http://www.nearingzero.net (nz014.jpg)

1. http://www.nearingzero.net (nz014.jpg) This relates (sort of) to a demo I’ll do later.

2. Today’s lecture is brought to you by… Physics Man

3. Not to be confused with… Electro-Man (http://www.thinkgeek.com).

4. Today’s agenda: Emf, Terminal Voltage, and Internal Resistance. You must be able to incorporate all of these quantities in your circuit calculations. Electric Power. You be able to calculate the electric power dissipated in circuit components, and incorporate electric power in work-energy problems. Examples For you to work through outside of lecture.

5. circuit components in series In lecture 7 (the first capacitors lecture) I suggested using conservation of energy to show that the voltage drop across circuit components in series is the sum of the individual voltage drops: Vab C1 C2 C3 a b V2 V3 V1 + - V Vab = V = V1 + V2 + V3

6. circuit components in series In general, the voltage drop across resistors in series (or other circuit components) is the sum of the individual voltage drops. R1 R2 R3 a b V1 V2 V3 - + V Vab = V = V1 + V2 + V3 I “derived” this in lecture 7. Here’s what your text means by Vab: Vab=Va-Vb=Vba You may use this in tomorrow’s homework. It “is”* on your starting equations sheet, and is a consequence of conservation of energy. Use this in combination with Ohm’s Law, V=IR. * V = 0 around closed loop

7. Starting at point a and following the current in a clockwise path around the circuit and back to point a… I R1 R2 R3 a b V1 V2 V3 - + V - V1 - V2 - V3 + V = 0 - IR1 - IR2 - IR3 + V = 0

8. Again, starting at point a and following the current in a clockwise path around the circuit and back to point a… I R1 R2 R3 a b V1 V2 V3 - - + + VB VA - V1 - V2 - V3 + VA - VB = 0 - IR1 - IR2 - IR3 + VA - VB = 0

9. Example: calculate I, Vab,and Vba (to be worked at the blackboard). I 5  b 10  - + a 9 V

10. Example: calculate I, Vab,and Vba (to be worked at the blackboard). I 5  b 10  - - + + a 9 V 6 V Also discuss what happens if you guess wrong current direction.

11. DC Currents In Physics 24, whenever you work with currents in circuits, you should assume (unless told otherwise) “direct current.” Current in a dc circuit flows in one direction, from + to -. We will not encounter ac circuits much in this course. For any calculations involving household current, which is ac, assuming dc will be “close enough” to give you “a feel” for the physics. If you need to learn about ac circuits, you’ll have courses devoted to them. The mathematical analysis is more complex. We have other things to explore this semester.

12. emf, terminal voltage, and internal resistance We have been making calculations with voltages from batteries without asking detailed questions about the batteries. Now it’s time to look inside the batteries. http://www.energizer.com We introduce a new term – emf – in this section. Any device which transforms a form of energy into electric energy is called a “source of emf.” “emf” is an abbreviation for “electromotive force,” but emf is not a force! The emf of a source is the voltage it produces when no current is flowing.

13. The voltage you measure across the terminals of a battery (or any source of emf) is less than the emf because of internal resistance. Here’s a battery with an emf. All batteries have an “internal resistance:” emf is the zero-current potential difference - + The “battery” is everything inside the green box. a b Hook up a voltmeter to measure the emf: emf - + The “battery” is everything inside the green box. a b Getting ready to connect the voltmeter (it’s not hooked up yet).

14. Measuring the emf???  (emf) - + The “battery” is everything inside the green box. a b As soon as you connect the voltmeter, current flows. I You can’t measure voltage without some (however small) current flowing, so you can’t measure emf directly. You can only measure Vab. Homework hint: an ideal voltmeter would be able to measure . For example, problem 5.33.

15. We model a battery as producing an emf, , and having an internal resistance r: - + The “battery” is everything inside the green box. a b  r  Vab The terminal voltage, Vab, is the voltage you measure across the battery terminals with current flowing. When a current I flows through the battery, Vab is related to the emf  by Not recommended for use by children under 6. Do not continue use if you experience dizziness, shortness of breath, or trouble sleeping. Do not operate heavy machinery after using. An extinct starting equation.

16. - +  r To model a battery, simply include an extra resistor to represent the internal resistance, and label the voltage source* as an emf instead of V (units are still volts): a b If the internal resistance is negligible, simply don’t include it! If you are asked to calculate the terminal voltage, it is just Vab = Va – Vb, calculated using the techniques I am showing you today. *Remember, all sources of emf—not just batteries—have an internal resistance.

