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

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Presentation on theme: "DC Circuits: Ch 32 Voltage – Starts out at highest point at “+” end of battery Voltage drops across lightbulbs and other sources of resistance. Voltage."— Presentation transcript:

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)

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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)

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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

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37 Junction Rule I 1 = I 2 + I 3

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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

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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.

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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

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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)


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