2.  Cells are units that contain stored electric potential  2 or more cells is called a “battery of cells” or just “battery” for short.  The “voltage” of a battery is a measurement of the amount of electric potential difference there is between the cell’s terminals

3.  If one of the cell’s terminals is “grounded” it has a potential difference of 0 volts  The cell’s electrical potential is then equal to the electric potential of the ungrounded terminal  Cells (or batteries) do work when the terminals are connected to conductors and charge is moved by the electric field of the battery  As charge moves, its electric potential is given up to the external device that it is moving through

4. + - 0.0 V 3.0 V As 1 C of electric charge leaves the negative terminal, it is driven out by an electric field that has a potential difference of 3 V This charge has the ability to provide 3 J worth of energy to the light bulb e-e- Potential Difference Drives Electrons through circuits

5. + e-e- - 0.0 V 3.0 V When the charge enters the + terminal all of its energy has been used and its electric potential is zero In the case of a chemical battery, electrolytes in the battery produce charge separation and electric potential is restored to the electron, and it is able to do work once more. e-e- Restoration of Electric Potential in a Battery

6.  Batteries don't supply the electrons that flow through the wire.  Instead, they create an electric field that 'push on' the electrons that are already in the wire.

7.  When electric charges move from one place to another  Current powers our society  Controls biological motion  The amount of charge that moves past a given point per second Potential difference e-e- e-e- e-e- e-e- e-e- e-e-

8. Electric Current When electric charges move from one place to another Current powers our society Controls biological motion The amount of charge that moves past a given point per second Potential difference e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e-

9.  - stream of electrically charged particles passing through a given area (movement of electrons in a wire  A) Conductivity of Solids  1) Metals - contain many free electrons (good conductors)  2) Glass, fused quartz, rubber - good insulator  Few free electrons  Insulators have a High Resistivity (Structural resistance to electrons flow)

10.  3) Semiconductor - a material with a resistance between conductor and an insulator  B) Conductivity of Liquids  1) pure liquids don’t conduct  2) liquid with dissolved ions conduct (electrolyte solution)  (ion movement transfers charge not electrons)

11.  C) Conductivity of Gases  Gases conduct when ionized (charged) when exposed to: a) high energy radiation b) OR electric field c) OR with collision with high speed particles  Charge in a gas is transferred by ions and electrons Plasma–very hot ionized gas 4th state of matter make up stars

12.  D) Conditions for passing current ◦ 1) complete circuit ◦ 2) potential difference (voltage) supplied by some source  ex. Battery, generator ◦ 3) conductor (wires)

14.  I. Important Circuit Terms ◦ A) Current (I) - measure of charge flow  I = q/t  Where: q = charge (C)  t = time (sec)  I= current (C/sec or Amperes)

15.  = 1 Coulomb / 1 second  Electric current is measured with an ammeter.  Ammeter must be placed in the circuit A

16.  If 10. coulombs of charge are transferred through an electric circuit in 5.0 seconds, then the current in the circuit is  I = 10. C/5.0 sec  I = 2.0 Amps

17.  1 Coulomb = 6.25 x 10 18 electrons  I = (#e-)(1.6 x 10 -19 C) t Ex) A wire carries a current of 2.0 amperes. How many electrons pass a given point in this wire in 1.0 sec?  I = (#e-)(1.6 x 10 -19 C) / t  2.0 A = (#e-)(1.6 x 10 -19 C) /1.0 sec  #e - = 1.3 x 10 19

18.  9.0 mC of charge pass through a cross sectional area of a wire in 3.5 s ◦ What is the current in the wire? ◦ How many electrons pass through the cross- sectional area in 10 s? ◦ If the number of charges that pass through the wire in 10 s doubles, what is the current in the wire?  I = Q/  t Q = Ne e = 1.6 x 10 -19 C

19.  B) Resistance (Ohms) Symbol – Ω

20.  1) R = V/I (at constant T)  V - Potential difference (voltage) between ends of conductor (volts)  I - current  Cross Multiply: V = IR  V directly related to I

21. Ohmic: Slope is constant. Non-Ohmic: Slope is not constant; resistance varies. E.g. diodes – resistance of a diode varies depending on the direction current goes through it.

22.  R = V/I  Slope of V, I plot = ???  Slope = Resistance Answer (1)

23.  2) The resistance in a wire depends on:  a) resistivity - ρ a metals structural resistance to electron flow

24.  b) Length (L) of the wire  c) Thickness (Cross Sectional Area) of the wire (A)  Why do powerful speakers require thick speaker wire?  "Large speakers draw a lot of current."

