Recommeded "workflow": > Alt-Print Scrn (screen capture of active window) [Tech Tip: "Print Scrn" captures entire screen] > Open MS Word template (circuit-template.doc) > View/Header & Footer [edit title of document (as necessary), name(s), period] > File/Save As... (e.g., one-bulb-series-circuit.doc) > Save CCK workspace == Notes re: Circuit Construction Kit* (DC Only) (Ver. 3.15, JAN/FEB 2009) -- Lessons Learned from Day 1 of "Go With The Flow" (GWTF) *a.k.a. CCK TECH TIPS: [Save] button: Use filenames with .cck extension, e.g., one-bulb-series-circuit.cck "Tools" aren't saved. [Load] button: By default [Files of type: (.cck)], only files with .cck filename extension appear in list (switch to "All Files") Problem: CCK doesn't feature an option for "File/Print." Solution: Print Screen or Alt-Print Screen (copies active window only) plus "circuit-template.doc" http://www.wsanford.com/~wsanford/gr8ps/02_purple/04_go_with_the_flow/ 00_resources.html TROUBLESHOOTING: If an object is moved accidentally off the "workspace," then change the "Size" of the workspace to "Small" and drag objects toward center of workspace. If circuit is complete but light bulbs don't glow, then you may need to press the ">" (Play) button located at the bottom of the window. CONCEPTUAL REPRESENTATIONS (ABSTRACTIONS): <-- annotate title of lab > voltage (symbol = V) = battery > current, a.k.a. amperage (symbol = I) = value(s) shown by non-contact ammeter > resistance, a.k.a. the "load" (symbol = R) = light bulb [Note: By default, light bulbs and resistors have the same resistance (10 Ohms); when a "Schematic" "Visual" is selected, light bulbs aren't displayed using the symbol for resistors.] SERIES CIRCUITS (points of focus) voltage varies; amperage constant Be sure to measure V and A: across the battery; and across the resistance (load). For example: >> three-bulb series circuit - amperage measures 0.30 Amps everywhere along circuit - measuring voltage across the battery shows 9.00 V; measuring voltage across part of the resistance/load (ONE light bulb) shows 3.00 V >> two-bulb series circuit - amperage measures 0.45 Amps everywhere along circuit - measuring voltage across the battery shows 9.00 V; measuring voltage across part of the resistance/load (ONE light bulb) shows 4.50 V >> one-bulb series circuit - amperage measures 0.90 Amps everywhere along circuit - measuring voltage across the battery shows 9.00 V; measuring voltage across part of the resistance/load (ONE light bulb) shows 9.00 V 1-bulb: amperage is 1/10 of voltage* 2-bulb: amperage is 1/20 of voltage* 3-bulb: amperage is 1/30 of voltage* PARALLEL CIRCUITS (points of focus) voltage constant; amperage varies Be sure to measure V and A: across the battery; and across the resistance (load). For example: >> three-bulb parallel circuit - voltage measures 9.00 V everywhere along circuit - measuring current/amperage across part of the resistance/load (ONE light bulb) shows 0.90 Amps >> two-bulb parallel circuit - voltage measures 9.00 V everywhere along circuit - measuring current/amperage across part of the resistance/load (ONE light bulb) shows 0.90 Amps; measuring current/amperage across entire circuit shows 2.70 Amps 2-bulb: outside battery, amperage is 1/5 voltage; across "load" amperage is 1/10 of voltage* 3-bulb: outside battery, amperage is ~1/3 voltage; across "load" amperage is 1/10 of voltage* -- *Activity Extension/Enrichment: Use Ohm's Law Magic Triangle to discover the mathematical interrelationships among V, I, and R across the resistance (load). For example, at a constant voltage, current and resistance are ? related. http://www.sengpielaudio.com/calculator-ohmslaw.htm == Use permanent markers (Sharpie) to label positive (+/red) and negative (-/black) terminals on volt- and amp meters. On page 26 [p. 28?, latest edition], PSLG, sketch & label the following VA SOL graphics [mark (+) & (-) terminals]: Series Circuit - 0259.gif Parallel Circuit - 2329.gif current (I, amps) = voltage (V, volts) / resistance (r, Ohms) V I = --- r Note: current & voltage are directly related (proportional); current & resistance are indirectly related (proportional) or... V = Ir -- Misc. Notes, discussion w. PSW: - voltage influences current (see mathematical relationship, shown above) - Notes re: water analogy: assume pipes are filled w. water (no air inside pipes); can't compress water (or electricity) <-- prerequisite understandings - SERIES CIRCUIT. Observation: If you add resistance (e.g., more light bulbs), all bulbs will shine equally brightly, but all of them will be dimmer. [from TRG] (a) voltage and current: as voltage is added, current increases (b) as resistance increases, current decreases Mathematical Explanation: Current is inversely proportional to resistance: Assuming voltage is constant, then increasing resistance (more operating light bulbs in circuit), then amperage must be lower. Analogy: Imagine fat pipes (1 ft. diam.) connected to 6" segments of garden hose: flow out = flow in; restrictions (resistance) cause water to take longer to flow thru pipe/hose, but the flow rate is the same throughout the entire system. - PARALLEL CIRCUIT. Observation: Bulbs burn equally brightly, despite more amperage. [from TRG] (a) voltage and current: as voltage increases, current increases (b) as resistance increases, current increases <-- MAKES NO SENSE!!! Mathematical Explanation: ?is the following statement true? Current is inversely proportional to resistance: Assuming voltage is constant, then decreasing resistance (fewer operating light bulbs in circuit), then amperage must be higher. Analogy: [need to write analogy extension] So, why don't bulbs burn brighter? Devices (like light bulbs) are rated for a given voltage & amperage; they may use less than rated power, but not more. [power (Watts) = voltage x amperage] == Comparing a DC Circuit to the Flow of Water