Electricity BHS Physical Science K Warne Grade 9 Revision Revision PresentationRevision PresentationRevision PresentationRevision Presentation.

Electricity

BHS Physical ScienceK Warne

Grade 9 Revision• Revision Presentation

Electrical Circuits

A

V1

VoltmeterAmmeter

Resistor

The Ammeter measures the ……………flowing in the circuit. (…….. A)

The Voltmeter Measures potential difference or …………….. in volts. (V)

The Resistance of the Resistor is given in …………… (Ω).

V1 = V2 V2

Voltage across …………..= voltage in ……………

Electric Current in a Conductor

Conventional current - __________________

Direct current - moves in _______________.

Alternating current - ___________________ continuously

Maintaining a current

Conductor - ___________ _____________________ _____________________ -

SOURCE

+ -

+ -

+ +

+

+

+

+

+ ++ +

+

+

+

+

+ ++ +

+

+

+

+

+ ++ +

+

+

+

+

+ ++ +

+

+

+

+

+ +

e-

e-

e-

e- e- e-

e-

e- e-

e-

e-

e- e- e-

e-

e- e-

e- e-

e- e-+ - e-

e-

e-

<------- electrons

“_____________” ------>Conventional current is the movement of from + to - in a conductor.

Conventional current e-

Electric Current & Voltage

The VOLTAGE is how much ENERGY they have.

+ -

+ -

Current is the number of soldiers moving past a point.

Conventional current e-

Potential Difference

The Voltmeter:

• is connected in ……………. to another component in the circuit (the current does NOT flow through it)

• has a …………. resistance• Is connected positive to positive - negative to

negative.

+_

VoltmeterV1.50

Resistor

V

The DIFFERENCE in POTENTIAL (energy) per unit ………………. of the current flowing between two points in the circuit. Measured by a ………………...

Volts =………….(J)

………….(C )

R

V =W or E (J)

Q (C )

?A

Resistance R

Example 1:

Calculate the potential difference between two points if 20 J of work are required to move a charge of 2 C.

Example 2:

Calculate the work done in moving a charge of 5 C through a potential difference of 2 V.

POTENTIAL DIFFERENCE

20J Energy lost

2 coulombs of charge

2VA

Resistance R

?? Energy lost or work done??

5 coulombs of charge

Vex

A

Resistance R

VCell

Emf 400V

I = 0A

Vcrt.= ….. v

Emf = …..cell

• Emf is the …………… amount of …………… that the cell can produce (per unit charge).

• Measured when the current in the circuit is ………..

EMF - Electro Motive Force

Open Circuit!!Vcell = EMF

Emf = V….. + V……

External Potential Difference…………Circuit!!

(EMF = 400V) 350V

Vex= VcellVex

A

Resistance R

Vcell

I = IA

Energy …………….. by battery is lost by resistance in the circuit.

The EMF of the cell is equal to the sum of the ………………..cell voltage and the ………… voltage.

This continues until the cells have no more energy.

400V

50V

350V

50V

Measuring Current

The Ammeter:• is connected in ……….in the circuit (the current flows

through it)• has a …………resistance • Is connected positive to positive - negative to negative.

An Ammeter measures the ……………… flowing through the circuit.

+_

A

AmmeterA1

The current is the number of charges passing a point in one second.

I = Q/t 1A = …..C/……s

Calculating CurrentCalculate the current flowing through the circuit.

+

_Ammeter

A0.53

If 160 C of charge flow through the ammeter in 3s what current is flowing?

I = Q/t

=

Parallel Circuits• Adding resistors in parallel…

decreases the total resistance.1 1 1

Rt R1 R2

1 1 1 Rt … …

1 1 1 1 Rt 2 2 2

Rt = …./…. = ……. Ω

= +

R1 2 Ω

2 Ω

R1

R2

2 Ω

Total R = …. Ω

Total R = … Ω

Total R = …… Ω

R1 2 Ω

2 ΩR2

R2 2 Ω

= +

= + +

Parallel Circuits• The voltage is EQUAL over the

resistances .

VT = …………….

