1. DC Circuits: Ch 32 Voltage – Starts out at highest point at “+” end of battery Voltage drops across lightbulbs and other sources of resistance. Voltage increases again at battery.

2. +- I Voltage highestVoltage zero

3. The following circuit uses a 1.5 V battery and has a 15 W lightbulb. a.Calculate the current in the circuit (P = IV) b.Calculate the voltage drop across the lightbulb. c.Sketch a graph of voltage vs. path (battery, top wire, resistor, bottom wire)

5. Resistors in Series Same current (I) passes through all resistors (bulbs) All bulbs are equally bright (energy loss, not current loss) Voltage drop across each resistor (V 1,V 2, V 3 )

6. V = V 1 + V 2 + V 3 V = IR 1 + IR 2 + IR 3 V = I(R 1 + R 2 + R 3 ) R eq = R 1 + R 2 + R 3 V = IR eq

7. Resistors in Parallel Current splits at the junction Same Voltage across all resistors

8. I = I1 + I2 + I3 I1 = V R 1 I = V R eq 1=1+1+1 R eq R 1 R 2 R 3

9. Which combination of auto headlights will produce the brightest bulbs? Assume all bulbs have a resistance of R.

10. For the Bulbs in Series: R eq = R + R = 2R For the Bulbs in Parallel 1=1+11=1+1 R eq RR 1=21=2 R eq R R eq = R/2 The bulbs in parallel have less resistance and will be brighter

11. What current flows through each resistor in the following circuit? (R = 100  ) R eq = R 1 + R 2 R eq = 200  V = IR eq I = V/R eq I = 24.0 V/ 200  = 0.120 A

12. Calculate the current through this circuit, and the voltage drop across each resistor. R eq = 400  + 290  R eq = 690  V = IR I = V/R eq I = 12.0 V/690  I = 0.0174 A

13. V ab = (0.0174A)(400  ) V ab = 6.96 V V bc = (0.0174A)(290  ) V bc = 5.04 V

14. What current flows through each of the resistors in this circuit? (R = 100  ) 1=1 + 1 R eq 100  100  1/R eq = 2/100  R eq = 50  I = V/R eq = 24.0 V/50  = 0.48 A

15. DC Circuits: Ex 4 What current will flow through the circuit shown? 1=1+1 R p = 500  700  R p = 290 

16. R eq = 400  + 290  R eq = 690  V = IR I = V/R I = 12.0 V/690  I = 0.017 A or 17 mA

17. Example 4 Calculate the equivalent resistance in the following circuit.

18. DC Circuits: Ex 5 What current is flowing through just the 500  resistor? First we find the voltage drop across the first resistor: V = IR = (0.017 A)(400  ) V = 6.8 V

19. The voltage through the resistors in parallel will be: 12.0 V – 6.8 V = 5.2 V To find the current across the 500  resistor: V = IR I = V/R I = 5.2 V/500  = 0.010 A = 10 mA

20. DC Circuits: Ex 6 Which bulb will be the brightest in this arrangement (most current)?

21. Bulb C (current gets split running through A and B) What happens when the switch is opened? –C and B will have the same brightness (I is constant in a series circuit)

22. DC Circuits: Ex 5 What resistance would be present between points A and B? (ANS: 41/15 R)

24. EMF and Terminal Voltage Batteries - source of emf (Electromotive Force), E (battery rating) All batteries have some internal resistance r V ab = E – Ir V ab = terminal(useful)voltage E = battery rating r = internal resistance

25. EMF: Example 1 A 12-V battery has an internal resistance of 0.1 . If 10 Amps flow from the battery, what is the terminal voltage? V ab = E – Ir V ab = 12 V – (10 A)(0.10  ) V ab = 11 V

26. EMF: Example 2 Calculate the current in the following circuit. 1/R eq = 1/8  + ¼  R eq = 2.7 

27. R eq = 6  R eq = 8.7 

28. 1/R eq = 1/10  R eq = 4.8 

29. Everything is now in series R eq = 4.8  R eq = 10.3  V = IR I = V/R I = 9.0 V/10.3  I = 0.87 A

30. EMF: Example 2a Now calculate the terminal(useful)voltage. V = E – Ir V = 9.0 V – (0.87 A)(0.50  ) V = 8.6 V

31. Grounded Wire is run to the ground Houses have a ground wire at main circuit box Does not affect circuit behavior normally Provides path for electricity to flow in emergency

