1. Electricity and Circuits
Unit 11
Electricity and Circuits

2. Starter If you are given a light bulb,
What other equipment would you need to light it up?
How would you set up the equipment? 2

3. I. Current
Battery acts as the pump to push the charges (electrons) around the circuit 3

4. I. Current B. Current (definition): HOW QUICKLY CHARGE MOVES
Equation (see ref. tabs.)
I = Current
Δq = change in charge
t = time
I = Δq t
Units:
C/s = Ampere/amp (A) 4

5. “It’s the volts what jolts, but it’s the mills that kills.”
I. Current
Note: High currents through your body can cause serious injury or death. Here are a few of the typical consequences of different currents
“It’s the volts what jolts, but it’s the mills that kills.” 5

6. II. Conditions Necessary for Electric Current
CHANGE IN
POTENTIAL
What is needed? A ___________ between ______ points in the circuit 2
Provided by:
Battery or chemical cell
Most types of 9 V batteries consists of V cells added in series 6

7. III. Ohm's Law
A. Potential Difference (V): Change in electric potential energy between two positions
“PUSH”
Provides the electrical __________
Units:
Volts (V)
OPPOSES
B. Resistance: ____________ the ________ of charges
FLOW
Units:
Ohm (Ω) 7

8. C. Ohm’s Law (see ref. tabs.)
R = Resistance (Ω)
V = Potential Difference (V)
I = Current (A)
R = V I 8

9. IV. How to Measure Current:
Device used: ____________
AMMETER
SERIES (LINE)
How to set it up: Place in ______________ with circuit 9

10. V. How to Measure Potential Difference (voltage):
Device used:
VOLTMETER
ACROSS
How to set it up: Place it _________ the device you want to measure 10

11. Whiteboard Problems:
20 coulombs of charge pass a given point in a conductor in 4 s. Calculate the current in this conductor.
In a television set, an electron beam with a current of 5 x ampere is directed at the screen. Approximately how many electrons are transferred to the screen in 60 seconds?
A potential difference of 12 volts is applied across a circuit which has a 4 ohm resistance. What is the magnitude of the current in the circuit?
A lamp has a current of 2 milliamperes at 6 volts. What is the resistance of the lamp?
5 A
3 x C 
1.88 x electrons
3 A
3000 Ohm 11

12. VI. Resistivity
How well a specific conductor allows a current to move through it
Factors that affect the resistance:
1. Length (↑ L, ↑ R)
2. Area (↑ A, ↓ R)
3. Material (resistivity – ↑ ρ, ↑ R) (see ref. tabs)
4. Temperature (↑ T, ↑ R) 12

13. VI. Resistivity
Equation (see ref. tabs) ↑ 13

14. VI. Resistivity
Example: Determine the resistance of a copper wire that has a cross-sectional area of 2 x m 2 and 40 m long. 14

15. VI. Resistivity
Example: Determine the resistance of a 15 gauge (radius = 7.25x m) gold wire that is 5.0 m long. 15

16. VI. Resistivity 3 A decrease 0.5 A 2.44 x 10 -6 ohms 2.13 x 10 6 m
Whiteboard:
A potential difference of 12 volts is applied across a circuit having a 4 ohm resistance. What is the current in the circuit?
If the temperature of a metal conductor is reduced, its resistance will increase, decrease, or remain the same.
A 60 W light bulb has a resistance of 240 ohms. If the lamp in which the bulb is used is plugged into a 120 volt outlet, what current does the bulb draw?
Calculate the resistance of 3 m gold wire with a cross sectional area of 0.03 m2.
Calculate the length of an aluminum wire that has a cross-sectional area of 3 x m2 and a resistance of 20 ohms.
An aluminum wire has a resistance of 48 ohms. What is the resistance of a second piece of wire of the same composition, same diameter, and at the same temperature, but with one half the length of the first wire?
Sketch a graph that shows the relationship between the resistance of a copper wire (y-axis) of uniform cross-sectional area and the wire’s length (x-axis) at constant temperature.
3 A
decrease
0.5 A
2.44 x ohms
2.13 x 10 6 m
24 ohms 16

17. VII. Electric Power Mechanics’ Equation for Power: P = W t Units:
J/s = Watt (W)
Electric Power:
P = W = IV t
Alternate Equations (see ref. tabs.):
P = IV = I2R = V2 R 17

18. VII. Electric Power
Example 1: A microwave draws 12.5 amperes of current and resistance of 9.6 ohms. What is the power dissipated in the resistor? 18

19. VII. Electric Power
Example 2: A 60 W and 100 W light bulb are plugged into the wall that provides a potential difference of 120 V. Which bulb has the greater resistance? What is the resistance of both? 19

20. VIII. Electrical Energy
Equations (see Reference Tables):
W = Pt = IVt = I2Rt = V2t R
Units:
Joule (J) 20

21. VIII. Electrical Energy
Example 1: A television set draws 2 A when operated on 120V.
How much power does the set use?
Calculate how much energy is used if the television is on for 2 hours a day. 21

22. VIII. Electrical Energy
Example 2: An electric dryer consumes 1.0 x 106 J of energy when operating at 220 volts for 2 minutes. During operation, how much current does the dryer draw? 22

23. All parts are connected to provide a ______________ for the current
IX. Series Circuit
All parts are connected to provide a ______________ for the current
SINGLE PATH
Schematic Diagram: Draw a circuit diagram with three resistors set up in series connected to a 12 V battery.
PHET CIRCUIT SIMULATION 23

