1. ELECTRICAL SYSTEMS

2. Chapter Twenty One: Electrical Systems  21.1 Series Circuits  21.2 Parallel Circuits  21.3 Electrical Power

3. Chapter 21.1 Learning Goals  Build and analyze series circuits.  Apply Ohm’s law to calculate the current in a series circuit.  Explain how energy conservation applies to electric circuits.

4. Investigation 21A  Key Question: What are the different types of circuits? Electric Circuits

5. 21.1 Electrical Systems  In a series circuit, current can only take one path, so the current is the same at all points in the circuit.

6. 21.1 Electrical Systems  Inexpensive strings of holiday lights are wired with the bulbs in series.  If you remove one of the bulbs from its socket, the whole string of mini bulbs will go out.

8. 21.1 Current and resistance in series circuits  If you know the resistance of each device, you can find the total resistance of the circuit by adding up the resistance of each device.

10. 21.1 Current and resistance in series circuits  Think of adding resistances like adding pinches to a hose.  Each pinch adds some resistance.

11. 21.1 Current and resistance in series circuits  Everything has some resistance, even wires.

12. Solving Problems  A series circuit contains a 12-V battery and three bulbs with resistances of1 Ω, 2 Ω, and 3 Ω.  What is the current in the circuit?

13. 1.Looking for:  …current (amps) 2.Given  …Voltage = 12V; resistances = 1 Ω, 2 Ω, 3 Ω. 3.Relationships:  R tot = R 1 +R 2 +R 3  Ohm’s Law I = V ÷ R 4.Solution  R tot = 6 Ω  I = 12 V ÷ 6 Ω = 2 amps Solving Problems

14. 21.1 Voltage drop  As each device in series uses power, the power carried by the current is reduced.  As a result, the voltage is lower after each device that uses power.  This is known as the voltage drop.

15. 21.1 Voltage drop  The law of conservation of energy also applies to a circuit.  In this circuit, each bulb has a resistance of 1 ohm, so each has a voltage drop of 1 volt when 1 amp flows through the circuit.

17. 21.1 Kirchhoff’s Voltage Law  Kirchhoff’s voltage law states that the total of all the voltage drops must add up to the battery’s voltage.

18. Solving Problems  The circuit shown contains a 9-volt battery, a 1-ohm bulb, and a 2- ohm bulb.  Calculate the circuit’s total resistance and current.  Then find each bulb’s voltage drop.

19. 1.Looking for:  …total resistance; voltage drop each bulb 2.Given  …Voltage = 9V; resistances = 1 Ω, 2 Ω. 3.Relationships:  R tot = R 1 +R 2 +R 3  Ohm’s Law I = V ÷ R 4.Solution- part 1  R tot = 3 Ω  I = 9 V ÷ 3 Ω = 3 amps Solving Problems

20. 4.Solution- part 2  Use resistance to find current I = 9 V ÷ 3 Ω = 3 amps  Solution- part 3  Rearrange Ohm’s law to solve for voltage  Use current to find each voltage drop V = I x R V 1 = (3 A) x (1 Ω ) = 3 volts V 2 = (3 A) x (2 Ω ) = 6 volts (3 + 6 ) = 9 V Solving Problems

21. Chapter Twenty One: Electrical Systems  21.1 Series Circuits  21.2 Parallel Circuits  21.3 Electrical Power

22. Chapter 21.2 Learning Goals  Build and analyze parallel circuits.  Compare and contrast series and parallel circuits.  Discuss advantages for using parallel circuits in homes.

23. 21.2 Parallel Circuits  In parallel circuits the current can take more than one path.

24. 21.2 Kirchhoff’s Current Law  All of the current entering a branch point must exit again.  This is known as Kirchhoff’s current law.

26. 21.2 Voltage and parallel circuits  If the voltage is the same along a wire, then the same voltage appears across each branch of a parallel circuit.

27. 21.2 Voltage and parallel circuits  Parallel circuits have two advantages over series circuits. 1.Each device in the circuit has a voltage drop equal to the full battery voltage. 2.Each device in the circuit may be turned off independently without stopping the current in the other devices in the circuit.

28. 21.2 Current and parallel circuits  Each branch works independently so the total current in a parallel circuit is the sum of the currents in each branch.

29. 21.2 Calculating in circuits  In a series circuit, adding an extra resistor increases the total resistance of the circuit.  In a parallel circuit, more current flows so the total resistance decreases.

31. 21.2 Parallel vs. Series  Remember: series/same/current; parallel/same/voltage.  Use Ohm’s law for both.