http://faraday.physics.utoronto.ca/IYearLab/Intros/DCI/Flash/WaterAnalogy.html (x) >> snag related resources: http://faraday.physics.utoronto.ca/IYearLab/Intros/DCI/Flash/ [the following resource automatically resizes itself to fill window:] http://faraday.physics.utoronto.ca/IYearLab/Intros/DCI/Flash/WaterAnalogy.swf == http://en.wikipedia.org/wiki/Ohm's_law http://www.sengpielaudio.com/calculator-ohm.htm http://www.sengpielaudio.com/calculator-ohmslaw.htm <-- magic triangle e.g., Type GE 131 light bulb 1.3V 0.3A = ~4.3 ohms ***verified by "Circuit Construction Kit"*** GE 222 (Flash)Light Bulbs 2.25V .25A/TL-3 Mini Screw Base Light Bulb Manufactured by: Eiko SKU 40492 Brand Eiko Volts 2.25 Base Miniature Screw Bulb TL-3 Filament C-2R Average Life (hours) 5 Boxed & Bulk Miniature Lamps MOD (Diameter) 0.38in/9.6mm -- Google: define: resistance Electrical resistance is a measure of the degree to which a body opposes the passage of an electric current. The SI unit of electrical resistance is the ohm. Its reciprocal quantity is electrical conductance measured in siemens. en.wikipedia.org/wiki/Resistance_(electricity) Google: define: resistivity Electrical resistivity (also known as specific electrical resistance) is a measure indicating how strongly a material opposes the flow of electric current. A low resistivity indicates a material that readily allows the movement of electrons. The SI unit for electrical resistivity is the ohm metre. en.wikipedia.org/wiki/Resistivity == (New) Mobile PC Lab - Cart No. 7 #106 (upper-left) not working (Old) Mobile PC Lab - Cart No. ? Java installed & tested [left side] #53 #54 #55 #56 #57 #58 #59 XP problems (solved?) #60 SLOW!!! boots with power brick [right side] #68 #67 #66 #65 SLOW!!! #64 SLOW!!! #63 SLOW!!! boots with power brick #62 SLOW!!! boots with power brick #61 SLOW!!! == misc. Notes re: volts, amperes, and ohm - discussions w. psw 05 APR 2006 The ammeter shows the amount of current passing that particular point. If you put an ammeter in series with the power supply (battery), you'll be measuring the amount of current flowing through the entire circuit. But if, say, you put an ammeter in series with a single resistive load (bulb), you'll be measuring the current draw of THAT BULB ONLY. Think about water meters in a system where you have a large pipe that then flows into three garden hoses. If you meter the water going into the system, you'll be measuring everything. If you put your meter on just one of the three parallel hoses, you'll record only the water that flowed through that particular path. Phil Walter Sanford wrote: > One last point of confusion: If I put multiple ammeters at various points > along the parallel circuit, it shows different readings! Now I'm REALLY > CONFUSED!!! > W -- 05 APR 2006 Walter, You added resistive loads in parallel, which REDUCED the overall resistance of the circuit, not increased. Go back to the garden hose example: If you have constant pressure and add additional exit hoses in parallel, you won't move any more water through a given hose per unit of time--but because there are more hoses, water will flow more easily through the whole system. Try putting an ammeter in series with a single bulb in your parallel circuit and you'll see that the per-bulb current doesn't change. I think that the problem here is once again terminology: It sounds like the lab is using the term "resistance" to refer both to overall resistance of the circuit AND resistive loads (the light bulbs). Phil -- 05 APR 2006 Walter, Statement "B" is incorrect. Adding resistance will REDUCE the current that flows at a given voltage. But keep in mind when you put together resistive loads in parallel, you're actually reducing the overall resistance of the system by providing more paths for the current to follow. The resistance of any one bulb doesn't change (and therefore nor does the current through that bulb), but the larger number of paths causes the aggregate resistance of the whole circuit to drop and therefore current will increase. Phil Walter Sanford wrote: > Phil, > > O.K., I get series circuits, which are easily explained by the > mathematical relations among the variables in the formula, I = V/r. It's > parallel circuits I don't get! > > The following statements appear in the Tchr. Resource Guide: > > (a) voltage and current: as voltage increases, current increases > (b) as resistance increases, current increases <-- MAKES NO SENSE!!! > > Statement "A" makes sense, assuming I and V are directly proportional. > > Statement "B" makes no sense to me! If voltage is constant, then adding > resistance should decrease current, right? After all, I & r are > inversely proportional, right? Why should a parallel circuit defy the > fundamental mathematical relationships among the three variables? -- 04 APR 2006 Q. Second, I'm SO lost re: volts, amps, & Ohms! I know the water analogy. Problem is, I can't apply it to explain everyday phenomena, like why light bulbs in series all burn equally brightly (e.g., an older string of x-mas lights). Each additional bulb adds resistance, right? So how could the current be the same for the next light in the series? A. From the water analogy, you know that voltage is like pressure and that current is like flow rate. Resistance is like the size of the pipe. So in a system of wires (low resistance == big pipe) and lamps (high resistance == small pipe), the flow of current through the system is going to be limited by the resistance of that system. Since there are no tanks or other storage devices (in electrical terms, capacitance or inductance) in the system, you know that the amount of water entering the system is going to exactly equal the amount of water exiting at the other end, and that the amount of water flowing past any given point during a given unit of time in the system (i.e., the rate of flow) will exactly equal the same measurement taken at another point.