2. 19 Current, Resistance, and Directed-Current Circuits Lectures by James L. Pazun Copyright © 2012 Pearson Education, Inc. publishing as Addison-Wesley

3. 2 Chapter 19: Electric Current and Circuits Topics Electric current EMF Current & Drift Velocity Resistance & Resistivity Kirchhoff’s Rules Series & Parallel Circuit Elements Applications of Kichhoff’s Rules Power & Energy Ammeters & Voltmeters RC Circuits

4. Electric Current & Circuits Electric Current & Circuits Electric Current: steady flow of electric charge.Electric Current: steady flow of electric charge. Potential Difference: work done by the electric force to move 1 Coulomb of charge between two points.Potential Difference: work done by the electric force to move 1 Coulomb of charge between two points. Convention: current has the direction of the flow of positive charges.Convention: current has the direction of the flow of positive charges.

5. Current Direction Direct current (DC): electrons flowing in one direction only.Direct current (DC): electrons flowing in one direction only. Alternating current (AC): electrons continuously reverse their direction of motion.Alternating current (AC): electrons continuously reverse their direction of motion.

6. Definition of I Definition of I  Current (I) =amount of charge that passes through a cross sectional area of a conductor in one unit of time. Charge Cross-sectional area Charge Time I = Unit of measurement: Ampere (A) = ΔQ/Δt

7. Copyright © 2012 Pearson Education, Inc. publishing as Addison-Wesley Current defined – Figures 19.1 and 19.2

8. Microscopic View of Current in a Metal The electrons moving in a wire collide frequently with one another and with the atoms of the wire This results in the zigzag motion shown When no electric field is present, the average electron displacement is zero, v drift = 0. There is no net movement of charge. There is no current. Section 19.3

9. Current, cont. With a battery connected, an electric potential is established There is an electric field in the wire: E = V / L The electric field produces a force that gives the electrons a net motion The velocity of this motion is the drift velocity, v drift  0. Typically, v drift < 1 mm/sec Section 19.3

10. 9 Calculate the number of charges (N e ) that pass through the shaded region in a time  t: The current in the wire is: l

11. 10 Example : A copper wire of cross- sectional area 1.00 mm 2 has a constant current of 2.0 A flowing along its length. What is the drift speed of the conduction electrons? Copper has 1.10  10 29 electrons/m 3.

12. Electric Current Example How many electrons flow through a cross sectional area of a conductor in 5s, if it is traveled by a current of 3.2 A 1. Find the total charge: 2. Find the number of electrons N using the charge of an electron e=1.6 x 10 -19 C

13. Current and Voltage The current is directed from a region of higher potential to a region of lower potential Or higher to lower voltage The direction of I is always from high to low potential, regardless if the current is carried by positive or negative charges The potential difference may be supplied by a battery Section 19.2

14. Analogy  A ball rolling down a hill High Low Difference of height causes the ball to roll down the hill.

15. Sources High Potential Low Potential The positive electrode is called the anode; the negative electrode is the cathode. In a complete circuit, electrons flow from the negative electrode to the positive one.

16. Sources  Batteries (Voltage sources, sources of emf): Purpose is to provide a constant potential difference (voltage) between two points. +- V + - OR The potential difference across the terminals in an open circuit is called electromotive force or “emf” ξ.

17. Voltage Voltage A useful meter is a multimeter, whichA useful meter is a multimeter, which can measure voltage or current, and sometimes resistance. can measure voltage or current, and sometimes resistance. To measure voltage, the meter’s probes are touched to two places in a circuit or across a battery.To measure voltage, the meter’s probes are touched to two places in a circuit or across a battery.

18. Measuring current I f you want to measure current you must force the current to pass through the meter. I f you want to measure current you must force the current to pass through the meter. Multimeters can measure two types of current: alternating current (AC) and direct current (DC).Multimeters can measure two types of current: alternating current (AC) and direct current (DC).

19. Construction of a Battery Alessandro Volta and his contemporaries developed the first batteries Batteries convert chemical energy to electrical energy In drawing an electric circuit the terminals of any type of battery are labeled with + and - Section 19.2

20. Thermocouples  Device that produces electric energy due to temperature difference at the junction b/w 2 different metals (Seebeck).  Copper  Iron HotCold Can be used to measure the temperature. Ex.:cars, gas furnaces.

21. Photoelectric Cell  Light energy is converted into electric energy. Light Electrons Metallic Surface Emitted electrons can be used in a circuit.