• The current flowing is divided between the resistances and would increase as more resistances are added- more routes for the current to flow.

AT= ………………..

V1

R1

VT

V2

R2

2 V

….. V

….. V

R1A

R2

A1

A2

…. A

….. A

4 A

Parallel Circuits• Adding resistors in parallel…

decreases the total resistance.1 1 1

Rt R1 R2

• The voltage is EQUAL over the resistances .

VT = V1 = V2

• The current flowing is divided between the resistances and would increase as more resistances are added- more routes for the current to flow.

= + V1

R1

AVT

V2

R2

A1

A2

AT= A1 + A2

2 Ω

2 Ω

The current will divide in such a way that the potential lost by both all branches of current will be the same.

Series Circuits• Adding resistors in series…increases

the total resistance - because all the current flows through all the resistors.

Rt = R1 + R2

• The total potential difference (voltage) is the sum of the potential differences of the resistors – the total potential loss must equal the all the potential lost along the way.

Vt = V1 + V2The potential differences will be

proportional to the resistances.

• The current flowing is the same all over the circuit and would decrease as more resistances are added -

A = A2 = A3

VTA

V1 V2

R1 R2

VT

V1 v2

AVT

A2

A3

3Ω1Ω

12v

Worked Example

A

4 Ω

2 Ω

8 Ω

12 Ω

6 Ω

12v

4.1

Calculate I

Current & Resistance

RESISTANCE

• Electrical charge experiences ……………………as it moves through a conductor.

• The resistance is due to ……………….. with particles in the metal atoms and ions.

• The moving charges lose ……………………in the collisions which …………….up the conductor.

+_

CURRENT: An electrical current is a movement of ……………. through a conducting material from positive to negative. (?!)

Ohm’s Law - PracticalAIM:

– Investigate the relationship between the …………………….. across a resistor and the ………………..flowing through it.

– Determine the …………………..of a resistor.

V

A

METHOD:

1. Set up the circuit as shown.

2. Using the rheostat vary the current in the circuit, obtain a range of readings for the potential difference across R for different currents.

3. RESULTS>>Resistance R

rheostat

Ohm’s Law - Results

Draw a graph of your results.

ResultsAnalysis - Graph

I (A) V (V)

Ohm’s Law - Analysis

I (A) V (V) V/I0.80 3.40 4.30.81 3.50 4.30.85 3.70 4.40.90 3.90 4.30.95 4.20 4.4

The ratio V/I produces a constant value - for any resistor

This is the Resistance of the resistor.

The Unit of measurement for resistance is the Ohm - symbol (Ω)

The SLOPE of the graph gives the RESISTANCE.

Current vs Voltage

3.00

3.20

3.40

3.60

3.80

4.00

4.20

4.40

0.75 0.80 0.85 0.90 0.95 1.00Current (A)

Volta

ge (V

)

Riserun

Slope = DY/ DX =

Current, voltage & resistance

We define the unit of resistance; one …….() is one volt per ampere.

R = V/I 1 = …V/…A

+ -

+ -

A V

The relationship between the …………… through a resistor, the …………... drop across the resistor and the resistance of the resistor is expressed by the following equation:

R

V

Rx ..

Calculate the voltage drop across a 2 resistor when a current of 1.5 A is flowing..

Worked Example

A

4 Ω

2 Ω

8 Ω

12 Ω

6 Ω

12v

4.1

Calculate I = V/R = 12/6 = 2 A

Ohm’s Law+ -

+ -

A V

Factors affecting Resistance

1. Material

2. Length

3. Temperature

Effects of CurrentElectric current generates heat in a conductor.

+_A small current (0.1A) would have only a few charges flowing.

+_A large current (15A) would have a large number of charges flowing and generate far more heat.

As a conductor heats up the RESISTANCE INCREASES.

Effects of electric current An electric current that flows in a conductor has a number of

effects:

1. HEATING The friction caused by the current causes the conductor to heat up. The greater the current the more heat is generated.

2. MAGNETIC EFFECT - A magnetic field is generated around any conductor when an electric current flows through it.