24. IX. Series Circuit
Things to Remember for Series Circuit (See Ref. Tabs)
1. Current
Remains constant through each resistor
Equation:
I = I1 = I2 = I3 = … 24

25. IX. Series Circuit
Things to Remember for Series Circuit (See Ref. Tabs)
2. Equivalent Resistance
Total resistance is equal to the sum of all resistors
Equation:
Req = R1 + R2 + R3 + … 25

26. IX. Series Circuit
Things to Remember for Series Circuit (See Ref. Tabs)
3. Voltage Drops
Applied (total) voltage from the power source equals the sum of the voltages across each device (resistor)
Equation:
V = V1 + V2 + V3 + … 26

27. IX. Series Circuit
Example: Draw a circuit diagram for series circuit that contains a 5 ohm, 10 ohm, and 15 ohm resistor that are connected to a 20 V battery. Include an ammeter to read the total current and three voltmeters to measure the potential difference across each resistor. 27

28. IX. Series Circuit
Example: Draw a circuit diagram for series circuit that contains a 5 ohm, 10 ohm, and 15 ohm resistor that are connected to a 20 V battery. Include an ammeter to read the total current and three voltmeters to measure the potential difference across each resistor.
Calculate the total resistance of the circuit.
Calculate the total current in the circuit.
Calculate the potential difference across each resistor.
Resistor V I R P 1 2 3
Total 28

29. X. Parallel Circuit TWO OR MORE PATHS
Allows ____________________ for the current to flow
PHET CIRCUIT SIMULATION
Schematic Diagram: Draw a circuit diagram with three resistors set up in parallel connected to a 12 V battery. 29

30. X. Parallel Circuit
Things to Remember for Parallel Circuit (See Ref. Tabs)
1. Current
Sum of the currents in each of the branches is equal to total current
Equation:
I = I1 + I2 + I3 + … 30

31. X. Parallel Circuit
Things to Remember for Parallel Circuit (See Ref. Tabs)
2. Voltage
Voltage remains constant across each device
Equation:
V = V1 = V2 = V3 = … 31

32. X. Parallel Circuit
Things to Remember for Parallel Circuit (See Ref. Tabs)
3. Equivalent Resistance
As more resistors are added in parallel, total resistance decreases
Equation:
= …
Req R1 R2 R3 32

33. X. Parallel Circuit
Example: Draw a circuit diagram for parallel circuit that contains a 10 ohm, 15 ohm, and 20 ohm resistor that are connected to a 20 V battery. Include ammeters to measure the current through each and voltmeters to measure the potential difference across each resistor. 33

34. X. Parallel Circuit
Calculate the equivalent resistance of the circuit.
ii. What is the potential difference across each resistor?
iii. Calculate the reading on each of the ammeters.
Res. V I R P 1 2 3
Total 34

35. X. Parallel Circuit
iv. Calculate the total current (the current leaving the source).
v. What would happen to the total current if more resistors are added in parallel?
Res. V I R P 1 2 3
Total 35

36. XI. Conservation of Charge
Like energy, there is a conservation of __________ in circuits
CHARGE
Junction Rule:
_______ of the currents entering a junction must ________ the _______ of the currents leaving
SUM
EQUAL
SUM 36

37. XI. Conservation of Charge37

38. XII. Compound (Complex) Circuit
Steps to calculate total (equivalent) resistance:
1. Break circuit into series and parallel parts
2. Find equivalent resistance of parallel devices
3. Find equivalent resistance of series devices
Break the following circuit down to a single resistor: 38

39. XII. Compound (Complex) Circuit
Conclusion Question
Two 60 ohm resistors are connected in parallel. This parallel arrangement is connected in series with a 30 ohm resistor. The combination is then placed across a 120 V battery.
A) Draw a diagram of the circuit 39

40. XII. Compound (Complex) Circuit
B) What is the equivalent resistance of the parallel portion of the circuit?
C) What single resistance could replace the three original resistors?
D) What is the current in the circuit? 40

41. DC (Direct Current) (video clip/animation)
XIII. AC vs. DC (BONUS)
DC (Direct Current) (video clip/animation)
AC (Alternating Current)
current (charges) flows in the same direction between the + and - terminals
used in many electronic devices (ie – computers)
can be produced by batteries, solar cells, fuel cells
the direction of the current (charges) reverses (alternates)
in the US it does this at a rate of 60/sec or 60 Hz
advantage is that power companies save a lot of money transmitting power at very high voltages over long distances
they convert AC to high voltages for transmission (above 100,000 V) then use transformers to step down to lower volts 41

42. Thomas Edison (DC – Low Voltage) vs. The high current problem
XIII. AC vs. DC (BONUS)
Thomas Edison (DC – Low Voltage) vs.
Nikola Tesla (AC – High Voltage)
Video (6 min)
The high current problem
P = IR2 42

43. XIII. AC vs. DC (BONUS) Sending AC Electricity
Probably the biggest advantage of AC is that you can use high voltages with small currents to reduce losses when you transmit power. Remember that lost energy increases the more collisions you have, and reducing current decreases the amount of collisions (and reduces heating in the wires). You can send power with DC, but the DC power transmission loses a lot of energy. You would have to put much more effort into sending DC power over the same distance 43