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Home End HolisticTuition CashPlants Chapter 2: Chapter 2: Electricity Electricity Form 5 Form 5 1 Physics Next > The study of matter The study of matter
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Page 1: Home End HolisticTuition CashPlants Chapter 2: Electricity Form 5 1 Physics Next > The study of matter.

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Chapter 2: Chapter 2: ElectricityElectricity

Form 5Form 5

1

PhysicsNext >

The study of The study of mattermatter

Page 2: Home End HolisticTuition CashPlants Chapter 2: Electricity Form 5 1 Physics Next > The study of matter.

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Objectives: Objectives: ((what you will learnwhat you will learn)) 1) electric fields & charge flow

2) electric current & potential difference

3) series & parallel circuits

4) electromotive force & internal resistance

5) electrical energy & power

Physics: Chapter Physics: Chapter 22

2

< Back

Next >

Page 3: Home End HolisticTuition CashPlants Chapter 2: Electricity Form 5 1 Physics Next > The study of matter.

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Electric FieldsElectric FieldsElectric field: region where a charged body

experiences a force

It is shown by a field pattern that are lines of forces.

line of force = path of a test charge in the field

direction = motion of a free positive charge

+

Positive point charge

Negative point charge

electric field pattern

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Electric FieldsElectric Fields

Between a positive and a negative point

charge

Between two positive point

charges

Electric lines of force

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Electric FieldsElectric FieldsElectric field between two parallel metal plates that are oppositely charged.

Electric field between two opposite charges.

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Electric FieldsElectric Fields

+ –

+ –

+ –F F

Ball coated with conductor hangs vertically in the centre because it is neutral.

Ball oscillating between 2 plates, after it touches one side causing a force, F to repel the ball due to like charges.

+ –

Negative ions

Positive ions

Candle flame spreading sideways between 2 plates due to attraction between oppositely charged ions and metal plates.

Experiments to show existence of electric fields.

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Electric FieldsElectric FieldsElectric fields cause charges to move.

Net movement of charges = electric current

In the late 1700s scientists chose the direction of electric current to be the direction in which positive charges move in an electric field. They did not know that electrons and protons were the negative and positive charge particles, and that the electron moved much more easily.

In a copper wire, the outer electrons of the copper atom move relative to the nucleus of the atom. + -

Current, I

electrons

So, the charge carriers (electrons) move in the opposite direction to the current.

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Electric ChargeElectric Charge

8

Electric current = Rate of flow of electric charge

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Next > Electric charge, Q = It

units Q in Coulomb, I in Ampere, t in second

I =

Q

t , t =

time

C = A s

Basic unit of electric charge = Coulomb (C)

Charge of a proton or electron =  ± 1.60 10-19 C

A Coulomb of charge is a lot, at 6.25 x 1018 electrons – most objects have charges in the µC (10-6 C) range.

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Potential Potential DifferenceDifference

V =

W Q

Work done Charg

e

=

Potential difference (V) between 2 points in an electric field = work done (W) in moving 1 coulomb of charge (Q) between the 2 points.

Unit of potential difference:

Volt (V) = = J C-1

J

C

A BMoving 1 coulomb of

charge

Potential difference between 2 points

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Electric CurrentElectric CurrentOhm’s LawThe current (I) in a conductor is directly proportional to the potential difference (V) across the conductor if the temperature is constant.V

I = constant

Ohmic conductorA conductor that obeys Ohm’s Law.

I

V0

A

V

I

Conductor

Switch

Rheostat Circuit used to find the relationship between current I and potential difference V for a conductor.

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Electric CurrentElectric CurrentNon-ohmic conductorA conductor that does not obey Ohm’s Law.

I

V0

I

V0

I

V0Dilute sulphuric

acidFilament

lampJunction diode

Examples

A circuit element is non-ohmic if the graph of current versus voltage is nonlinear.

A filament lamp is a non-ohmic conductor since its resistivity, like most materials, varies with temperature. As the filament gets hot, the resistance increases quickly.

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ResistanceResistanceThe resistance, R of a conductor is defined as the ratio of the potential difference V across the conductor to the current I in the conductor.

V I

Resistance, R =

The unit of resistance is the ohm (Ω).

conductor

V

I I

Potential difference, V = IR

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ResistanceResistanceFactors that affect the resistance of a conductor:a. length of wire, lb. cross-sectional area, Ac. type of material with resistivity, pd. temperature, T

pl A

Resistance, R =

Based on a constant temperature:

R

T/oC0Metal Semi-conductor

R

0 T/oC

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Series CircuitSeries Circuit

IR1 R2 R3

V1 V2 V3

V

V1 = IR1

V2 = IR2

V3 = IR3

When resistors are connected in series:

a. Same current I is in all the resistors

b. Potential difference,

c. V = V1 + V2 + V3

d. Effective resistance,

R = R1 + R2 + R3

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Parallel CircuitParallel Circuit

IR1 I1

R2 I2

R3 I3

V

When resistors are connected in parallel:

a. Same potential differences across all resistors, V

1 R

1 R1

1 R3

1 R2

= + +

c. I = I1 + I2 + I3

d. Effective resistance,

b. Current in the resistors,V R1 I1 =

V R2 I2 =

V R3 I3 =

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Electromotive Electromotive ForceForceElectromotive force (e.m.f.), E

Work done to drive a unit charge (1 C) around circuit – where the unit is

volt, V = J C-1

Using a high resistance voltmeter

Potential difference V < e.m.f. E because work is done to drive a charge through a cell with internal resistance, r.

E = 1.5 V

V

I

r

V

RI

E = V + Ir = I(R + r)

E V

R + r R

r R= = 1

+

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Electrical EnergyElectrical EnergyThe potential difference V across a conductor is the work done in moving a charge of 1 C across the conductor. The work done is transformed into heat which is dissipated from the conductor.

From volt, V = J C-1 =

Energy dissipated, ECharge, Q

Energy dissipated, E = QV Q = It = IVt V = IR

= I2Rt I = V/R

V2t

RE =

substitutions

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Electrical PowerElectrical PowerElectrical power, P =

Energy dissipatedTime, t

V2

RP =

= I2R

I = V/R= IV

V = IRsubstitutions

E = IVt

Power rating of an electrical appliance is the power consumed by it when the stated voltage is applied. V2

PResistance of the appliance, R =1 unit of electrical energy consumed = 1 kW h= (1000 Js-1)(3600 s) = 3.6 x 106 J

Cost of electrical energy = units x cost per unit

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SummarySummary

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What you have learned:What you have learned:

1.1. Electric fields & charge flowElectric fields & charge flow

Thank YouThank You

2.2. Electric current & potential Electric current & potential differencedifference

3.3. Series & parallel circuitsSeries & parallel circuits

4.4. Electromotive force & internal Electromotive force & internal resistanceresistance

5.5. Electrical energy & powerElectrical energy & power


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