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http://www.lab-initio.com/ (nz345.jpg). my office!. Announcements.  Special Homework #4 is due tomorrow. You can download it here, if necessary: http://campus.mst.edu/physics/courses/24/Handouts/special_homework.html.

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  1. http://www.lab-initio.com/ (nz345.jpg) my office!

  2. Announcements  Special Homework #4 is due tomorrow. You can download it here, if necessary: http://campus.mst.edu/physics/courses/24/Handouts/special_homework.html Today is our last day on circuits. By the end of today, you should have mastered the elementary aspects of dc circuits, including debugging and fixing broken ones.

  3. Announcements • Exam 2 is two weeks from yesterday. Contact me by the end of next Wednesdays lecture if you have special circumstances different than for exam 1. • A note on homework…

  4. Today’s agenda: Measuring Instruments: ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations. Leftovers. Optional (not for test) material, if time permits.

  5. r Measuring Instruments: Ammeter You know how to calculate the current in this circuit: R If you don’t know V or R, you can measure I with an ammeter. V Any ammeter has a resistance r. Any ammeter has a resistance r. The current you measure is To minimize error the ammeter resistance r should very small.

  6. Example: an ammeter of resistance 10 m is used to measure the current through a 10  resistor in series with a 3 V battery that has an internal resistance of 0.5 . What is the percent error caused by the nonzero resistance of the ammeter? R=10  Actual current: r=0.5  V=3 V You might see the symbol  used instead of V.

  7. RA Current with ammeter: R=10  r=0.5  V=3 V Not bad in a Physics 24 lab!

  8. A Galvanometer When a current is passed through a coil connected to a needle, the coil experiences a torque and deflects. See the link below for more details. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1

  9. An ammeter (and a voltmeter—coming soon) is based on a galvanometer. OK, everything is electronic these days, but the principles here still apply. We’ll learn about galvanometers later. For now, all you need to know is that the deflection of the galvanometer needle is proportional to the current in the coil (red). A typical galvanometer has a resistance of a few tens of ohms. Hold it right there. Didn’t you say an ammeter must have a very small resistance. Is there a physics mistake in there somewhere?

  10. RG IG RSHUNT I G A ISHUNT A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel:  I Everything inside the blue box is the ammeter. The resistance of the ammeter is

  11. RG IG RSHUNT I G ISHUNT A B Homework hint: “the galvanometer reads 1A full scale” means a current of IG=1A produces a full-scale deflection of the galvanometer needle. The needle deflection is proportional to the current IG. If you want the ammeter shown to read 5A full scale, then the selected RSHUNT must result in IG=1A when I=5A. In that case, what are ISHUNT and VAB (=VSHUNT)?

  12. RG IG RS I G IS A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel: Example: what shunt resistance is required for an ammeter to have a resistance of 10 m, if the galvanometer resistance is 60 ? (actually 0.010002 ) The shunt resistance is chosen so that IG does not exceed the maximum current for the galvanometer and so that the effective resistance of the ammeter is very small.

  13. To achieve such a small resistance, the shunt is probably a large-diameter wire or solid piece of metal. Web links: ammeter design, ammeter impact on circuit, clamp-on ammeter (based on principles we will soon be studying).

  14. G Measuring Instruments: Voltmeter You can measure a voltage by placing a galvanometer in parallel with the circuit component across which you wish to measure the potential difference. RG R=10  a b Vab=? r=0.5  V=3 V

  15. Example: an galvanometer of resistance 60  is used to measure the voltage drop across a 10 k resistor in series with a 6 V battery and a 5 k resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the galvanometer? R1=10 k First calculate the actual voltage drop. a b R2=5 k V=6 V

  16. G The measurement is made with the galvanometer. 60  and 10 k resistors in parallel are equivalent to an 59.6  resistor. The total equivalent resistance is 5059.6 , so 1.19x10-3 A of current flows from the battery. RG=60  R1=10 k a b The voltage drop from a to b is then measured to be 6-(1.19x10-3)(5000)=0.07 V. R2=5 k The percent error is. V=6 V I=1.19 mA Your opinions? Would you pay for this voltmeter?

  17. G V To reduce the percent error, the device being used as a voltmeter must have a very large resistance, so a voltmeter can be made from galvanometer in series with a large resistance. RG RSer  a a b b Vab Vab Everything inside the blue box is the voltmeter. Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A when Vab=10V.

  18. Example: a voltmeter of resistance 100 k is used to measure the voltage drop across a 10 k resistor in series with a 6 V battery and a 5 k resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the voltmeter? R1=10 k a b We already calculated the actual voltage drop (3 slides back). R2=5 k V=6 V

  19. G The measurement is now made with the voltmeter. 100 k and 10 k resistors in parallel are equivalent to an 9090  resistor. The total equivalent resistance is 14090 , so 4.26x10-3 A of current flows from the battery. RG=100 k R1=10 k a b The voltage drop from a to b is then measured to be 6-(4.26x10-3)(5000)=3.9 V. R2=5 k The percent error is. V=6 V I=4.26 mA Not great, but much better. Larger Rser is needed for high accuracy.

  20. G  Measuring Instruments: Ohmmeter An ohmmeter measures resistance. An ohmmeter is made from a galvanometer, a series resistance, and a battery. RG RSer Everything inside the blue box is the ohmmeter. The ohmmeter is connected in parallel with the unknown resistance with external power off. The ohmmeter battery causes current to flow, and Ohm’s law is used to determine the unknown resistance. R=?