22. Sources Sources  Electromagnetic induction: conversion of electromagnetic energy into electricity.  Piezoelectric: mechanical stress on certain crystals creates a difference of potential.

23. Electric Circuit  Electric Circuit: complete path through which a current can flow. Made of a source of current (battery), connecting wires, and an electricity user. S User Wires I- +

24. Electric Circuits  When drawing a circuit diagram, symbols are used to represent each part of the circuit.

25. Electrical Symbols Electrical Symbols  Electrical symbols are quicker and easier to draw than realistic pictures of the components.

26. Resistance Resistance (R): opposition to the flow of the electric current. Caused by collisions b/w electrons and the atoms of the metal.Resistance (R): opposition to the flow of the electric current. Caused by collisions b/w electrons and the atoms of the metal. Measured in ohms Ω.Measured in ohms Ω. Depends upon type of material (resistivity ρ), proportional to length L, inversely proportional to thickness (cross sectional area A).Depends upon type of material (resistivity ρ), proportional to length L, inversely proportional to thickness (cross sectional area A). R = ρL/AR = ρL/A

27. Resistivity The resistivity, ρ, depends only on the material used to make the wire The resistance of a wire of length L and cross- sectional area A is given by Section 19.3 The structure of this relation is identical to heat flow through materials … think of a window for an intuitive example How thick? How big? What’s it made of? or

28. Resistance Continuation R is proportional to the temperature. At low temperatures conductors exhibit low resistance (superconductors).R is proportional to the temperature. At low temperatures conductors exhibit low resistance (superconductors). Causes electric energy to turn into heat.Causes electric energy to turn into heat.

29. Simple Electric Circuit Analogy

31. Example Deformation of Wire A very thin metal wire patterned as shown is bonded to some structure. As the structure is deformed slightly, this stretches the wire (slightly). –When this happens, the resistance of the wire: (a) decreases (b) increases (c) stays the same Because the wire is slightly longer, is slightly increased. Also, because the overall volume of the wire is ~constant, increasing the length decreases the area A, which also increases the resistance. By carefully measuring the change in resistance, the strain in the structure may be determined.

32. Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. Compare the resistance of the two cylinders. a) R 1 > R 2 b) R 1 = R 2 c) R 1 < R 2 Larger cross sectional area decreases the resistance. Exercise :Compare Resistances

34.  Determined experimentally the relationship between current and voltage in a circuit. Ohm’s Law

35.  I=V/R S V A V - voltmeter: measures voltage (parallel) Source: provides electric energy. A - ammeter: measures electric current (series) R

36. Ohm’s Law Experimental: Vary applied voltage V. Measure current I Ratio remains constant. V I I R V I slope = R V=RI or

37. Solving Problems  A toaster oven has a resistance of 12 ohms and is plugged into a 120-volt outlet.  How much current does it draw?

38. 1. Looking for: …current in amps …current in amps 2. Given …R = 12  ; V = 120 V …R = 12  ; V = 120 V 3. Relationships: I = V I = V R 4. Solution I = 120 V I = 120 V 12  12  Solving Problems = 10 A

39. Example Compare Currents The resistivity of both resistors is the same (  ). Therefore the resistances are related as: The resistors have the same voltage across them; therefore  Two cylindrical resistors, R 1 and R 2, are made of identical material. R 2 has twice the length of R 1 but half the radius of R 1. These resistors are then connected to a battery V as shown: These resistors are then connected to a battery V as shown: V I1I1 I2I2 What is the relation between I 1, the current flowing in R 1, and I 2, the current flowing in R 2 ? What is the relation between I 1, the current flowing in R 1, and I 2, the current flowing in R 2 ? (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2

40. Electric Energy  The work done by the electric field to move the charge ΔQ through the potential difference V is a measure of the electric energy expanded in this case. Therefore the electric energy is given by the formula: Since: V=IR, we can also write: W= I 2 Rt (thermal or Joule effect) Also:

41. Power Batteries & Resistors Energy expended chemical to electrical to heat What’s happening? Assert: Rate is: Charges per time Potential difference per charge Units okay? For Resistors: or you can write it as

42. Electric Power  Electric Power: is the time rate at which electricity does work or provides electric energy.  P=W/t, therefore: Power = Voltage x Current Power = Voltage x Current  or P = VI, P= I 2 R, P= V 2 /R  Power is measured in Watts.  Watts = Volts x Amperes  1 kilowatt = 1000 watts.  1kwh=1000 W x 3600 s= 3,600,000 J

