pass through the device, and arrive at the positive
terminal.

Conventional current originate from the positive

terminal. The direction of conventional current is
always from a point of higher potential toward a point
of lower potential, thus from the +ve toward the –ve
terminal.
 INTRODUCTION TO ELECTROSTATICS
ELECTRIC CURRENT
Electric current is the flow of electric charge through a
material.

It is measured in amperes (A) and is the rate of flow of

electric charge through a conductor.

Electric current can be direct current (DC) or alternating

current (AC).

DC flows in one direction and AC changes direction

periodically.
 INTRODUCTION TO ELECTROSTATICS
USES OF ELECTRIC CURRENT
Electric current is used to power many everyday appliances
and devices. It is used to power lights, motors, and other
electronics.

Electric current is also used in industrial processes, such as

welding and electroplating.

It is also used in medical applications, such as

electrocardiograms and defibrillators.
 ELECTRIC CURRENT
The purpose of a battery is to produce a Potential
Difference(PD), which can then make charges move. When
a continuous conducting path is connected between the
terminals of a battery, we have an ELECTRIC CIRCUIT,

A simple electric circuit

On any diagram of a circuit we use the symbol below

to represent a battery.
 ELECTRIC CURRENT
The device connected to the battery could be a lightbulb, a
heater, a radio, or some other device.

When such a circuit is closed, charge can move (or flow)

through the wires of the circuit, from one terminal of the
battery to the other, as long as the conducting path is
continuous.

Any flow of charge such as this is called an electric current.

 CONTD.
Hence , 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.

Thus, the current I is defined as

Where ∆Q is the amount of charge that passes through the

conductor at any location during the time interval ∆t.
 CONTD
Electric current is measured in coulombs per second; this is
given a special name, the ampere (abbreviated amp or A),
after the French physicist André Ampère.

Thus 1A = 1C/s. smaller units of current are often used.

Example 1: A steady current of 2.5 A exists in a wire for 4.0

minutes. How much total charge passes by a given point in
the circuit during those 4.0 minutes?
SOLUTION
 DEFINATION OF SOME COMMON TERMS
Note the following terms associated with electricity

CIRCUIT: It is a path along which an electric current flows in an

electrical network.

OPEN CIRCUIT: This is a circuit which has a break at one or more

points so that no current flows.

SHORT CIRCUIT: A bypass for current; it provides an easier or

shorter path for current to flow

POTENTIAL DIFFERENCE (P.D.) : P.d. is defined as the work done

in moving a coulomb of charge from one point to the other of a
conductor. It is the energy used in driving charges through
external resistances. The unit of p.d. is J/C or volts.
 CONTD
ELECTROMOTIVE FORCE (E.M.F.): P.d. is established in a
circuit using a source of E.M.F. This (E.M.F.) is the driving
force which causes current to flow in a circuit.

 It is defined as the energy supplied by a cell (or battery) to

drive charges through both internal and external
resistances.

 It is the p.d. between the terminals of a cell when it is not

delivering any current. E.M.F. is the sum of the p.d.s across
all components in a closed circuit, including that required to
drive the current through the cell itself. It is also measured
in volts
 CONTD
CELL: It is a device that converts chemical energy directly
into electrical energy.

In a cell chemical reactions take place, releasing energy.

This energy creates a potential difference (p.d.) across the
ends of a conductor to drive electrons through it as current.

BATTERY: It is a collection of two or more cells connected

together

VOLTAGE: It is the amount energy per charge move of

electrons from one point to another in a circuit. Its
measured in volts.
 THE OHM’S LAW
In the early 19th century, Georg Ohm, a German physicist,
discovered the relationship between electric current and
voltage.

Ohm's Law states that the current flowing through a

conductor between two points is directly proportional to
the voltage applied across the two points.

This relationship between current and voltage, which is

now known as Ohm's Law, was the foundation of the study
of electricity and electronics.
 THE OHM’S LAW
Today, Ohm's Law is used in a variety of applications. It is
used to calculate the amount of current flowing through a
circuit, the voltage needed to power a circuit, and the
resistance of a circuit.

Ohm's Law is also used to calculate the power dissipated in

a circuit, which is important for designing efficient and safe
circuits
 THE OHM’S LAW
The Ohm’s law states that at constant temperature the
current passing through a conductor is directly proportional
to the applied voltage at its ends.

Current through an ideal conductor is directly proportional

to the applied voltage. Implies
𝒗∝𝑰

A plot of 𝒗 against 𝑰 gives a straight line passing passing

through the origin. The gradient of the line is the resistance
of the conductor through which the current is passing.