  21. V A To measure a really small resistance, an ohmmeter won’t work. Solution: four-point probe. Measure current and voltage separately, apply Ohm’s law. reference: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movcoil.html#c4

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

  23. Today’s agenda: Measuring Instruments: ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations. Leftovers. Optional (not for test) material, if time permits.

  24. Q t Q t RC Circuits RC circuits contain both a resistor R and a capacitor C (duh). Until now we have assumed that charge is instantly placed on a capacitor by an emf. The approximation resulting from this assumption is reasonable, provided the resistance between the emf and the capacitor being charged/discharged is small. If the resistance between the emf and the capacitor is finite, then the charge on the capacitor does not change instantaneously.

  25. Charging a Capacitor I Switch open, no current flows. C Close switch, current flows.  R Apply Kirchoff’s loop rule* (green loop) at the instant charge on C is q. switch t<0 t>0 This equation is deceptively complex because I depends on q and both depend on time. *Convention for capacitors is “like” batteries: negative if going across from + to -.

  26. Limiting Cases I C  When t=0, q=0 and I0=/R. R When t is “large,” the capacitor is fully charged, the current “shuts off,” and Q=C. switch

  27. Math:

  28. More math:

  29. Still more math: Why not just solve this for q and I? RC is the “time constant” of the circuit; it tells us “how fast” the capacitor charges and discharges.

  30. recall that this is I0, also called Imax Charging a capacitor; summary: Sample plots with =10 V, R=200 , and C=1000 F. RC=0.2 s

  31. In a time t=RC, the capacitor charges to Q(1-e-1) or 63% of its capacity… …and the current drops to Imax(e-1) or 37% of its maximum. RC=0.2 s =RC is called the time constant of the RC circuit

  32. Discharging a Capacitor I Capacitor charged, switch open, no current flows. +Q +q C -Q -q Close switch, current flows. Apply Kirchoff’s loop rule* (green loop) at the instant charge on C is q. R switch t<0 t>0 *Convention for capacitors is “like” batteries: positive if going across from - to +.

  33. Math: negative because charge decreases

  34. More math: same equation as for charging

  35. Disharging a capacitor; summary: Sample plots with =10 V, R=200 , and C=1000 F. RC=0.2 s

  36. In a time t=RC, the capacitor discharges to Qe-1 or 37% of its capacity… …and the current drops to Imax(e-1) or 37% of its maximum. RC=0.2 s

  37. Notes This is for charging a capacitor. /R = I0 = Imax is the initial current, and depends on the charging emf and the resistor. This is for discharging a capacitor. I0 = Q/RC, and depends on how much charge Q the capacitor started with. I0 for charging is equal to I0 for discharging only if the discharging capacitor was fully charged.

  38. Homework Hints This is always true for a capacitor. Qfinal = CV, where V is the potential difference of the charging emf. Q0 is the charge on the capacitor at the start of discharge. Q0 = C only if you let the capacitor charge for a “long time.” Ohm’s law applies to resistors, not capacitors. Sometimes you can get away with using this. Better to take dq/dt.

  39. Example: For the circuit shown C = 8 μF and ΔV = 30 V. Initially the capacitor is uncharged. The switch S is then closed and the capacitor begins to charge. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. (From a prior test.) Also determine the current through the capacitor and voltage across the capacitor terminals at that time. To be worked at the blackboard (if time permits).

  40. Demo Charging and discharging a capacitor.

  41. make your own capacitor circuits http://phet.colorado.edu/en/simulation/circuit-construction-kit-ac For a “pre-built” RC circuit that lets you both charge and discharge (through separate switches), download this file, put it in your “my documents” folder, run the circuit construction applet (link above), maximize it, then select “load” in the upper right. Click on the “capacitor_circuit” file and give the program permission to run it. You can put voltmeters and ammeters in your circuit. You can change values or R, C, and V. Also, click on the “current chart” button for a plot of current (you can have more than one in your applet) or the “voltage chart” button for a plot of voltage.

  42. more applets http://webphysics.davidson.edu/physlet_resources/bu_semester2/c11_RC.html http://subaru.univ-lemans.fr/AccesLibre/UM/Pedago/physique/02/electri/condo2.html http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31.0

  43. Today’s agenda: Measuring Instruments: ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations. Leftovers. Optional (not for test) material, if time permits.

  44. Real quickly… some leftover circuits material…if time permits… EMF’s in Series and in Parallel: Charging a Battery If you put batteries in series the “right way,” their voltages add: + = 6 V 3 V 9 V If you put batteries in series the “wrong way,” their voltages add algebraically: + = magnitudes only  6 V 3 V 3 V  algebraically, using chosen loop direction -6 V +3 V -3 V chosen loop direction

  45. Why would you want to connect emf’s in series? More voltage! Brighter flashlights, etc. Chemical reactions in batteries yield a fixed voltage. Without changing the chemical reaction (i.e., inventing a new battery type), the only way to change voltage is to connect batteries in series. Why would you want to connect batteries in series the “wrong” way? You could connect a source of emf – like the alternator in your car – so that it charges a battery. Rechargeable batteries use an ac to dc converter as a source of emf for recharging.

  46. Go to www.howstuffworks.com to see how batteries work. They even expose the secret of the 9 volt battery! Click on the picture above only if you are mature enough to handle this graphic exposé.

  47. Go to www.howstuffworks.com to see how batteries work. They even expose the secret of the 9 volt battery! Shocking! Six 1.5 V batteries in series!

  48. Could you connect batteries (or sources of emf) in parallel? Sure! a b 3 V 3 V You would still have a 3 V voltage drop across your resistor, but the two batteries in parallel would “last” longer than a single battery. You could use Kirchhoff’s rules to analyze this circuit and show that Vab = 3 V.

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