32. Kirchoff’s Rules 1.Junction Rule - The sum of the currents entering a junction must equal the sum of currents leaving 2.Loop Rule - The sum of the changes in potential around any closed path = 0

33. Kirchoff Conventions The “loop current” is not a current. Just a direction that you follow around the loop.

34. Kirchoff Conventions

35. Kirchoff’s Rule Ex 1

37. Junction Rule I 1 = I 2 + I 3

39. Loop Rule Main Loop 6V – (I 1 )(4  ) – (I 3 )(9  ) = 0 Side Loop (-I 2 )(5  ) + (I 3 )(9  ) = 0

40. I 1 = I 2 + I 3 Eqn 3 6V – (I 1 )(4  ) – (I 3 )(9  ) = 0Eqn 2 (-I 2 )(5  ) + (I 3 )(9  ) = 0Eqn 3 Solve Eqn 1 (-I 2 )(5  ) + (I 3 )(9  ) = 0 (I 3 )(9  ) = (I 2 )(5  ) I 3 = 5  I 2

41. Substitution into Eqn 2 6V – (I 2 + I 3 )(4  ) – (I 3 )(9  ) = 0 6 – 4I 2 -4I 3 - 9I 3 = 0 6 – 4I 2 - 13I 3 = 0 I 3 = 5  I 2 (from last slide) 6 – 4I 2 - 13(5  I 2 ) = 0 6 = 101/9 I 2 I 2 = 0.53 A I 3 = 5/9 I 2 = 0.29 A I 1 = I 2 + I 3 = 0.53 A + 0.29 A = 0.82 A

42. Kirchoff’s Rule Ex 2

43. I 1 = I 2 + I 3

45. Loop Rule Main Loop 9V – (I 3 )(10  ) – (I 1 )(5  ) = 0 Side Loop (-I 2 )(5  ) + (I 3 )(10  ) = 0

46. (I 2 )(5  ) = (I 3 )(10  ) I 2 = 2I 3 9V – (I 3 )(10  ) – (I 1 )(5  ) = 0 9V – (I 3 )(10  ) – (I 2 + I 3 )(5  ) = 0 9 –10I 3 – 5I 2 – 5I 3 = 0 9 –15I 3 – 5I 2 = 0 9 –15I 3 – 5(2I 3 ) = 0 9 –25I 3 = 0 I 3 = 9/25 = 0.36 A

47. I 2 = 2I 3 = 2(0.36 A) = 0.72 A I 1 = I 2 + I 3 = 0.36A + 072 A = 1.08 A

48. Kirchoff’s Rules: Ex 3 Calculate the currents in the following circuit. I 1 + I 2 = I 3

49. Bottom Loop (clockwise) 10V – (6  )I 1 – (2  )I 3 = 0 Top Loop (clockwise) -14V +(6  )I 1 – 10 V -(4  )I 2 = 0 Work with Bottom Loop 10V – (6  )I 1 – (2  )I 3 = 0 I 1 + I 2 = I 3 10 – 6I 1 – 2(I 1 + I 2 ) = 0 10 – 6I 1 – 2I 1 - 2I 2 = 0

50. 10 - 8I 1 - 2I 2 = 0 10 = 8I 1 + 2I 2 5 = 4I 1 + I 2 I 2 = 5 - 4I 1 Working with Top Loop -14V +(6  )I 1 – 10 V -(4  )I 2 = 0 24 = 6I 1 - 4I 2 12 = 3I 1 - 2I 2 12 = 3I 1 - 2(5 - 4I 1 ) 22 = 11I 1

51. I 1 = 22/11 = 2.0 Amps I 2 = 5 - 4I 1 I 2 = 5 – 4(2) = -3.0 Amps I 1 + I 2 = I 3 I 3 = -1.0 A

52. Batteries in Series If + to -, voltages add (top drawing) If + to +, voltages subtract (middle drawing = 8V, used to charge the 12V battery as in a car engine) Batteries in Parallel Provide large current when needed

53. Extra Kirchoff Problems I 1 = -0.864 A I 2 = 2.6 A I 3 = 1.73 A

54. A Strange Example Calculate I (-0.5 Amps (we picked wrong direction))

55. a.Calculate the equivalent resistance. (2.26  ) b.Calculate the current in the upper and lower wires. (3.98 A) c.Calculate I 1, I 2, and I 3 (0.60 A, 0.225 A, 1.13 A) d.Sketch a graph showing the voltage through the circuit starting at the battery.

