1. 22.1 Current and Circuits

2. Producing Electric Current When 2 conducting spheres touch, charges flow from the sphere at a higher potential difference to the one at a lower potential difference until the potential differences of both spheres are equal. A flow of charged particles is an electric current. The flow of positive charge is called conventional current. (open books to pg. 508) The conventional current will stop when the potential differences are equal. If you wanted to keep the flow going you would need to maintain a potential difference between the two conductors. You would need to pump the charged particles from one conductor to the other. The electric potential energy of the charges have to be increased by the pump, so it would require external energy to run. A battery is a group of several voltaic or galvanic cells connected together to convert chemical energy to electrical energy. A photovoltaic cell is a solar cell that changes light energy into electric energy. A generator can be used to convert kinetic energy into electric energy.

3. Electric Circuits Electric circuit – a continuous path through which electric charges can flow. The potential energy lost by the charges, qV, in moving through the device is usually converted into some other form of energy. Ex. A motor converts electric energy to kinetic energy, a lamp changes electric energy into light, and a heater converts electric energy into thermal energy. The charge pump creates the flow of charged particles, or current. Page 509 No generator is 100% efficient, some energy gets converted into thermal energy. Charges can’t be created or destroyed,, only separated.

4. Rates of Charge Flow and Energy Transfer Power measures the rate at which energy is transferred. Power is measured by watts, 1 J/s = 1 watt The unit used for quantity of electric charge is the coulomb. Electric current, I, is measured in coulombs per second. An ampere is one coulomb per second, 1 C/s = 1 A. Power P = IV

5. Resistance and Ohm’s Law Resistance is the property that determines how much current will flow. Resistance R = V/I Resistance is measured in ohms. The resistance for most conductors does not vary as the magnitude or the direction of the potential applied to it changes. Resistors are devices designed to have a specific resistance. Superconductors are materials that have zero resistance, therefore they can conduct electricity w/o the loss of energy. Resistors are used to control the current in circuits. Potentiometer- a variable resistor that allows continuous, rather than step-by-step, changes in current in an electric circuit; also called a rheostat.

6. Diagramming Circuits Schematic- a diagram of an electric circuit that uses standard symbols for circuit elements. Current is shown out of the positive terminal of the battery. Parallel connection- the arrangement of electric devices in a current where there is more than one current path. The ammeter measures the current through a circuit component. Series connection- The connection of electric devices in a circuit where there is only one current path.

7. 22.2 Using Electric Energy Each appliance in your home converts electrical energy into another form: kinetic, sound or thermal energy

8. Energy Transfer in Electric Circuits Energy supplied to a circuit can be used in many different ways. Not all energy delivered to a motor or light bulb ends up in a useful form. Some energy is converted into thermal energy.

9. Heating a Resistor Household appliances act like resistors when devices that convert electric energy into thermal energy are in a circuit.

10. P = I 2 R Rearrange R = V/I to V = IR to solve for V Substitute V into P = IV to get P = I 2 R In other words: you’re given 2 quantities (out of P, I, V and R). You find the 2 unknown quantities using the above equations. E = Pt = I 2 Rt combination of P = E/t and the above equation the total energy that’ll be converted to thermal energy

11. Transmission of Electric Energy How can the transmission of (hydroelectric) energy take place with as little loss to (thermal) energy as possible? P = I 2 R (represents production rate of thermal energy) Unwanted thermal energy (the joule heating or I 2 R loss) To reduce that loss, either the current (I), or the resistance (R) must be reduced. All wires have some resistance

13. The Kilowatt-Hour … Electric companies provide energy. Electric energy used by a device is rate of energy consumptions [in joules per second (watts)] times the number of seconds it’s operated. Joules per second times seconds (J x s/s) equals total joules of energy.

14. … continued Joule is a watt-second Too small for commercial sales use –Electric company’s measure their energy sales in a large number of joules called a kilowatt-hour (kWh) »1 kWh = 1000 watts delivered continuously for 3600 seconds (1 hour) Not many devices require more than 1000 watts of power.

15. Physics Final-Electricity on Ohm's Law and Power Ohms Law Voltage versus Current Relationship

16. Example Problems Chapter 22

17. Electric Power A 6.0 V battery delivers a 0.50 A current to an electric motor that is connected to its terminals A.) what power is consumed by the motor? B.) if the motor runs for 5.0 min., how much electric energy is delivered?

18. P=IV E=Pt P=(0.50 A)(6.0) 3.0 W E=(3.0 W)(5.0 min.)(60 s/1 min.) 9.0 x 10 2

19. Current Through a Resistor A 30.0 V battery is connected to a 10.0 Ω resistor. What is the current in the circuit?

20. I= 30.0 V/ 10.0 Ω 3.00 A

21. Chapter 22 Current Electricity

22. Without electricity our current lifestyle would be impossible It is indispensable in our daily lives because it can be easily changed into other forms of energy easily such as sound, thermal, light, and motion Electric energy is easily transported over long distances

23. Current and Circuits In this section, you will learn about. Producing Electric Current Electric Circuits Rates of Charge Flow and Energy Transfer Resistance and Ohm’s Law and Diagramming Circuits

24. Producing Electric Current An electric current is a flow of charged particles If two conductors with potential difference are connected by a wire conductor energy will flow until the potential differences are zero (a very short time) To sustain such a current a fourth conductor must be introduced with a charge pump (power source) to complete a circuit

25. Rate of Change Flow and Energy Transfer Power measures the rate at which energy is transferred One watt equals one joule per second or W=J/s One coulomb per second is an ampere, A, represented by 1 C/s=1 A

26. Electric Circuits A closed loop, includes a charge pump that increases the potential energy of the charges moving from A to B Energy lost in a circuit qV is usually converted to some other type of energy Ex.) lamp to light, and engine into thermal,

27. Resistance and Ohm’s Law If 2 conductors with a potential difference are connected with a copper rod a large current is made on the other hand if the same conductors are connected with a glass rod virtually no current is produced The propertu that determines how much current will flow is called resistance.