𝒗 = 𝑰𝑹 … … … . 𝑶𝒉𝒎′ 𝒔𝒍𝒂𝒘
 CONTD
 Where R is the resistance, The unit of resistance is Ω
(ohm).

The ohm is defined as the resistance of a conductor

through which a current of one ampere flows when a P.D. of
1 volt is maintained across it.

Not all materials obey Ohm’s law. Those that obey are
referred to as OHMIC, while those which don’t are NON
OHMIC.
 VERIFICATION OF OHM’S LAW

The key in the above circuit is closed. The rheostat is used to vary the
current flowing through the resistance wire R and hence the P.D.
across it.

Values of currents are measured using the ammeter and the

corresponding P.D.s across R measured with the potentiometer.

P.D. (V) values are plotted on the y-axis against current (I) on the x-
axis. For an Ohmic conductor, this gives a straight line passing through
the origin. The gradient of the line is found to be the resistance of the
conductor. This verifies Ohm’s law.
•
 CONTD
A graph of voltage(P.d) against current(I)
 RESISTANCE AND RESISTIVITY
Resistance and Resistivity
 In a water pipe, the length and x-sectional area of the
pipe determine the resistance that the pipe offers to
the flow of water.
 Longer pipes with smaller x-sectional areas offer
greater resistance. Analogous effect are found in the
elctrical case. For a wide range of materials, the
resistance of a piece of material of length L and x-
sectional area A is
 R = ρL/A,
 where ρ is a proportionality constant known as the
resistivity of the material. SI unit of resistivity is the
ohm-meter (Ωm).
 RESISTANCE AND RESISTIVITY
The resistance of a conductor is a property of the conducting material.
Resistance, R, is directly proportional to length, 𝒍, and inversely
proportional to the cross-sectional area, A, of the conductor.
𝒍
𝑹∝
𝑨

𝒍
∴ 𝑹=𝝆
𝑨
where ρ is the constant of proportionality, known as the resistivity of
the material.
𝑹𝑨
𝝆=
𝒍
Resistivity is defined as the resistance of a material of unit cross-
sectional area and unit length. It has units of ohm-meter (Ωm).
 RESISTANCE AND RESISTIVITY
All the conductors (metals) have small resistivities.
Insulators (rubber) have large resistivities. Materials like
germanium and silicon have intermediate resistivity
values and are called semiconductors
Resistivity is an inherent property of a material, inherent
in the same sense that density is an inherent property.
Resistance, on the other hand, depends on both the
resistivity and the geometry of the material.
A short wire with a large x-sectional area has a smaller
resistance than does a long, thin wire. Wires that carry
large currents, such as main power cables, are thick
rather than thin so that the resistance of the wires is kept
as small as possible.
 ARRANGEMENT OF RESISTORS
PARALLEL ARRANGEMENT
When current approaches a system of resistor connected in
parallel, it splits up to flow through each of them.
The magnitude of the current passing through each resistor
depends on the size of the resistance.
The p.d.s across them are however the same. From Ohm’s
law
 ARRANGEMENT OF RESISTORS
PARALLEL ARRANGEMENT
R is the resultant resistance or the effective resistance due
to all three resistors. For n resistors,
 ARRANGEMENT OF RESISTORS
SERIES ARRANGEMENT
For a series connection, the same current passes through
the resistors. The p.d. established across each is however
different. From Ohm’s law, 𝑣 = 𝐼𝑅
 ARRANGEMENT OF RESISTORS
EXAMPLES 1
A charge of 3 × 10−6 𝑪 passes through a wire in 5 seconds.
(a)What is the current in the wire?
(b)If the wire carries a current of 5A, how much charge moves
through it in 4 seconds?
SOLUTION
(a)
𝑄 3 × 10−6
𝐼= = = 6 × 10−7 𝐴
𝑡 5
(b)
𝑄 = 𝐼𝑡 = 5 × 4 = 20𝐶
 ARRANGEMENT OF RESISTORS
EXAMPLES 2
A silver wire, 1mm in diameter, carries a charge of 90C in 2
hours 15 minutes. If silver contains 5.8 × 1028 electrons per
metre cube,
(a)What is the current in the wire?