25.  R = ρL/A  R, A relationship?  Inverse relationship 4  3) Resistance and Temperature  (many electrical devices have fans to decrease T)

26.  Nonmetals – When temperature increases, resistance decreases  Superconductor - conductor with no resistance  Achieved only at very low temperatures (expensive now)

27.  1) What is the resistance of.30 m length of copper wire that has a cross-sectional area of 5.0 x 10 -5 m 2 ?  given  R = ρL/A  L =.30 m  A = 5.0 x 10 -5 m 2  ρ = 1.72 x 10 -8 Ωm  Work:  R = 1.72 x 10 -8 Ωm(.30 m) 5.0 x 10 -5 m 2  R = 1.0 x 10 -4 Ω

28.  a) How could you alter the dimensions of this wire to reduce the resistance?  R = ρL/A  To reduce the resistance in a wire you could:  L ?  Shorten  A?  Increase

29.  2) If the length of a wire were halved, how would that change the wires resistance?  R = ρL/A  R, L direct R would be halved  Cross-sectional Area (A) doubled?  R, A inverse R, halved

30.  3) Which material on your reference table would produce a wire that would allow current to flow the best?  Silver, lowest resistivity

31.  R, L plot for resistor?  Answer # 1

32. "Why does the bulb get brighter when the rod is put in the liquid nitrogen?" “"Lowering the temperature of a wire reduces resistance and increases current"

33.  Conservation of charge and energy for electric current  Gustav Robert Kirchhoff

34.  In truth, current (as electrons) flows from the negative terminal to the positive terminal  This is not the model of “Conventional Current”  Artifact of Benjamin Franklin’s ideas of current  Franklin said electricity was of two types that he called “positive” and “negative”

35. “ Amps In = Amps Out”

36. 1)Kirchhoff's 1st Law – Total current arriving at a point in a circuit = Total current leaving the point. (Conservation of Charge)

37. Ex. Amps In = Amps Out 5 Amps = 3 Amps + x X = 2 amps

38. Ex 2) Current in bottom branch? 2 Amps + X = 12 Amps X = 10 amps Ex 3) 22 Amps = x Amps + 6 Amps X = 16 amps

39. What's Wrong With This Picture? Current In (10 A), does not equal current out (4 A)

40. 2) Kirchhoff's 2nd Law - the algebraic sum of all the voltage drops and applied voltages around a circuit = 0 (conservation of energy) Amount of PE gained when you go up a ski slope = energy released going down a ski slope (KE, Work against friction and snow)

41. Applied Voltage ( + ) Voltage Drop ( - ) 0 = 12 V + -2 V + x + -6 V 0 = 12 V + -8 V + x 0 = 4 V + x x = - 4 volts V T = V 1 + V 2 +...

42.  Why is it a good idea to connect Christmas lights 'in parallel'?

43.  1) Series Circuit - circuit that has only one current path

44. a)I T = I 1 = I 2 = I 3 =... Total Current coming from source = Current going through each resistor.

45. b) R T = R 1 + R 2 + R 3 + …… R T = 2 + 4 R T = 6.0 Ω c) V T = V 1 + V 2 + V 3 + …… d) V T = I T R T or R T = V T /I T Find total current (I T ) V T = I T R T 20. V = I T (6.0 Ω) I T = 3.3 amps = I 1 = I 2

46.  "Which kind of current do we have in our homes?“  "Alternating Current"

47. Series or Parallel?? 2) Parallel Circuit - circuit in which each component has its own circular path back to the source.

48. a)I T = I 1 + I 2 + I 3 +... b)If I 1 = 2 amps, I 2 = 3 amps I 3 = 4 amps c)(ignore picture above) Then... I T = 9 amps b) The voltage drops across each component in parallel are equal V T = V 1 = V 2 = V 3 V T = Source Voltage

49. c) V 1 = I 1 R 1 V 2 = I 2 R 2 V 3 = I 3 R 3 etc. d) Current in 1st & 2nd resistor? e) What’s R 3 ? V T = V 1 = V 2 = V 3 = 20 V I 1 = V 1 / R 1 = 20V / 5Ω = 4A I 2 = V 2 / R 2 = 20V / 5 Ω I t = I 1 + I 2 + I 3 18 A = 4A +4A + I 3

