Your SlideShare is downloading. ×
0
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

# Classification of Substances

29,287

Published on

A set of slides created to teach Classification of Substances to learners following the NSC Caps curriculum in Cape Town.

A set of slides created to teach Classification of Substances to learners following the NSC Caps curriculum in Cape Town.

Published in: Education, Technology, Business
0 Comments
1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

No Downloads
Views
Total Views
29,287
On Slideshare
0
From Embeds
0
Number of Embeds
10
Actions
Shares
0
Downloads
118
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Transcript

• 1. Electricity BHS Physical Science K Warne
• 2. Grade 9 Revision
• Revision Presentation
• 3. Electrical Circuits A Voltmeter Ammeter 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). V 1 = V 2 V 2 Voltage across …………..= voltage in …………… V 1
• 4. Electrical Circuits A Voltmeter Ammeter Resistance The Ammeter measures the current flowing in the circuit. (Amps A) The Voltmeter Measures potential difference or voltage in volts. (V) The Resistance of the Resistor is given in Ohms (Ω). V 1 = V 2 V 2 Voltage across battery = voltage in circuit V 1
• 5. 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 -
• 6. Electric Current in a Conductor
• Conventional current - positive to negative
• Direct current - moves in one direction.
• Alternating current - changes direction continuously
• Maintaining a current
• Conductor - closed circuit
• Potential difference
• Replacement of charges - 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 “ Positive spaces” ------> Conventional current is the movement of from + to - in a conductor. Conventional current e -
• 7. Electric Current & Voltage + - + - Conventional current e -
• 8. Electric Current & Voltage + - + - Conventional current e -
• 9. Electric Current & Voltage
• You can think of an electric CURRENT as a flow of charges – almost like people moving through a corridor.
• The CURRENT is the NUMBER of charges ( people ) passing any point in a second.
• The VOLTAGE is how much ENERGY they have.
+ - + - Conventional current e -
• 10. 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.
+ _ Voltmeter V 1.50 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 ) Resistor
• 11. Potential Difference
• The Voltmeter :
• is connected in parallel to another component in the circuit (the current does NOT flow through it)
• has a HIGH resistance
• Is connected positive to positive - negative to negative.
+ _ Voltmeter V 1.50 V The difference in POTENTIAL ENERGY per unit charge of the current flowing between two points in the circuit. Measured by a Voltmeter. Volts = Energy (J) charge (C ) R V = W or E (J) Q (C ) ENERGY/WORK FLOW OF CHARGE Resistor
• 12.
• 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.
•
? A Resistance R POTENTIAL DIFFERENCE 20J Energy lost 2 coulombs of charge 2V A Resistance R ?? Energy lost or work done?? 5 coulombs of charge
• 13.
• Example 1:
• Calculate the potential difference between two points if 20 J of work are required to move a charge of 2 C.
•
• V = W/Q
• = 20/2 = 10 V
• Example 2:
•   Calculate the work done in moving a charge of 5 C through a potential difference of 2 V.
•
• W = QV
• = 5 x 2 = 10 J
V ex A Resistance R V Cell POTENTIAL DIFFERENCE
• 14.
• Emf is the …………… amount of …………… that the cell can produce (per unit charge).
• Measured when the current in the circuit is ………. .
I = 0A V crt. = ….. v Emf = ….. cell EMF - Electro Motive Force Open Circuit!! V cell = EMF V ex A Resistance R V Cell Emf 400V
• 15.
• Emf is the maximum amount of energy that the cell can produce per unit charge (voltage).
• Measured when the current in the circuit is ZERO .
I = 0A V crt = 0 v Emf = V cell EMF - Electro Motive Force Open Circuit!! V cell = EMF V ex A Resistance R V Cell Emf 400V
• 16. Emf = V ….. + V …… External Potential Difference ………… Circuit!! (EMF = 400V) 350V V ex = V cell 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 V ex A Resistance R V cell I = IA
• 17. Emf = V crt +V int External Potential Difference Closed Circuit!! EMF = 400V 350V V crt = V cell Energy supplied by battery is lost by resistance in the circuit. The EMF of the cell is equal to the sum of the internal cell voltage and the circuit voltage This continues until the cells have no more energy. 50V 350V 50V V ex A Resistance R V cell I = IA
• 18. 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. + _ Ammeter A 1 The current is the number of charges passing a point in one second . I = Q / t 1A = …..C/……s A
• 19. Measuring Current
• The Ammeter:
• is connected in series in the circuit (the current flows through it)
• has a low resistance
• Is connected positive to positive - negative to negative.