57. RC Circuits Capacitors store energy (flash in a camera) Resistors control how fast that energy is released Car lights that dim after you shut them Camera flashes

58.  V c +  V r = 0 Q - IR = 0(Divide by R) C Q - I = 0(I = -dQ/dt) RC Q + dQ = 0 RC dt

60.  = time constant (time to reach 63% of full voltage)  = RC

61. A versatile relationship V = V o e -t/RC I = I o e -t/RC Q = Q o e -t/RC I generally find voltage, then use V=IR and Q=VC

62. RC Circuits: Ex 1 What is the time constant for an RC circuit of resistance 200 k  and capacitance of 3.0  F?  = (200,000  )(3.0 X 10 -6 F)  = 0.60 s (lower resistance will cause the capacitor to charge more quickly)

63. RC Circuits: Ex 2 What will happen to the bulb (resistor) in the circuit below when the switch is closed (like a car door)?

64. Answer: Bulb will glow brightly initially, then dim as capacitor nears full charge.

65. RC Circuits: Ex 3 An uncharged RC circuit has a 12 V battery, a 5.0  F capacitor and a 800 k  resistor. Calculate the time constant.  = RC  = (5.0 X 10 -6 F)(800,000 W)  = 4.0 s

66. What is the maximum charge on the capacitor? Q = CV Q = (5.0 X 10 -6 F)(12 V) Q = 6 X 10 -5 F or 60  F What is the voltage and charge on the capacitor after 1 time constant? V = (0.632)(12 V) = 7.584 V Q = (0.632)(60  F) = 38  F

67. Consider the circuit below. Calculate: a.The time constant (6 ms) b.Maximum charge on the capacitor (3.6  C) c.Time to reach 99% of maximum charge (28 ms) d.Current when charge = ½ Q max (300  A) e.Maximum current (600  A) f.The charge when the current is 20% of the maximum value. (2.9  C)

68. Discharging the RC Circuit V = V o e -t/RC

69. RC Circuits: Ex 4 An RC circuit has a charged capacitor C = 35  F and a resistance of 120 . How much time will elapse until the voltage falls to 10 percent of its original (maximum) value? V = V o e -t/RC 0.10V o =V o e -t/RC 0.10 =e -t/RC ln(0.10) = ln(e -t/RC ) -2.3 = -t/RC

70. 2.3 = t/RC t = 2.3RC t = (2.3)(120  )(35 X 10 -6 F) t = 0.0097 s or 9.7 ms

71. RC Circuits: Ex 5 If a capacitor is discharged in an RC circuit, how many time constants will it take the voltage to drop to ¼ its maximum value? V = V o e -t/RC 1.39 = t/RC 0.25V o =V o e -t/RC t = 1.39RC 0.25 =e -t/RC ln(0.25) = ln(e -t/RC ) -1.39 = -t/RC

72. A fully charged 1.02 mF capacitor is in a circuit with a 20.0 V battery and a resistor. When discharged, the current is observed to decrease to 50% of it’s initial value in 40  s. a.Calculate the charge on the capacitor at t=0 (20.4  C) b.Calculate the resistance R (57  ) c.Calculate the charge at t = 60  s (7.3  C)

73. The capacitor in the drawing has been fully charged. The switch is quickly moved to position b (camera flash). a.Calculate the initial charge on the capacitor. (9  C) b.Calculate the charge on the capacitor after 5.0  s. (5.5  C) c.Calculate the voltage after 5.0  s (5.5 V d.Calculate the current through the resistor after 5.0  s (0.55 A)

74. Meters Galvanometer –Can only handle a small current Full-scale Current Sensitivity (I m ) –Maximum deflection Ex: –Multimeter –Car speedometer

