1. Chapters 18 - 19

2.  The Electric Battery  Electric Current  Ohm’s Law: Resistance and Resistors  Resistivity  Electric Power

3.  EMF and Terminal Voltage  Resistors in Series and in Parallel  Kirchhoff’s Rules  EMFs in Series and in Parallel; Charging a Battery  Circuits Containing Capacitors in Series and in Parallel  RC Circuits – Resistor and Capacitor in Series  Ammeters and Voltmeters

4.  Question:  Given a light bulb, a wire, and a battery. Can you get the light bulb to light?  Procedure: 1. Try to find as many ways as possible to get the light bulb to light. 2. Diagram two ways in which you are able to get the bulb to light. Be sure to label the battery, the bulb, and the wire. 3. Diagram three ways in which the bulb does not light. Again, label!

5. 1. How did you know if electric current was flowing? 2. What do your diagrams of the lit bulb have in common? 3. What do your diagrams of the unlit bulb have in common? 4. From your observations, what conditions seem necessary in order for the bulb to light? 5. Critical Thinking: What causes electricity to flow through the bulb? 6. Draw a bulb. (Show the inside of the metal base.) Answer in complete sentences.

6.  Volta discovered that electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte.  This is a simple electric cell.

7.  A battery transforms chemical energy into electrical energy.  Chemical reactions within the cell create a potential difference between the terminals by slowly dissolving them.  This potential difference can be maintained even if a current is kept flowing, until one terminal is completely dissolved.

8.  Several cells connected together make a battery.  (Although we now refer to a single cell as a battery as a well.)

9.  The purpose of a battery is to produce a potential difference, which can then make charges move.  When a continuous conducting path is connected between the terminals of a battery, we have an electric circuit.  On a diagram of a circuit, the symbol for a battery is:

10.  When such a circuit is formed, charge can flow through the wires of the circuit from one terminal of the battery to the other, as long as the conducting path is continuous.  A flow of charged particles is an electric current.

11.  Current is measured in coulombs per second.  This unit is given a special name, the ampere (amp or A), after French physicist André Ampére.  A current can flow in a circuit only if there is a continuous conducting path. We then have a complete circuit.  If there is a break in the circuit we call it an open circuit and no current flows.

12.  The electric current in a wire is defined as the net amount of charge that passes through the wire’s full cross section at any point per unit time.  Where:  I = current (amps)  ΔQ = the amount of charge  Δt = time interval

13.  In any single circuit, with only a single path for current to follow, a steady current at any instant is the same at one point as at any other point.  This is because of the law of conservation of electric charge.  A battery does not create charge and a light bulb does not absorb or destroy charge.

14. What is wrong with each of the schemes for lighting a bulb with a battery and a single wire as shown on this and the next two slides? There is no closed path for a charge to flow through. This scheme will not light the bulb. There is no closed path for a charge to flow through. This scheme will not light the bulb.

15.  There is a closed path passing to and from the light bulb, however, the wire is only connected to one terminal.  No potential difference in the circuit to make the charge move.  There is a closed path passing to and from the light bulb, however, the wire is only connected to one terminal.  No potential difference in the circuit to make the charge move.

16.  Nothing is wrong here.  This is a complete circuit: charge can flow out from one terminal of the battery, through the wire and the bulb, and into the other terminal.  Nothing is wrong here.  This is a complete circuit: charge can flow out from one terminal of the battery, through the wire and the bulb, and into the other terminal.

17. A steady current of 2.5 A exists in a wire for 4.0 min. a) How much total charge passed by a given point in the circuit during those 4 minutes? b) How many electrons would this be?

18. Homework: Practice Problems  p.515 #1-3 Homework: Practice Problems  p.515 #1-3

19.  In many real circuits, wires are connected to a common conductor to provide continuity.  This common conductor is called ground.  This is the symbol for ground in a circuit diagram:

20.  By convention, current is defined as flowing from positive to negative.  Electrons actually flow in the opposite direction, but not all currents consist of electrons.

21.  Conceptual Questions: p.514 #1-4

22.  George Ohm (1787-1854) studied the relationship between the potential difference and the current.

23.  Experimentally, it is found that the current in a wire is proportional to the potential difference between its ends:  The ratio of voltage to current is called the resistance:

24.  Resistance is the property that determines how much current will flow.  Unit of resistance: the ohm, Ω.  1 Ω = 1 V / A.  Ex. Suppose two conductors have a potential difference between them.  When connected by copper rod, a large current is created.  However, connecting them with a glass rod creates almost no current.

25.  A resistor is a device designed to have a specific resistance.  Resistors may be made of graphite, semiconductors, or wires that are long and thin. This is the symbol for a resistor in a circuit diagram.

26.  Standard resistors are manufactured for use in electric circuits; they are color-coded to indicate their value and precision.

28.  Some clarifications:  Batteries maintain a (nearly) constant potential difference; the current varies. (Details in the next chapter.)  Resistance is a property of a material or device.  Current is not a vector but it does have a direction. In a wire, the current is always parallel to the wire and the direction of conventional (positive) current is from high potential (+) toward lower potential (-).  Current and charge do not get used up. Whatever charge goes in one end of a circuit comes out the other end.

29.  The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area:  The constant ρ, the resistivity, is characteristic of the material.

30.  For any given material, the resistivity increases with temperature:  Semiconductors are complex materials, and may have resistivities that decrease with temperature.  The Greek letter alpha represents the temperature coefficient.