43. Example Higher Resistance  Two appliances operate at the same voltage. Appliance “A” has a higher power rating than “B”. How does the resistance of “A” compare with that of “B”: a) largerb) smallerc) the same a) largerb) smallerc) the same P = VI, I = P/V, Since P A > P B it results that I A > I B, and R A < R B

44. Example Numerical A computer monitor has a power requirement of 200 W, whereas a toaster is rated at 1500 W. Find and compare: a) the electric currents flowing through each device, and b) the resistances of the two devices if both are connected at 120 V. Monitor: I m = P m /V = 200W/120V = 1.67A P = VI, I = P/V Toaster: I t = P t /V = 1500W/120V = 12.5A I= V/R, R = V/I Monitor: R m = V/I m = 120V/1.67A = 71.9Ω Toaster: R t = V/I t = 120V/12.5A = 9.6Ω a) b)

45. Example Repair A hair dryer is rated @ 1200 W for 115 V operating voltage. A repair done by the owner shortens the wire of the filament by 10%. What will be the power output after the repair? P = V 2 /R, R = V 2 /P= (115V) 2 / 1200W = 11.02Ω R = ρL/A so R’=0.9 R =.9 x 11.02Ω = 9.92Ω P’= V 2 /R’= (115V) 2 / 9.92Ω =1333.15 W Change + 133W, do not perform the repair!

46. Consumer Electric Energy Consumer Electric Energy  Energy = Power x Time; W = P x t = V x I x t  Measured in Kilowatt-hours (kWh).  1 kilowatt-hour =1kilowatt x 1 hour.  Electric bill: # of kWh x price of 1 kWh.  Electric safety. NO: 1. Wet hands, 2. Wires under carpets, 3. Too many appliances, 4. Worn or frayed wires, 5. Touch electric sockets without protection, 6. Close to wires or power poles.

47. Example Cost  A toaster rated at 1500 W. If you pay $0.10/KWh, how much does it cost to use the appliance for 15 min.? W = Pt = 1500 W x 900s = 1,350,000 J # of kwh = 1,350,000 J/3,600,000 J=0.375 Cost= 0.375 x 0.1= $ 0.0375

48. Review Review  I=V/R; P=V x I; W=P x t; W = V x I x t  Cost = # of kWh x Price of kWh

49. ConcepTest 17.1Connect the Battery Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 3) 2)

50. continuous connection Current can only flow if there is a continuous connection from the negative terminal through the bulb to the positive terminal. This is only the case for Fig. (3). ConcepTest 17.1Connect the Battery Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 3) 2)

51. Ohm’s Law is obeyed since the current still increases when V increases 1) Ohm’s Law is obeyed since the current still increases when V increases Ohm’s Law is not obeyed 2) Ohm’s Law is not obeyed this has nothing to do with Ohm’s Law 3) this has nothing to do with Ohm’s Law ConcepTest 17.2Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude?

52. Ohm’s Law is obeyed since the current still increases when V increases 1) Ohm’s Law is obeyed since the current still increases when V increases Ohm’s Law is not obeyed 2) Ohm’s Law is not obeyed this has nothing to do with Ohm’s Law 3) this has nothing to do with Ohm’s Law V = I R linear Ohm’s Law, V = I R, states that the relationship between voltage and current is linear. Thus for a conductor that obeys Ohm’s Law, the current must double when you double the voltage. ConcepTest 17.2Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? Follow-up: Where could this situation occur?

53. ConcepTest 17.3aWires I Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? d A = 4 d B 1) d A = 4 d B d A = 2 d B 2) d A = 2 d B d A = d B 3) d A = d B 4) d A = 1/2 d B 5) d A = 1/4 d B

54. area is less arearadiussquared diameter of A must be two times less than B The resistance of wire A is greater because its area is less than wire B. Since area is related to radius (or diameter) squared, the diameter of A must be two times less than B. ConcepTest 17.3aWires I Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? d A = 4 d B 1) d A = 4 d B d A = 2 d B 2) d A = 2 d B d A = d B 3) d A = d B 4) d A = 1/2 d B 5) d A = 1/4 d B

55. ConcepTest 17.3bWires II A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? it decreasesby a factor 4 1) it decreases by a factor 4 it decreasesby a factor 2 2) it decreases by a factor 2 it stays the same 3) it stays the same 4) it increasesby a factor 2 4) it increases by a factor 2 5) it increasesby a factor 4 5) it increases by a factor 4

56. doubledhalved Keeping the volume (= area x length) constant means that if the length is doubled, the area is halved. four Since, this increases the resistance by four. ConcepTest 17.3bWires II A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? it decreasesby a factor 4 1) it decreases by a factor 4 it decreasesby a factor 2 2) it decreases by a factor 2 it stays the same 3) it stays the same 4) it increasesby a factor 2 4) it increases by a factor 2 5) it increasesby a factor 4 5) it increases by a factor 4

57. ConcepTest 17.4Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3)

58. increases the resistancedecreases the current The voltage is provided at 120 V from the outside. The light dimmer increases the resistance and therefore decreases the current that flows through the lightbulb. ConcepTest 17.4Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3) Follow-up: Why does the voltage not change?