32.  All of the electrical outlets in Jonah’s living room are on one parallel circuit.  The circuit breaker cuts off the current if it exceeds 15 amps.  Will the breaker trip if he uses a light (240 Ω ), stereo (150 Ω ), and an air conditioner (10 Ω )? Solving Problems

33. 1.Looking for:  whether current exceeds 15 amps 2.Given:  ……resistances = 240 Ω ; 150 Ω ; 10 Ω 3.Relationships:  Assume voltage for each branch = 120 V  Ohm’s Law I = V ÷ R  Kirchhoff’s Current Law I total = I 1 +I 2 +I 3 4.Solution:  I light = 120 V ÷ 240 Ω = 0.5 amps  I stereo = 120 V ÷ 150 Ω = 0.8 amps  I a/c = 120 V ÷ 10 Ω = 12 amps Solving Problems 0.5 0.8 +12.0 13.3 Breaker will not trip

34. 21.2 Short circuits  A short circuit is a parallel path in a circuit with very low resistance.  A short circuit can be created accidentally by making a parallel branch with a wire.

36. 21.2 Short circuits  Each circuit has its own fuse or circuit breaker that stops the current if it exceeds the safe amount, usually 15 or 20 amps  If you turn on too many appliances in one circuit at the same time, the circuit breaker or fuse cuts off the current.  To restore the current, you must FIRST disconnect some or all of the appliances.

37. 21.2 Fuses  In newer homes, flip the tripped circuit breaker.  In older homes you must replace the blown fuse (in older homes).  Fuses are also used in car electrical systems and in electrical devices such as televisions or in electrical meters used to test circuits.

39. Investigation 21C  Key Question: How do you build and analyze a network circuit? Analyzing Circuits

40. Chapter Twenty One: Electrical Systems  21.1 Series Circuits  21.2 Parallel Circuits  21.3 Electrical Power

41. Chapter 21.3 Learning Goals  Define electric power and apply a formula to perform power calculations.  Distinguish direct current and alternating current.  Discuss applications of electricity in daily living.

42. Investigation 21B  Key Question: How much energy is carried by electricity? Electrical Energy and Power

43. 21.3 Electrical Power  Electrical power is measured in watts, just like mechanical power.  Power is the rate at which energy is changed into other forms of energy such as heat, sound, or light.  Anything that “uses” electricity is actually converting electrical energy into some other type of energy.

44. 21.3 Important review

45. 21.3 Electrical Power  The watt is an abbreviation for one joule per second.  A 100-watt light bulb uses 100 joules of energy every second.

46. 21.3 Power  Power is a “rate” and is measured using current and voltage.

47. 21.3 Different forms of the Power Equation

48. 21.3 Kilowatt  Most electrical appliances have a label that lists the power in watts (W) or kilowatts (kW).  The kilowatt is used for large amounts of power.

49.  A 12-volt battery is connected in series to two identical light bulbs.  The current in the circuit is 3 amps.  Calculate the power output of the battery. Solving Problems

50. 1.Looking for:  …power of battery 2.Given:  …voltage = 12 V; current = 3 amps 3.Relationships:  Power: P = I x V 4.Solution:  P = 3 A x 12 V = 36 watts Solving Problems

51. 21.3 Buying Electricity  Utility companies charge customers for the number of kilowatt-hours (kWh) used each month.  A kilowatt-hour is a unit of energy.  The number of kilowatt-hours used equals the number of kilowatts multiplied by the number of hours the appliance was turned on.

52. 21.3 Buying Electricity  There are many simple things you can do to use less electricity.  When added up, these simple things can mean many dollars of savings each month.

53.  How much does it cost to run a 3,000 kW electric stove for 2 hours?  Use an electricity cost of $0.15 per kilowatt-hour. Solving Problems 1.Looking for:  …cost to run stove for 2h 2.Given:  … P = 3,000W; T = 2h; price $0.15/kW

54. 3.Relationships:  1000 watts = 1 kW  Charge in kWh 4.Solution:  3000 W x 1 kW = 3 kW 1000 W  Charge = 3 kW x 2 h = 6 kWh  Cost = 6 kWh x $ 0.15 1 kWh Solving Problems = $ 0.90

55. 21.3 AC and DC  Although the letters “DC” stand for “direct current” the abbreviation “DC” is used to describe both voltage and current.  DC current flows in one direction as in a battery.

56. 21.3 AC and DC  The electrical system in your house uses alternating current or AC.  Alternating current constantly switches direction.

57. 21.3 Electricity in homes  Electricity comes into most homes or buildings through a control panel which protect against wires overheating and causing fires.

59. 21.3 Electricity in homes  Electrical outlets in bathrooms, kitchens, or outdoors are now required to have ground fault interrupt (GFI) outlets.  GFI outlets are excellent protection against electric shocks, especially in wet locations.

60. 21.3 Distributing electricity  Electricity is a valuable form of energy because electrical power can be moved easily over large distances.  Alternating current is easier to generate and transmit over long distances.

61. 21.3 Distributing electricity  Many electronic devices, like cell phones or laptop computers, use DC electricity.  An “AC adapter” is a device that changes the AC voltage from the wall outlet into DC voltage for the device.

62. Bright Ideas  What makes one bulb more efficient than another? How much more efficient are the LEDs? What kind of savings does this mean in terms of electricity?