32. FactorHow Resistance Changes LengthResistance increases as length increases Cross-sectional Area Resistance increases as cross-sectional area decreases Temperature Resistance increases as temperature increases Material Keeping length, cross-sectional area, and temperature constant, resistance varies with the material used. Platinum Aluminum Iron Gold Copper Silver Resistance increases

33.  Questions  p.514, #5+6  Practice Problems  p.515-516  #4-6, 9, 11-13

35.  Power, as in kinematics, is the energy transformed by a device per unit time: So:

36.  The unit of power is the watt, W.  For ohmic devices, we can make the substitutions (Using V = IR):

37.  What you pay for on your electric bill is not power, but energy – the power consumption multiplied by the time.  We have been measuring energy in joules, but the electric company measures it in kilowatt-hours, kWh.

38.  The wires used in homes to carry electricity have very low resistance.  However, if the current is high enough, the power will increase and the wires can become hot enough to start a fire.  To avoid this, we use fuses or circuit breakers, which disconnect when the current goes above a predetermined value.

39.  Fuses are one-use items.  If they blow, the fuse is destroyed and must be replaced.

40.  Circuit breakers, which are now much more common in homes than they once were, are switches that will open if the current is too high; they can then be reset.

41.  Questions:  p.515 #9, 10, 14  Problems:  p.516-517 #26, 27, 29, 30, 32, 33-37

42.  Be sure to draw labeled circuit diagrams for both series and parallel circuits.  Answer the following: 1. Compare and contrast the current flow in a series circuit with that in a parallel circuit. 2. Compare and contrast the voltage in a series circuit with that in a parallel circuit.  Both of the above responses need to be backed up with data.

43.  EMF and Terminal Voltage  Resistors in Series and in Parallel  Kirchhoff’s Rules  EMFs in Series and in Parallel; Charging a Battery  Circuits Containing Capacitors in Series and in Parallel  RC Circuits – Resistor and Capacitor in Series  Ammeters and Voltmeters

44.  To have current in an electric circuit, we need a device (such as a battery or an electric generator) that transforms one type of energy into electric energy.  Such a device is called a source of emf.  The potential difference given to the charges by a battery is called emf.  (Not an actual force, it is a potential difference measured in volts.)  The emf is the influence that makes current flow from a lower potential to a higher potential.

45.  Remember: A battery is not a source of constant current.  The current out of a battery varies according to the resistance in the circuit.  A battery is, however, a nearly constant voltage source.  It does have a small internal resistance, r, which reduces the terminal (actual) voltage from the ideal emf.  Terminal Voltage: V AB = V a - V b  Ideal emf: V AB = E

46.  V AB = terminal voltage  E = emf  I = current  r = internal resistance  Internal resistance increases as the batteries get older and the electrolyte dries out.  The internal resistance behaves as though it were in series with the emf. V AB = E - Ir

47. A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.500 Ω. Calculate: a) the current in the circuit b) the terminal voltage of the battery, V ab c) the power dissipated in the resistor R and in the battery’s internal resistance r.

48.  Start with Ohm’s law: V = IR  Voltage of emf with internal resistance: V ab = E – Ir  V = V ab for this circuit, Ohm’s Law again: V ab = IR  Substitute E – Ir for V ab : E – Ir = IR  Rearrange to solve for I :  E = IR + Ir  E = I(R + r)  I = E /(R + r) I = E /(R + r) = 12.0 V/(65.0 Ω + 0.500 Ω ) = 0.183 A

49.  Terminal Voltage = V ab = E – Ir V ab = E – Ir = 12.0 V – (0.183 A)(0.500 Ω) = 11.9 V

50. Power Dissipated in RPower Dissipated in r P R = I 2 R P R = (0.183 A) 2 (65 Ω) = 2.18 W P r = I 2 r P R = (0.183 A) 2 (0.5 Ω) = 0.0167 W Power dissipated = P = I 2 R

51.  In much of what follows, unless otherwise stated, we assume that the internal resistance of a battery is negligible.  Therefore, the battery voltage that is given is its terminal voltage, which we will write as V instead of V ab.

52.  A series connection has:  two or more resistors connected end to end.  a single path from the battery, through each circuit element in turn, then back to the battery.

53.  The current through each resistor is the same.  The voltage depends on the resistance.  The sum of the voltage drops across the resistors equals the battery voltage.  Note that when you add more resistance to a circuit, the current through the circuit will decrease.

54.  From this we get the equivalent resistance (that single resistance that gives the same current in the circuit).

55.  A parallel connection splits the current into different paths.  The voltage across each resistor is the same.  The devices in houses are wired in parallel.  If one device is disconnected, the current to the other devices is not interrupted.

56.  The total current is the sum of the currents across each resistor:  This gives the reciprocal of the equivalent resistance:

57. equal evenly 3 V drop Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R

58. 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V The voltage drop across R 1 has to be twice as big as the drop across R 2. The voltage drop across R 1 has to be twice as big as the drop across R 2. This means that V 1 = 8 V V 1 = 8 V and V 2 = 4 V. Or you could find the current I = V/R = (12 V)/(6  = 2 A, then use Ohm’s Law to get the voltages.

59. voltagesame V 1 = I 1 R 1 I 1 = 2 A The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm’s Law, V 1 = I 1 R 1 to find the current I 1 = 2 A. In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5  R 2 = 2  1) 10 A 2) zero 3) 5 A 4) 2 A 5) 7 A

60. 1) increases 2) remains the same 3) decreases 4) drops to zero resistance of the circuit drops resistance decreasescurrent must increase As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit?

61. Terminal Voltage and emf Resistors in Series and Parallel  Practice Problems:  p.547 #1-3  Practice Problems:  p.547 #5-7, 9, 11-13, 17, 18