59. ConcepTest 17.5aLightbulbs Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance? 1) the 25 W bulb 2) the 100 W bulb 3) both have the same 4) this has nothing to do with resistance

60. P = V 2 / R, lower power ratinghigher resistance Since P = V 2 / R, the bulb with the lower power rating has to have the higher resistance. ConcepTest 17.5aLightbulbs Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance? 1) the 25 W bulb 2) the 100 W bulb 3) both have the same 4) this has nothing to do with resistance Follow-up: Which one carries the greater current?

61. ConcepTest 17.5bSpace Heaters I Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? 1) heater 1 2) heater 2 3) both equally

62. P = V 2 / R, smaller resistance larger power Using P = V 2 / R, the heater with the smaller resistance will have the larger power output. Thus, heater 2 will give off more heat. ConcepTest 17.5bSpace Heaters I Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? 1) heater 1 2) heater 2 3) both equally Follow-up: Which one carries the greater current?

63. (Basic) Electric Circuits  Series Connections:  Total Resistance = Sum of individual Resistances. Current the same through every load.

64. Resistors in Series a c R effective a b c R1R1 R2R2 I The Voltage “drops”: Hence: Whenever devices are in SERIES, the current is the same through both ! This reduces the circuit to: In general:

65. Another (intuitive) way… Consider two cylindrical resistors with lengths L 1 and L 2 V R1R1 R2R2 L2L2 L1L1 Put them together, end to end to make a longer one...

66. Two resistors are connected in series to a battery with emf E. The resistances are such that R 1 = 2R 2. 1) Compare the current through R 1 with the current through R 2 : a) I 1 > I 2 b) I 1 = I 2 c) I 1 < I 2 2) What is the potential difference across R 2 ? a) V 2 = E b) V 2 = 1/2 E c) V 2 = 1/3 E Exercise: Series

67. Parallel Connection  Different parts of an electric circuit are on separate branches. Voltage: the same Total current: sum of currents in each branch

68. Resistors in Parallel Parallel a d I I R1R1 R2R2 I1I1 I2I2 V I a d I RV Current Devices in parallel have the same voltage drop V. How is I related to I 1 & I 2 ? Current (Charge) is conserved!   In general:

69. Another (intuitive) way… Consider two cylindrical resistors with cross-sectional areas A 1 and A 2 V R1R1 R2R2 A1A1 A2A2 Put them together, side by side … to make one “fatter”one, 

70. Example Series & Parallel Given: R 1 =4Ω, R 2 =6 Ω, R 3 =12Ω, find: a) R s b) R p

71. Example 2 Series Given: Three bulb s with the resistances R 1 =4Ω, R 2 =6 Ω, R 3 =12Ω, are connected in series with a 6 V battery. Required to find: a) The equivalent resistance of the bulbs R s b) The current flowing through each bulb c) The voltage drop across each bulb: d) The power dissipated by each bulb:

72. Example 3 Parallel Given: Three bulb s with the resistances R 1 =4Ω, R 2 =6 Ω, R 3 =12Ω, are connected in parallel with a 6 V battery. Required to find: a) The equivalent resistance of the bulbs R p b) The current flowing through each bulb and the total current: c) The voltage drop across each bulb: d) The power dissipated by each bulb:

73. Batteries (“Non-ideal” = cannot output arbitrary current)  R I I r V With "internal resistance“, terminal voltage V: 

74. ConcepTest 18.1aSeries Resistors I 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

75. 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. ConcepTest 18.1aSeries Resistors I 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 Follow-up: What would be the potential difference if R=  Follow-up: What would be the potential difference if R= 1 

76. ConcepTest 18.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  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

77. ConcepTest 18.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  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.V 1 = 8 V 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 and V 2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6  = 2 A, then use Ohm’s Law to get voltages. Follow-up: What happens if the voltage is doubled?

78. ConcepTest 18.2aParallel Resistors I 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

79. 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. ConcepTest 18.2aParallel Resistors I 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 Follow-up: What is the total current through the battery?