28. Continued Resistance is measured by placing a potential difference across two points on a conductor and measuring equations Resistance,R, is defined to be the ratio of the potential difference to the current, I. R=V/I I is in amps V is in volts

29. Diagramming Circuits A simple circuit can be described in words. It can also be depicted by photographs or artist's drawings Most often a diagram is drawn using a standard symbols for the circuit elements

30. Using Electric Energy Among your home appliances you probably have lamps, a hair dryer, a stereo, microwave, and a refrigerator. Each converts electrical energy into another form such as light thermal or kinetic. How much is converted, and at what rate?

31. Energy Transfer in Electric Circuits Energy supplied to a circuit can be used in many different ways. A motor to mechanical, A lamp to light But, all of energy delivered to the motor and lamp does not end up in its useful form Lights and motors get hot, some of the electric is turned into thermal energy

32. Heating a Resistor A space heater, a hot plate and the heating element in a hair dryer are designed to convert almost all of their electric energy into thermal energy. Most household appliances act as resistors when in a circuit. Charge,q, moves through a resistorm its potential difference is reduced by V P=I 2 R

33. Transmission of Electric Energy Niagara Falls and the Hoover Dam can produce electric energy with little pollution. But, this hydroelectric energy often must be transmitted long distances. How can this distance be traveled with as little loss to thermal energy as possible?

34. Thermal energy is produced at a rate represented by P=I 2 R this is called joule heating or I 2 R loss To reduce loss Current, I, or the resistance, R, must be recuced

35. The Kilowatt-Hour While electric companies are often called “power” companies they really provide energy. When consumers pay their electric bills they actually pay for electric energy not power. The unit which is used to measure energy consumption is the kilowatt-hour 1 kWh=(1000 J/s)(3600 s) = 3.6E 6 J

37. Sample Problem One A 6.0-V battery delivers a 0.50-A current to an electric motor that is connected across its terminals –1.What is the power consumed by the motor? –2. If the motor runs for 5.0 minutes, how much electric energy is delivered?

38. How to find the answer To solve this problem we will use the equation P=IV Where P is Power, is the rate at which energy is transferred, I is current, the rate of flow of an electric charge, and V which is the potential difference or voltage. Now we right down what we know –V=6.0 V –I=0.50 A –T=5.0 t We also write down what we don’t know –P=? –E=?

39. Solving the questions Plug in what we know into the equation P=IV for question one –P=(0.50 A)(6.0 V)= 3.0 Watts “Power is defined in watts” For question two we use the equation P=E/t, which comes from a concept that the class learned in chapter 10 that says “power is the rate at which energy is transferred.” Plug in what we know –E=(3.0)(5.0 min)… remember that time has to be seconds so we can just add (60 s/1 min) to the end of our equation giving us E=(3.0)(5.0 min) (60 s/1 min) = 90 × joules “Units for energy”

40. Sample problem two The current through a light bulb connected across the terminals of a 120-V outlet is 0.50 A. At what rate does the bulb convert electric energy to light?

41. To solve we write down what we know –V= 120 V –I = 0.50 A Unknow –P = ? To solve we just plug in P= (120 V)(0.50 A)=60 watts

42. Sample problem Three A 30.0v battery is connected to a 10.0 Ω resistor. What is the current in the circuit? –To solve this problem we will use the equation R=V/I Where R is the resistance “property that determines how much current will flow”, V is potential difference, and I is current Write down we know –V = 30.0 V –R = 10.0 Ω Unknown –I = ?

43. Sample Problem Three Continued Now plug in the quantities into the equation –10.0 Ω= –I = = 3.00 A

44. Sample Problem Four A Motor with an operating resistance of 32 Ω is connected to a voltage source. The current in the circuit is 3.8 A. What is the voltage of the source? To solve this problem we will use the equation R=V/I Write down what we know –I = 3.8 A –R = 32 Ω Unknown –V = ?

45. Sample Problem Four Continued Just plug in what is known into the equation to solve –32 Ω= –(32 Ω)(3.8 A)= 120 joules

46. Sample Problem Five A heater has a resistance of 10.0 Ω. It operates on 120 V –What thermal energy is supplied by the heater in 10.0 seconds? To solve we use the equation E=, which is derived from combining the equation for power dissipated in a resistor, P=IV, the equation for resistance, V=IR, and the equation E=Pt which is the rate at chich energy is being transferred.

47. Sample Problem Five continued Write down what we know –R = 10.0 Ω –V = 120.0 V –T = 10.0 seconds Unknown –E = ? –I = ?

48. Sample Problem Five Continued First thing to do is figure out what I is, so we use the equation for resistance R=V/I –10.0 Ω= –I= = 12.0 A Plug I into E= along with all the other know nquantities. –E= = 14400 J

49. Sample problem 6 A television set draws 2.0 A when operated on 120 V and the power is 2400 W. The set consumes 50000 Wh in 30 days, what is the cost per 30 days if it is a 11 cents per kWh? –First convert Wh to kWh by dividing 50000 by 1000 –Then just multiply 50 by 11 cents, which is $5.50

50. Book problems Turn to page 511 and look and practice problem 3 Turn page 522 example problem 17 “a” and “b” only Turn page 525 example problem 19