17. Summary of procedures for tomorrow’s homework: Draw the current in a circuit so that it flows from – to + through the battery. The sum of the potential changes around a circuit loop is zero.  + V is + - Potential decreases by IR when current goes through a resistor. I I V is - Potential increases by  when current passes through an emf in the direction from the - to + terminal. loop Treat a battery internal resistance like any other resistor. If I flows through a battery + to -, potential decreases by .

18. Example: a battery is known to have an emf of 9 volts. If a 1 ohm resistor is connected to the battery, the terminal voltage is measured to be 3 volts. What is the internal resistance of the battery? R=1  Because the voltmeter draws “no” current, r and R are in series with a current I flowing through both. I emf - + a b IR, the potential drop across the resistor, is just the potential difference Vab. internal resistance r terminal voltage Vab the voltmeter’s resistance is so large that approximately zero current flows through the voltmeter

19. R=1  I emf - + a b A rather unrealistically large value for the internal resistance of a 9V battery.

20. By the way, the experiment described in the previous example is not a very good idea. I’ll do a demo on this in a bit.

21. Today’s agenda: Emf, Terminal Voltage, and Internal Resistance. You must be able to incorporate all of these quantities in your circuit calculations. Electric Power. You must be able to calculate the electric power dissipated in circuit components, and incorporate electric power in work-energy problems. Examples For you to work through outside of lecture.

22. Electric Power Last semester you defined power in terms of the work done by a force. We’d better use the same definition this semester! So we will. We focus here on the interpretation that power is energy transformed per time, instead of work by a force per time. The above equation doesn’t appear on your equation sheet, but it should appear in your brain.

23. However, we begin with the work aspect. We know the work done by the electric force in moving a charge q through a potential difference: The work done by the electric force in moving an infinitesimal charge dq through a potential difference is: The instantaneous power, which is the work per time done by the electric force, is

24. Let’s get lazy and drop the  in front of the V, but keep in the back of our heads the understanding that we are talking about potential difference. Then But wait! We defined I = dQ/dt. So And one more thing… the negative sign means energy is being “lost.” So everybody writes and understands that P<0 means energy out, and P>0 means energy in.

25. Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R. I can’t believe it, but I got soft and put P = I2R = V2/R on your starting equation sheet. Truth in Advertising I. The V in P=IV is a potential difference, or voltage drop. It is really a V. Truth in Advertising II. Your power company doesn’t sell you power. It sells energy. Energy is power times time, so a kilowatt-hour (what you buy from your energy company) is an amount of energy.

26. “Quiz” time (maybe for points, maybe just for practice!)

27. Demo (Remember the terminal voltage example?) How your professor almost burned down Mark Twain Elementary School (on three different occasions). (And Truman Elementary School another time, just for good measure.) Note: no equipment set-up is needed for this demo.

28. Possible Homework Hint A 3 volt and 6 volt battery are connected in series, along with a 6 ohm resistor. The batteries* are connected the “wrong” way (+ to + and - to -). What is the power dissipated in the resistor? In the 3 volt battery? To be worked at the blackboard. (If time allows.) *Assume zero internal resistance unless the problem suggests otherwise.

29. Very Handy Homework Hint Remember: Vab = Va – Vb.

30. Today’s agenda: Emf, Terminal Voltage, and Internal Resistance. You must be able to incorporate all of these quantities in your circuit calculations. Electric Power. You must be able to calculate the electric power dissipated in circuit components, and incorporate electric power in work-energy problems. Examples Mostly for you to work through outside of lecture.

31. Example: A 12 V battery with 2  internal resistance is connected to a 4  resistor. Calculate (a) the rate at which chemical energy is converted to electrical energy in the battery, (b) the power dissipated internally in the battery, and (c) the power output of the battery. (a) Rate of energy conversion. Start at negative terminal of battery… R=4 - +  - I R2 - I R4 = 0  = 12 V I r=2 I =  / (R2 + R4) =12 V / 6  = 2 A Energy is converted at the rate Pconverted=I=(2 A)(12 V)=24W. Rate of energy conversion example.

32. (b) Power dissipated internally in the battery. R=4 - +  = 12 V r=2 I=2A Pdissipated = I2r = (2 A)2 (2 ) = 8 W. (c) Power output of the battery. Poutput = Pconverted - Pdissipated = 24 W - 8 W = 16W. Rate of energy conversion example.

33. (c) Power output of the battery (double-check). I=2A R=4 - +  = 12 V r=2 I=2A The output power is delivered to (and dissipated by) the resistor: Poutput = Presistor = I2 R = (2 A)2 (4 ) = 16W. Rate of energy conversion example.