75. Measuring I and V Measuring Current –Anmeter is placed in series –Current is constant in series Measuring Voltage –Voltmeter placed in parallel –Voltage constant in parallel circuits –Measuring voltage drop across a resistor

76. Anmeter (Series)Voltmeter (parallel)

77. DC Anmeter Uses “Shunt” (parallel) resistor Shunt resistor has low resistance Most of current flows through shunt, only a little through Galvanometer I R R = I G r

78. Meters: Ex 1 What size shunt resistor should be used if a galvanometer has a full-scale sensitivity of 50  A and a resistance of r= 30  ? You want the meter to read a 1.0 A current. Voltage same through both (V=IR) I R R = I g r Since most of the current goes through the shunt I R ~ 1 A

79. I R R = I g r (1 A)(R) = (50 X 10 -6 A)(30  ) R = 1.5 X 10 -3  or 1.5 m 

80. Meters: Ex 2 Design an anmeter that can test a 12 A vacuum cleaner if the galvanometer has an internal resistance of 50 W and a full scale deflection of 1 mA. I R R = I g r (12 A)(R) = (1 X 10 -3 A)(50  ) R = 4.2 X 10 -3  or 4.2 m 

81. DC Voltmeter Resistor in series Large R for resistor (keeps current low in Galvanometer) V = I(R + r)

82. Meters: Ex 3 What resistor should be used in a voltmeter that can read a maximum of 15 V? The galvanometer has an internal resistance of 30  and a full scale deflection of 50  A. V = I(R + r) 12 V = (50 X 10 -6 A)(R + 30  ) R + 30  = 12 V 50 X 10 -6 A R + 30  = 300,000  R ~ 300,000 

83. Meters: Ex 4 Design a voltmeter for a 120 V appliance with and internal galvanometer resistance of 50  and a current sensitivity of 1 mA. (ANS: R = 120,000  )

84. Electric Power Watt 1 Watt = 1 Joule 1 second P = I 2 R “Twinkle, twinkle, little star. Power equals I 2 R”

85. Power: Ex 1 Calculate the resistance of a 40-W auto headlight that operates at 12 V. P = I 2 R V = IR (so I =V/R) P = V 2 R R 2 P = V 2 R

86. R R = V 2 = (12 V) 2 = 3.6  P40 W

87. Household Electricity Kilowatt-hour You do not pay for power, you pay for energy 1 kWh = (1000 J)(3600 s) = 3.60 X 10 6 J 1s

88. Power: Ex 2 An electric heater draws 15.0 A on a 120-V line. How much power does it use? P = I 2 R V = IR so R = V/I P = I 2 V I P = IV = (15 A)(120V) = 1800 W or 1.8 kW

89. Power: Ex 3 How much does it cost to run it for 30 days if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh? Hours = 30 days X 3.0 h/day = 90 h Cost = (1.80 kW)(90 h)($0.105/kWh) = $17

90. Power: Ex 4 A lightening bolt can transfer 10 9 J of energy at a potential difference of 5 X 10 7 V over 0.20 s. What is the charge transferred? V = PE/Q Q = E/V = 10 9 J/ 5 X 10 7 V = 20 C

91. What is the current? I =  Q/  t I = 20 C/0.2 s = 100 A What is the power? P = I 2 R V = IR so R = V/I P = I 2 V I P = IV = (100 A)(5 X 10 7 V ) = 5 X 10 9 W

92. Household Electricity Circuit breakers – prevent “overloading” (too much current per wire) Metal melts or bimetallic strip expands

93. Household Electricity: Ex 1 Determine the total current drawn by all of the appliances shown. P = IV I = P/V I light = 100W/120 V = 0.8 A

94. I heater = 1800W/120 V = 15 A I stereo = 350W/120 V = 2.9 A I hair = 1200W/120 V = 10.0 A I total = 0.8A + 15.0A + 2.9A + 10.0A = 28.7 A This would blow the 20 A fuse

95. DC vs AC DC Electrons flow constantly Electrons flow in only one direction Batteries AC Electrons flow in short burst Electrons switch directions (60 times a second) House current

96. DC vs AC http://www.ibiblio.org/obp/electricCircuits/AC/AC_1.html

97. Jump Starting a Car POSITIVE TO POSITIVE (or your battery could explode)