An Ammeter measures the current flowing through the circuit. + _ Ammeter A 0.0001 The current is the number of charges passing a point in one second . I = Q / t 1A = 1C/1s Q .. x .. A
• 20. Calculating Current Calculate the current flowing through the circuit. + _ Ammeter A 0.53 If 160  C of charge flow through the ammeter in 3  s what current is flowing? I = Q / t =
• 21. Calculating Current Calculate the current flowing through the circuit. + _ Ammeter A 0.053 If 160  C of charge flow through the ammeter in 3  s what current is flowing? I = Q / t = 160x10 -12 /3x10 -9 = 0.053 A
• 22. What is going to happen if other bulbs are switched on? Refer to both the current reading (Ammeter) and potential difference reading. (Volts) Write down your predictions then use the croc clips simulation (Parallel circuits – bulbs) to see if you are correct. Copy and paste the circuits into a blank slide in your presentation. Image: crocodile clip screenshot
• 23. X Refer to both the current reading (Ammeter) and potential difference reading. (Volts) What has caused these changes? Make a prediction for the observations if the third bulb was switched on. Image: crocodile clip screenshot Describe the changes you see (from previous slide).
• 24. Check your predictions. (from previous slide). Refer to both the current reading (Ammeter) and potential difference reading. (Volts) Try and state a general rule which therefore applies to this type of circuit. Where do you think this might be useful? Can you think of a situation where this type of circuit would not be suitable? Image: crocodile clip screenshot
• 25.
• As the number of resistors in parallel INCREASES
• The total CURRENT flowing
• .........................
• The total RESISTANCE must therefore
• ...........................
• the BRIGHTNESS of the bulbs
• .....................................
• The VOLTAGE across each bulb
• .........................................
• If the RESISTANCE is DOUBLED the CURRENT
• ........................
Parallel circuit Summary Images: crocodile clip screenshots
• 26.
• As the number of resistors in parallel INCREASES
• The total CURRENT flowing INCREASES
• The total RESISTANCE must therefore DECREASE
• the BRIGHTNESS of the bulbs remains the SAME
• The VOLTAGE across each bulb remains the SAME
• If the RESISTANCE is DOUBLED the CURRENT also DOUBLES
Parallel circuit Summary Images: crocodile clip screenshots
• 27.
• What do you think will happen as the number of resistors in SERIES INCREASES
• Predict what will happen to the:
• Brightness
• Total current
• Voltage across each bulb
Series circuits Images: crocodile clip screenshots
• 28. Series circuit Summary
• As the number of resistors in SERIES INCREASES
• The total CURRENT flowing ...........................
• The total RESISTANCE must therefore ...................
• the BRIGHTNESS of the bulbs ......................
• The VOLTAGE across each bulb ................ to the ..................................
• If the RESISTANCE is DOUBLED the CURRENT
• ......................
Images: crocodile clip screenshots
• 29.
• As the number of resistors in SERIES INCREASES...
• The total CURRENT flowing DECREASES
• The total RESISTANCE must therefore INCREASE
• the BRIGHTNESS of the bulbs DECREASES
• The VOLTAGE across each bulb ADDS UP to the TOTAL VOLTAGE
• If the RESISTANCE is DOUBLED the CURRENT HALVES
Images: crocodile clip screenshots Series circuit Summary
• 30. Parallel Circuits
• Adding resistors in parallel…decreases the total resistance .
• 1 1 1
• R t R 1 R 2
• 1 1 1
• R t … …
• 1 1 1 1
• R t 2 2 2
• R t = …./…. = ……. Ω
= + R 1 2 Ω 2 Ω R 1 R 2 2 Ω Total R = …. Ω Total R = … Ω Total R = …… Ω R 1 2 Ω 2 Ω R 2 R 2 2 Ω = + = + +
• 31. Parallel Circuits
• Adding resistors in parallel… decreases the total resistance !!!! .
• 1 1 1
• R t R 1 R 2
• 1 1 1
• R t 2 2
• 1 1 1 1
• R t 2 2 2
• R t = 2/3 = 0.67 Ω
= + R 1 2 Ω 2 Ω R 1 R 2 2 Ω Total R = 2 Ω Total R = 1 Ω Total R = 0.7 Ω R 1 2 Ω 2 Ω R 2 R 2 2 Ω = + = + +
• 32. Parallel Circuits
• The voltage is EQUAL over the resistances .
• V T = …………….
• The current flowing is divided between the resistances and would increase as more resistances are added- more routes for the current to flow .
A T = ……………….. V 1 R 1 V T V 2 R 2 2 V … .. V … .. V R 1 A R 2 A 1 A 2 … . A … .. A 4 A
• 33. Parallel Circuits
• The voltage is EQUAL over the resistances .
• V T = V 1 = V 2
• The current flowing is divided between the resistances and would increase as more resistances are added- more routes for the current to flow .
I T = I 1 + I 2 V 1 R 1 V T V 2 R 2 2 V 2 V 2 V R 1 A R 2 A 1 A 2 2 A 2 A 4 A
• 34. Parallel Circuits
• Adding resistors in parallel…decreases the total resistance .
• 1 1 1
• R t R 1 R 2
• The voltage is EQUAL over the resistances .
• V T = V 1 = V 2
• The current flowing is divided between the resistances and would increase as more resistances are added- more routes for the current to flow .