SOLUTION

𝑸 = 90𝐶, 𝒕 = 2ℎ𝑟𝑠 15𝑚𝑖𝑛 = 135𝑚𝑖𝑛 = 8100𝑠, 𝒏 =

5.8 × 1028
 ARRANGEMENT OF RESISTORS
(a)
𝑄 90
𝑰= = = 0.0111𝐴
𝑡 8100

(a)
𝜋𝑑2 𝜋 × 1 × 10−3 2
𝑨= = = 7.8540 × 10−7 𝑚2
4 4

𝐼 0.0111
𝒗= =
𝑛𝑒𝐴 5.8 × 1028 × 1.6 × 10−19 × 7.854 × 10−7

= 1.52 × 10−6 𝑚𝑠 −1
 ARRANGEMENT OF RESISTORS
EXAMPLES 3
Calculate the resistance of a 50cm wire with diameter 5mm
and resistivity 1.5 × 10−7 Ω𝑚.
SOLUTION
𝒍 = 50𝑐𝑚 = 0.5𝑚, 𝒅 = 5𝑚𝑚 = 5 × 10−3 𝑚, 𝝆 = 1.5 ×
10−7 Ω𝑚

𝜋𝑑 2 𝜋 × 5 × 10 −3 2
𝑨 = 𝝅𝒓𝟐 = = = 1.9635 × 10−5 𝑚2
4 4

𝝆𝒍 1.5 × 10−7 × 0.5 −3 Ω

𝑹= = = 3.8197 × 10
𝑨 1.9635 × 10−5
 ARRANGEMENT OF RESISTORS
EXAMPLE 4
Given three resistors 1Ω, 2Ω and 3Ω, what will the effective
resistance be if connected in (a) series (b) parallel
SOLUTION
𝑅1 = 1Ω, 𝑅2 = 2Ω, 𝑅3 = 3Ω
(a)

𝑅 = 𝑅1 + 𝑅2 + 𝑅3 = 1 + 2 + 3 = 6Ω
(a)
1 1 1 1 1 1 1 11
= + + = + + =
𝑅 𝑅1 𝑅2 𝑅3 1 2 3 6

6
∴ 𝑅= = 0.5455Ω
11
 ARRANGEMENT OF RESISTORS
EXAMPLE 5
The p.d. acrossthe resistors inthe circuit above is 240V.
Find
(a) the effective resistance in the circuit
(b) the current I, flowing
(c) the current I1, flowing through the 250Ω resistor
(d) the current I2, flowing through the 400Ω resistor
SOLUTION (c)
1 1 1 1 1 650
(a) = + = + = 𝑣 240
𝑅 𝑅1 𝑅2 250 400 1×105
𝐼1 = = = 0.96𝐴
𝑅1 250
∴ 𝑅 = 153.85Ω (d)
(a) 𝑣 240
𝑣 240 𝐼2 = = = 0.6𝐴
𝐼= = = 1.56𝐴 𝑅2 400
𝑅 153.85 Alternatively,
𝐼2 = 𝐼 − 𝐼1 = 1.56 − 0.96 = 0.6𝐴
 ARRANGEMENT OF RESISTORS
EXAMPLE 6
(a)What is the effective resistance in the circuit diagram
above?
(b)How much current is registered by the ammeter?
(c)Calculate the voltage across the parallel resistors.
 ARRANGEMENT OF RESISTORS
SOLUTION
(a) For the resistors in parallel, (c)
1 1 1 1 1 10
= + = + = 𝑣 = 𝑣12 + 𝑣3
𝑅𝑝 𝑅1 𝑅2 6 4 24

24 𝑣12 = 𝑣 − 𝑣3
⟹ 𝑅𝑝 = = 2.4Ω
10
But 𝑣3 = 𝐼𝑅3 = 3 × 1.6 =
Therefore effective resistance
4.8𝑉
𝑅 = 𝑅𝑝 + 𝑅3 = 2.4 + 1.6 = 4Ω
∴ 𝑣12 = 12 − 4.8 = 7.2𝑉
(b)
𝑣 12
𝐼= = = 3𝐴
𝑅 4
 ARRANGEMENT OF RESISTORS
EXAMPLE 7
Calculate the
(a)current I
(b)p.d. across the external resistors and
(c)current passing through each of the resistors.
 ARRANGEMENT OF RESISTORS
SOLUTION
(a)The effective external resistance (a) 𝑣1 = 𝐼1 𝑅1
𝑅1 𝑅2 1×1 1
𝑅= = =
𝑅1 + 𝑅2 1 + 1 2 𝑣 0.67
𝐼1 = = = 0.6667𝐴
𝑅1 1
𝐸 = 𝐼(𝑅 + 𝑟)
⟹ 2 = 𝐼 0.5 + 1 𝑣2 = 𝐼2 𝑅2

2
∴ 𝐼= = 1.3333𝐴 𝑣 0.67
1.5 𝐼2 = = = 0.6667𝐴
𝑅2 1

(a) 𝑣 = 𝐼𝑅 =
1.3333 × 0.5 = 0.6667𝑉