80. ConcepTest 18.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero 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?

81. ConcepTest 18.2bParallel Resistors II 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? Follow-up: What happens to the current through each resistor?

82. ConcepTest 18.3aShort Circuit I Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? all the current continues to flow through the bulb 1) all the current continues to flow through the bulb half the current flows through the wire, the other half continues through the bulb 2) half the current flows through the wire, the other half continues through the bulb all the current flows through the wire 3) all the current flows through the wire none of the above 4) none of the above

83. zeroALL The current divides based on the ratio of the resistances. If one of the resistances is zero, then ALL of the current will flow through that path. ConcepTest 18.3aShort Circuit I Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? all the current continues to flow through the bulb 1) all the current continues to flow through the bulb half the current flows through the wire, the other half continues through the bulb 2) half the current flows through the wire, the other half continues through the bulb all the current flows through the wire 3) all the current flows through the wire none of the above 4) none of the above Follow-up: Doesn’t the wire have SOME resistance?

84. ConcepTest 18.3bShort Circuit II Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will: glow brighter than before 1) glow brighter than before glow just the same as before 2) glow just the same as before glow dimmer than before 3) glow dimmer than before 4) go out completely 5) explode

85. total resistance of the circuit decreasescurrent through bulb A increases Since bulb B is bypassed by the wire, the total resistance of the circuit decreases. This means that the current through bulb A increases. ConcepTest 18.3bShort Circuit II Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will: glow brighter than before 1) glow brighter than before glow just the same as before 2) glow just the same as before glow dimmer than before 3) glow dimmer than before 4) go out completely 5) explode Follow-up: What happens to bulb B?

86. ConcepTest 18.4aCircuits I circuit 1 1) circuit 1 circuit 2 2) circuit 2 both the same 3) both the same it depends on R 4) it depends on R The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness  power)

87. ConcepTest 18.4aCircuits I circuit 1 1) circuit 1 circuit 2 2) circuit 2 both the same 3) both the same it depends on R 4) it depends on R The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness  power) parallel lowering the total resistance #1 willdraw a higher current P = I V In #1, the bulbs are in parallel, lowering the total resistance of the circuit. Thus, circuit #1 will draw a higher current, which leads to more light, because P = I V.

88. ConcepTest 18.4bCircuits II twice as much 1) twice as much the same 2) the same 1/2 as much 3) 1/2 as much 1/4 as much 4) 1/4 as much 4 times as much 5) 4 times as much 10 V A B C The three lightbulbs in the circuit all have the same resistance of 1  By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness  power)

89. ConcepTest 18.4bCircuits II twice as much 1) twice as much the same 2) the same 1/2 as much 3) 1/2 as much 1/4 as much 4) 1/4 as much 4 times as much 5) 4 times as much 10 V A B C We can use P = V 2 /R to compare the power: P A = (= 100 W P A = (V A ) 2 /R A = (10 V) 2 /1  = 100 W P B = (= 25 W P B = (V B ) 2 /R B = (5 V) 2 /1  = 25 W The three lightbulbs in the circuit all have the same resistance of 1  By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness  power) Follow-up: What is the total current in the circuit?

90. ConcepTest 18.5aMore Circuits I increase 1) increase decrease 2) decrease stay the same 3) stay the same What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: V R1R1 R3R3 R2R2 S

91. ConcepTest 18.5aMore Circuits I increase 1) increase decrease 2) decrease stay the same 3) stay the same What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: decreases the equivalent resistancecurrent from the battery increases increase in the voltage across R 1 With the switch closed, the addition of R 2 to R 3 decreases the equivalent resistance, so the current from the battery increases. This will cause an increase in the voltage across R 1. V R1R1 R3R3 R2R2 S Follow-up: ? Follow-up: What happens to the current through R 3 ?

92. ConcepTest 18.5bMore Circuits II increases 1) increases decreases 2) decreases stays the same 3) stays the same V R1R1 R3R3 R4R4 R2R2 S What happens to the voltage across the resistor R 4 when the switch is closed?

93. V R1R1 R3R3 R4R4 R2R2 S A B C increase in the voltage across R 1 V AB constant V AB increasesV BC must decrease We just saw that closing the switch causes an increase in the voltage across R 1 (which is V AB ). The voltage of the battery is constant, so if V AB increases, then V BC must decrease! What happens to the voltage across the resistor R 4 when the switch is closed? increases 1) increases decreases 2) decreases stays the same 3) stays the same ConcepTest 18.5bMore Circuits II Follow-up: ? Follow-up: What happens to the current through R 4 ?