= + V 1 R 1 A V T V 2 R 2 A 1 A 2 A T = A 1 + A 2 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.
• 35. Series Circuits
• Adding resistors in series…increases the total resistance - because all the current flows through all the resistors.
• R t = R 1 + R 2
• 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.
• V t = V 1 + V 2
• The 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 = A 2 = A 3
A 3 3 Ω 1 Ω 12v V T V 1 v 2 A V T A 2
• 36. Worked Example 4.1 Calculate I A 4 Ω 2 Ω 8 Ω 12 Ω 6 Ω 12v
• 37. Worked Example 4.1 R // =(1/6+1/12) -1 = 4  R top = 4+4 = 8  R //2 = (1/8+1/8) -1 = 4  R tot = 2 + 4 = 6  I = V/R = 12/2 = 6 A A 4 Ω 2 Ω 8 Ω 12 Ω 6 Ω 12v
• 38. 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 . (?!)
• 39. Current & Resistance
• RESISTANCE
• Electrical charge experiences resistance as it moves through a conductor.
• The resistance is due to collisions with particles in the metal atoms and ions.
• The moving charges lose kinetic energy in the collisions which heat up the conductor.
+ _ CURRENT: An electrical current is a movement of charges through a conducting material from positive to negative . (?!)
• 40. Ohm’s Law - Practical
• AIM:
• Investigate the relationship between the …………………….. across a resistor and the ……………….. flowing through it.
• Determine the ………………….. of a resistor.
V A
• METHOD:
• Set up the circuit as shown.
• Using the rheostat vary the current in the circuit , obtain a range of readings for the potential difference across R for different currents.
• RESULTS>>
Resistance R rheostat
• 41. Ohm’s Law - Practical
• AIM:
• Investigate the relationship between the potential difference across a resistor and the current flowing through it.
• Determine the resistance of a resistor.
V A
• METHOD:
• Set up the circuit as shown.
• Using the rheostat to vary the current in the circuit , obtain a range of readings for the potential difference across R for different currents.
• RESULTS>>
Resistance R rheostat
• 42. Ohm’s Law - Results Draw a graph of your results. Results Analysis - Graph I (A) V (V)
• 43. Ohm’s Law - Results The graph is a straight line showing that the current (I) is directly proportional to the voltage (V). Results Analysis 0.95/0.8 = 1.1875 4.2/3.4 = 1.2
• 44. Ohm’s Law - Analysis 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 . Rise run Slope =  Y/  X =
• 45. Ohm’s Law - Analysis 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 . Rise run Slope =  Y/  X = (4.2-3.2)/(0.95-0.78) = 5.88
• 46. Current, voltage & resistance
• We define the unit of resistance; one ……. (  ) is one volt per ampere.
• R = V / I 1  = …V / …A
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.. + - + - A V
• 47. Current, voltage & resistance
• We define the unit of resistance; one ohm (  ) is one volt per ampere.
• R = V / I 1  = 1V / 1A
The relationship between the current through a resistor, the voltage drop across the resistor and the resistance of the resistor is expressed by the following equation: R V R x I
• Calculate the voltage drop across a 2  resistor when a current of 1.5 A is flowing..
• V = IR = (1.5)(2) = 3 V
+ - + - A V
• 48. Worked Example 4.1 Calculate I A 4 Ω 2 Ω 8 Ω 12 Ω 6 Ω 12v
• 49. Worked Example A 4 Ω 2 Ω 8 Ω 12 Ω 6 Ω 12v 4.1 Calculate I = V/R R //1 R //1 = ((1/6)+(1/12)) -1 = ((2/12) +(1/12)) -1 = (3/12) -1 = 12/3 = 4 Ω 4 Ω Therefore large parallel combination becomes 8 Ω on top and 8 Ω on bottom which gives a total parallel resistance of 4 Ω Combining this with the 2 Ω series resistor gives a TOTAL RESISTANCE OF 6 Ω. CALCULATE CURRENT FOR HW
• 50. Ohm’s Law
• Factors affecting Resistance
• Material
• Length
• Temperature
+ - + - A V
• 51. Ohm’s Law
• Material - different materials have different resistances.
• Length – resistance increase with length – more resistance (collisions)
• Temperature – higher temp gives more resistance – particles move faster .: have more collisions
• Thickness (cross sectional area) – more space for charges to move through – less collisions – less resistance.
• Factors affecting Resistance
• Material
• Length
• Temperature
+ - + - A V
• 52. Effects of Current Electric current generates heat in a conductor. 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 . + _ A small current (0.1A) would have only a few charges flowing. + _
• 53. Effects of electric current
• An electric current that flows in a conductor has a number of effects:
• HEATING The friction caused by the current causes the conductor to heat up. The greater the current the more heat is generated.
• MAGNETIC EFFECT - A magnetic field is generated around any conductor when an electric current flows through it.
• 54. Hope you found this useful. Visit my site for more presentations & resources. Keith Warne www. Teach Bomb .com