51.  Since I T = I 1 + I 2 + I 3 + I 4 +……  Then …  V T = V 1 + V 2 + V 3 R T R 1 R 2 R 3 1 = 1 + 1 + 11 = 1 + 1 + 1  R T - Total Resistance  A.K.A.  "Equivalent Resistance" "Combined Resistance" "Effective Resistance"

52.  Ex 3) Find the total resistance for 3 resistors of 2.0 ohms, 5.0 ohms and 2.0 ohms connected in parallel. 1 = 1 + 1 + 11 = 1 + 1 + 1 R T R 1 R 2 R 3 1 = 1 + 1 + 1 R T 2.0Ω 5.0Ω 2.0 Ω 1/R T = (5 + 2 + 5)/10 1/R T = 12/10  R T = 10/12 ohms  R T =.83 ohms ** The more resistors connected in parallel the smaller the total resistance!!!

53. a) Find the equivalent resistance for this parallel circuit 1/R T = 1/2.0 Ω + 1/4.0 Ω 1/R T = 3.0 Ω/4.0 Ω R T =4.0 Ω/3.0 Ω = 1.3 ohms b) Find I T (Current coming out of the source) V T = I T R T 12. V = I T (1.3 ohms) I T = 9.0 amps c) Find the potential drop (V) across R 1 & R 2 V T = V 1 = V 2 = V 3 V T = V 1 = V 2 = 12V

54.  Ex 5) Series Circuit  a) Find the 'equivalent resistance‘  R T = R 1 + R 2 = 2.0 Ω + 1.0 Ω  = 3.0 ohms  b) Find the current (I T ) going through this circuit  V T = I T R T 12. Volt = I T (3.0 ohms)  I T = 4.0 amps  c) Find potential drop across R 1 & R 2  V = IR  V 1 =4.0 A (2.0 ohms)  = 8.0 Volts  V 2 =4.0 A(1.0 ohms)  = 4.0 Volts

55. V T = I T R T 60. volts = (10 amps)R T = 6.0 ohms b) Find the current in R 1 V 1 = I 1 R 1 60. Volts = I 1 20. ohms I 1 = 3.0 amps c) Find I 3 V 3 = V T = 60. volts Use Ohm's Law V 3 = I 3 R 3 60. volts = I 3 (30. ohms) I 3 = 2.0 amps d) Find R 2

56. Voltmeter - Measures Voltage (Potential Difference) (= Energy Change/Coulomb ) Voltmeters are connected 'in parallel‘ Ammeters are connected 'in series' Find Ammeter, Voltmeter

58.  Power - time rate of doing work or expending energy  P = W/t (watts) (J/seconds)  since V = W/q,  CROSS MULTIPLY  then W = Vq  P = Vq/t  P = VI  P = VI since V = IR …  THEN P = I 2 R  since I = V/R …  THEN P = V 2 /R III) Electric Power

59.  P = VI = I 2 R = V 2 /R  Ex) Series circuit w/ two lamps, the battery supplies a potential difference of 1.5 volts. Current in the circuit is 0.10 ampere, what rate does the circuit use energy?  a)15 W b)1.5 W d).15 W c).015W  V = 1.5 V I =.10 Amps P = ?

60.  Find the resistance of a space heater as it dissipates 1500 W of power (mostly at the infrared wavelength) if it is connected to a 120 V outlet.   V = 120 V  P = 1500 W  Select the appropriate equation to solve for R (hint: I is not known)

61. P =  V 2 /R, R =  V 2 /P = (120V) 2 / 1500 W = 14400 / 1500 = 9.6   Note that as power increases resistance decreases.

64.  Invented by Alessandro Volta in 1800  2 strips of metal in an electrolyte solution  Produces a separation of charge  The electrolyte breaks down into + and - ions  Ions react with electrodes

66.  Electrodes are Zinc and Carbon  Electrolyte is Ammonium Chloride paste  Excess electrons gather at one electrode, deficit at the other one  This sets up an electric potential  Electron flow occurs from excess to deficit (- to +)

67.  Connecting cells in series produces greater electrical potential  Resulting source is called a “Battery of Cells”  Batteries produce Direct Current ◦ current that flows in one direction only  Electrons flow in a current from an area of high potential (where electrons are concentrated) to one of low potential.

68.  Maintain a potential difference by input of chemical energy  Chemical energy moves electrons, concentrating them in a high potential area  As electrons collide with the atoms of a device, their energy is transferred to the device  Energy is continuously supplied until the battery’s chemical energy is used up. Current continues until there is no potential difference between the two sides, and the battery is “dead”