94. ConcepTest 18.6ircuits ConcepTest 18.6Even More Circuits Which resistor has the greatest current going through it? Assume that all the resistors are equal. V R1R1 R2R2 R3R3 R5R5 R4R4 1) R 1 and 2) both R 1 and R 2 equally and 3) R 3 and R 4 4) R 5 5) all the same

95. I 1 = I 2 R 5 R 3 + R 4 branch containing R 5 has less resistance The same current must flow through left and right combinations of resistors. On the LEFT, the current splits equally, so I 1 = I 2. On the RIGHT, more current will go through R 5 than R 3 + R 4 since the branch containing R 5 has less resistance. ConcepTest 18.6ircuits ConcepTest 18.6Even More Circuits 1) R 1 and 2) both R 1 and R 2 equally and 3) R 3 and R 4 4) R 5 5) all the same Which resistor has the greatest current going through it? Assume that all the resistors are equal. V R1R1 R2R2 R3R3 R5R5 R4R4 Follow-up: ? Follow-up: Which one has the smallest voltage drop?

96. Kirchhoff’s First Rule “Junction Rule” or “Kirchhoff’s Current Law (KCL)” In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchhoff's First Rule (the junction rule). "At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node." This is just a statement of the conservation of charge at any given node. The currents entering and leaving circuit nodes are known as “branch currents”. Each distinct branch must have a current, I i  assigned to it

97. Kirchhoff’s Second Rule “Loop Rule” or “Kirchhoff’s Voltage Law (KVL)” "When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero." KVL: This is just a restatement of what you already know: that the potential difference is independent of path!  R1R1  R2R2 I     IR 1  IR 2     0 0

98. Rules of the Road  R1R1  R2R2 I     IR 1  IR 2     0 0 Our convention: Voltage gains enter with a + sign, and voltage drops enter with a  sign. We choose a direction for the current and move around the circuit in that direction (cw or ccw). When a battery is traversed from the negative terminal to the positive terminal, the voltage increases, and hence the battery voltage enters KVL with a + sign. When moving across a resistor, the voltage drops, and hence enters KVL with a  sign.

99. Loop Demo a d b e c f  R1R1 I R2R2 R3R3 R4R4 I    KVL:

100. Example Switch (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 The key here is to determine the potential ( V a - V b ) before the switch is closed. From symmetry, V a - V b = +12V. Therefore, when the switch is closed, NO additional current will flow! Therefore, the current before the switch is closed is equal to the current after the switch is closed. Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –Just after the switch is closed, the current flowing through the bottom resistor is I 1. –What is the relation between I 0 and I 1 ? R 12V R I a b

101. Example Switch Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –After the switch is closed, the current flowing through the bottom resistor is I 1. –What is the relation between I 0 and I 1 ? (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 Write a loop law for original loop: 12V  I 1 R = 0 I 1 = 12V/R Write a loop law for the new loop: 12V +12V  I 0 R  I 0 R = 0 I 0 = 12V/R R 12V R I a b

102. Summary  When you are given a circuit, you must first carefully analyze circuit topology find the nodes and distinct branches find the nodes and distinct branches assign branch currents assign branch currents  Use KVL for all independent loops in circuit sum of the voltages around these loops is zero! sum of the voltages around these loops is zero!

103. a b  R C I I R  R I I r V

104. How to use Kirchhoff’s Laws A two loop example: Analyze the circuit and identify all circuit nodes and use KCL. (2)  1  I 1 R 1  I 2 R 2 = 0 (3)  1  I 1 R 1   2  I 3 R 3 = 0 (4) I 2 R 2   2  I 3 R 3 = 0  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 (1) I 1 = I 2 + I 3 Identify all independent loops and use KVL.

105. How to use Kirchoff’s Laws  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 Solve the equations for I 1, I 2, and I 3 : First find I 2 and I 3 in terms of I 1 :  From eqn. (2) From eqn. (3) Now solve for I 1 using eqn. (1):

106. Let’s plug in some numbers  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3  1 = 24 V  2 = 12 V R 1 = 5  R 2 =3  R 3 =4  Then, and I 1 =2.809 A I 2 = 3.319 A, I 3 = -0.511 A

107. Junction Demo  I1I1   R R R I2I2 I3I3 Outside loop:Top loop:Junction:

108. Potential Difference, SET 1 Consider the circuit shown: –What is the relation between V a - V d and V a - V c ? (a) (V a -V d ) < (V a -V c ) (b) (V a -V d ) = (V a -V c ) (c) (V a -V d ) > (V a -V c ) 12V I1I1 I2I2 a b d c 50  20  80  (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 1B – What is the relation between I 1 and I 2 ? 1B 1A