34. Example: an electric heater draws 15.0 A on a 120 V line. How much power does it use and how much does it cost per 30 day month if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh. For simplicity assume the current flows steadily in one direction. What’s the meaning of this assumption about the current direction? The current in your household wiring doesn’t flow in one direction, but because we haven’t talked about current other than a steady flow of charge, we’ll make the assumption. Our calculation will be a reasonable approximation to reality. Electric heater example.

35. An electric heater draws 15.0 A on a 120 V line. How much power does it use. How much does it cost per 30 day month if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh. Electric heater example.

36. How much energy is a kilowatt hour (kWh)? So a kWh is a “funny” unit of energy. K (kilo) and h (hours) are lowercase, and W (James Watt) is uppercase. Electric heater example.

37. How much energy did the electric heater use? Electric heater example.

38. That’s a ton of joules! Good bargain for $17. That’s about 34,000,000 joules per dollar (or 0.0000029¢/joule). OK, “used” is not an SI unit, but I stuck it in there to help me understand. And joules don’t come by the ton. One last quibble. You know from energy conservation that you don’t “use up” energy. You just transform it from one form to another. Electric heater example.

39. Example: a typical lightning bolt can transfer 109 J of energy across a potential difference of perhaps 5x107 V during a time interval of 0.2 s. Estimate the total amount of charge transferred, the current and the average power over the 0.2 s. learn about lightning at howstuffworks Numbers obtained from an “old” text. Actual current is likely far more. What kind of a problem is this? You are given energy transferred, potential difference, time. You need to calculate charge transferred, current, and average power. Equations for current and power are “obvious:” Lightning bolt example.

40. We could calculate power right now, but let’s do this in the order requested. Besides, we can’t get current without Q, charge transferred. We need to think in terms of energy transformations rather than work done by forces. The equation above tells us that potential energy stored in clouds can be transferred to the ground (at a different potential) by moving charge from cloud to ground. We are given energy transferred and potential difference, so we can calculate q. “Could I think of the cloud-earth system as a giant capacitor which stores energy?” You could, except our capacitor equation U=QV/2 assumes the same charge on both plates; that’s untrue here. Lightning bolt example.

41. Qtransferred = 20 C Continuing with our energy transformation idea: Etransferred= Qtransferred Vif Qtransferred = Etransferred / Vif Qtransferred = 109 J / 5x107 V That’s a lot of charge (remember, typical charges are 10-6 C. Once we have the charge transferred, the current is easy. This is probably less than the actual current in a lightning bolt by a factor of anywhere from 10 to 1000. See this link. Lightning bolt example.

42. Average power is just the total energy transferred divided by the total time. The numbers in this calculation differ substantially than the numbers in a homework problem(not necessarily assigned this semester). “This” lightning bolt carries relatively low current for a long time through a high potential difference, and transports a lot of energy. In reality, there is no such thing as a universally-typical lightning bolt, so expect different results for different bolts. See here. Holy ****, Batman. That’s the power output of five enormous power plants! Lightning bolt example.

43. VH I VT R Example: the electric utility company supplies your house with electricity from the main power lines at 120 V. The wire from the pole to your house has a resistance of 0.03 . Suppose your house is drawing 110 A of current… (a) Find the voltage at the point where the power wire enters your house. VHT= IR VT-VH= IR VH= VT-IR VH= (120 V) – (110 A) (0.03) = 116.7 V Household voltage example.

44. VH I VT R (b) How much power is being dissipated in the wire from the pole to your house? Three different ways to solve; all will give the correct answer. P=IV = I2R = (V)2/R P=I(VT-VH) = I2R = (VT-VH)2/R P=(110 A) (120 V -116.7 V) = 363 W or P= (110 A)2 (0.03) = 363 W or P= (120 V – 116.7 V)2 / (0.03) = 363 W Household voltage example.

45. VH I VT R (c) How much power are you using inside your house? You need to understand that your household voltage represents the potential difference between the “incoming” and “outgoing” power lines, and the “outgoing” is at ground (0 V in this case)…except… …because the “outgoing” power line is at 0 V, you can “accidentally” get this correct if you simply multiply the current by the voltage at the point where the power wire enters your house. Household voltage example.

46. VH I VT R (c) How much power are you using inside your house? P=IV P=(110 A) (116.7 V – 0 V) P=12840 W You don’t want to use the P=I2R=V2/R equations because you don’t know the effective resistance of your house (although you could calculate it). P=(110 A) (120 V) – (110 A)(3.3 V) is also a reasonable way to work this part. Household voltage example.