109. Example Multiloop Consider the circuit shown: –What is the relation between V a - V d and V a - V c ? (a) (V a -V d ) < (V a -V c ) (b) (V a -V d ) = (V a -V c ) (c) (V a -V d ) > (V a -V c ) 12V I1I1 I2I2 a b d c 50  20  80  1A Do you remember that thing about potential being independent of path? Well, that’s what’s going on here !!! ( V a - V d ) = ( V a - V c ) Point d and c are the same, electrically

110. Example Multiloop (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 1B – What is the relation between I 1 and I 2 ? 1B Consider the circuit shown: –What is the relation between V a - V d and V a - V c ? (a) (V a -V d ) < (V a -V c ) (b) (V a -V d ) = (V a -V c ) (c) (V a -V d ) > (V a -V c ) 12V I1I1 I2I2 a b d c 50  20  80  1A Note that: V b -V d = V b -V c Therefore,

111. Summary of Simple Circuits Resistors in series: Resistors in parallel: Current thru is same; Voltage drop across is IR i Voltage drop across is same; Current thru is V / R i

112. Household Circuits  Connections are made in parallel.  Ground wire.  Fuses: made of a strip of metal that melts and breaks the circuit if the current becomes to high.  Circuit breakers: a switch that flips open if the current becomes to high.

113. ConcepTest 18.7Junction Rule ConcepTest 18.7 Junction Rule 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A 5 A 8 A 2 A P What is the current in branch P?

114. red One exiting branch has 2 A other branch (at P) must have 6 A The current entering the junction in red is 8 A, so the current leaving must also be 8 A. One exiting branch has 2 A, so the other branch (at P) must have 6 A. 5 A 8 A 2 A P junction 6 A S 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A What is the current in branch P? ConcepTest 18.7Junction Rule

115. ConcepTest 18.7Kirchhoff’s Rules ConcepTest 18.7 Kirchhoff’s Rules The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes

116. the point between the bulbs is at 12 VBut so is the point between the batteries When the switch is open, the point between the bulbs is at 12 V. But so is the point between the batteries. If there is no potential difference, then no current will flow once the switch is closed!! Thus, nothing changes. The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes ConcepTest 18.8Kirchhoff’s Rules ConcepTest 18.8 Kirchhoff’s Rules 24 V Follow-up: Follow-up: What happens if the bottom battery is replaced by a 24-V battery?

117. ConcepTest 18.9Wheatstone Bridge 1) I 2) I/2 3) I/3 4) I/4 5) zero An ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. The current through the ammeter is: I Vab

118. resistors are identical voltage drops are the same potentialsab same Since all resistors are identical, the voltage drops are the same across the upper branch and the lower branch. Thus, the potentials at points a and b are also the same. Therefore, no current flows. ConcepTest 18.9Wheatstone Bridge 1) I 2) I/2 3) I/3 4) I/4 5) zero An ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. The current through the ammeter is: I Vab

119. ConcepTest 18.10Kirchhoff’s Rules ConcepTest 18.10 More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I 2 + 6 = 0 5) 2 – I 1 – 3I 3 – 6 = 0

120. ConcepTest 18.10Kirchhoff’s Rules ConcepTest 18.10 More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Equation 3 is valid for the left loop Equation 3 is valid for the left loop: The left battery gives +2V, then there is a drop through a 1  resistor with current I 1 flowing. Then we go through the middle battery (but from + to – !), which gives –4V. Finally, there is a drop through a 2  resistor with current I 2. Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I 2 + 6 = 0 5) 2 – I 1 – 3I 3 – 6 = 0

121. Kirchhoff’s Rules with Capacitors Kirchhoff’s Rules can be applied to all kinds of circuits The change in the potential around the circuit is +ε – I R – q / C = 0 Want to solve for I I and q will be time dependent Section 19.5

122. Capacitors, cont. When the switch is closed, there is a current carrying a positive charge to the top plate of the capacitor When the capacitor plates are charged, there is a nonzero voltage across the capacitor Section 19.5

123. Copyright © 2012 Pearson Education, Inc. publishing as Addison-Wesley Resistors and capacitors combine – Figures 19.32, 33 (optional) Combinations of resistors and capacitors form what are called RC Devices. A camera flash storing charge is a good example.

124. Charging Capacitors (optional) The current in the circuit is described by The voltage across the capacitor is The charge is given by τ = RC and is called the time constant Section 19.5

125. Time Constant (optional) Current At the end of one time constant, the current has decreased to 37% of its original value At the end of two time constants, the current has decreased to 63% of its original value Voltage and charge At the end of one time constant, the voltage and current have increased 63% of their asymptotic values Section 19.5

126. Capacitor at Various Times Just after the switch is closed The charge is very small V cap is very small I = ε / R When t is large The charge is very large V cap ≈ ε The polarity of the capacitor opposes the battery emf The current approaches zero Section 19.5

127. Capacitors at Various Times, Circuits

128. Discharging the Capacitor (optional) Current: Voltage: V cap = ε e -t/ τ Charge: q = C ε e -t/ τ Time constant: τ = RC, the same as for charging Section 19.5

129. Example A 10-V emf battery is connected in series with the following: a 2-µF capacitor, a 2-Ώ resistor, an ammeter, and a switch, initially open; a voltmeter is connected in parallel across the capacitor. Immediately after the switch is closed, what are the current and capacitor voltage readings, respectively? Short time capacitor acts like a wire, R = 0 ohm I = 10V/2ohm = 5 A V = O V What are the current and capacitor voltage readings, after a long time? V = (-) 10 V, I = 0 A What is the charge on the capacitor after a long time? Q = CV = 2μF x 10 V = 20 μC

130. Consider the circuit below. Switch S is closed at t = 0. What is the voltage across the capacitor C just after the switch is closed? 1.0 2.1 V 3.2 V 4.3 V 5.4 V 6.5 V 7.6 V 8.9 V 9.None of the above 10.Impossible to determine

131. Consider the circuit below. Switch S is closed at t = 0. What is the current through the capacitor C just after the switch is closed? 1.0 2.1 A 3.2 A 4.3 A 5.4 A 6.5 A 7.6 A 8.9 A 9.None of the above 10.Impossible to determine

132. Consider the circuit below. Switch S is closed at t = 0 (no clicker) What is the voltage across the capacitor C long after the switch is closed? 1.0 2.1 V 3.2 V 4.3 V 5.4 V 6.5 V 7.6 V 8.9 V 9.None of the above 10.Impossible to determine

133. Consider the following circuit. The capacitor is uncharged when the switch is closed at t = 0. Which circuit is equivalent to this circuit for the instant immediately after the switch is closed? 1.2. 3.4.

134. Consider the circuit below. Switch S is closed at t = 0. What is the voltage across the capacitor C just after the switch is closed? 1.0 2.1 V 3.2 V 4.3 V 5.4 V 6.5 V 7.6 V 8.9 V 9.None of the above 10.Impossible to determine

135. Consider the circuit below. Switch S is closed at t = 0. (no clicker) What is the current through capacitor C just after the switch is closed? 1.0 2.1 A 3.2 A 4.3 A 5.4 A 6.5 A 7.6 A 8.9 A 9.None of the above 10.Impossible to determine

136. Consider the circuit below. Switch S is closed at t = 0. What is the voltage across the capacitor C long after the switch is closed? 1.0 2.1 V 3.2 V 4.3 V 5.4 V 6.5 V 7.6 V 8.9 V 9.None of the above 10.Impossible to determine

137. Electric Currents in Nerves Many nerves are long and thin, much like wires The conducting solution inside the fiber acts as a resistor The lipid layer acts as a capacitor The nerve fiber behaves as an RC circuit Needs a small time constant Small R through large radius Small C through a layer of myelin increasing the distance between the capacitor plates

138. Electric Currents and Health Your body is a moderately good conductor of electricity The body’s resistance when dry is about 1500 Ω When wet, the body’s resistance is about 500 Ω Current is carried by different parts of the body Skin Internal organs Section 19.8

139. Household Circuits Connections are made in parallel. Ground wire. Fuses: made of a strip of metal that melts and breaks the circuit if the current becomes to high. Circuit breakers: a switch that flips open if the current becomes to high.

140. Household Currents The voltage in your home is an AC voltage oscillating at a certain frequency Frequency is 60 Hz in the US Most modern outlets and plugs have three connections Two are flat One is round and connects to ground Polarized plugs have flat connectors of different sizes The larger connects to the lower potential

141. Fuses and Circuit Breakers In a fuse, current passes through a thin metal strip This strip acts as a resistor with a small resistance If a failure causes the current to become large, the power causes the strip to melt and the current stops A circuit breaker also stops the current when it exceeds a predetermined limit The circuit breaker can be reset for continued use