PPRuNe Forums > PPRuNe Social > Jet Blast > Stoopid quiz questions PDA View Full Version : Stoopid quiz questions barit122nd May 2007, 14:54When in high school I once took a statewide science quiz with this question: 10 watts of power is drawn by a 100 ohm resistor. The resistance is then reduced to 50 ohms. If all other conditions remain constant, how much power is drawn by the 50 ohm resistor? :eek: Bus_Bar22nd May 2007, 14:5510 watts I believe c-bert22nd May 2007, 15:18P = I^2 R 10 = I^2 . 100 0.1 = I^2 I = root 0.1 therefore when R = 50 P = 0.1 x 50 P = 5 Watts But I'm sure I've done something wrong. :8 Spui1822nd May 2007, 15:37I guess about 1m 20. Right? UniFoxOs22nd May 2007, 16:01Depends what the other conditions are. Constant voltage supply would double power, constant current supply would halve it. Question meaningless. UFO kansasw22nd May 2007, 16:23E = I x R and P = I x I x R. E = volts, I = current in amperes, R = resistance in ohms, P = power in watts. In the first example, P = 10W and R = 100ohms, so P = I x I x R -----> 10watts = I x I x 100ohms. I x I = 10/100 = 0,1 I = square root of 0.1 = 0.316amperes. We needed to know this to determine system voltage by the next step. E = I x R -----> E = 0.316 x 100 = 31.6volts. Now we know the voltage drop across the 100ohm resistor. The structure of the question appears to require that this is the supply voltage, ie there are no other resistances in the circuit to confuse the issue. So now we can use E = I x R to determine current in the second case. E = I x R -----> 31.6 = I x 50. I = 31.6/50 = 0.63amperes in the second case, ie double the current in the first case, as is logical since the resistance is halved. Less resistance means more current will flow with the same applied voltage, or voltage drop across the resistor (a better way to say it). So P = I x I x R -----> P = 0.63amps x 0.63amps x 50ohms = 19.85watts which really means 20watts, the discrepancy is due to rounding error of 0.63 which is really 0.6333333333......... . I hope this makes sense. Also I hope it's right. Once the relationships in the two equations at the top are understood, it is quickly obvious that halving the resistance in a circuit, whose voltage is constant, doubles the power in watts that is dissipated. http://www.pprune.org/forums/images/infopop/icons/icon3.gif Lightbulb Why can't I get an icon to show? Tone22nd May 2007, 16:48Don't you just hate it when technical questions are asked by ejuts. Power is not 'drawn' by a resistor, it is more a function of the heat dissipated when current is drawn. In the circuit considered, if the resistance is reduced from 100 to 50 ohms then the other conditions cannot remain constant (not in this universe anyway) One would expect the current to increase from 0.316A to 0.632A (if the power supply is capable of delivering the increase) If it is then the resistor would dissipate 19.97W give or take a bead of sweat. Aaaaaaaaaaaaaaaargh!22nd May 2007, 16:52Watt was the question again? frostbite22nd May 2007, 16:58How do they pack all that smoke into resistors? Aaaaaaaaaaaaaaaargh!22nd May 2007, 17:04Resistors are a futile component RAC/OPS22nd May 2007, 17:20E = I x R and P = I x I x R. E = volts, I = current in amperes, R = resistance in ohms, P = power in watts. In the first example, P = 10W and R = 100ohms, so P = I x I x R -----> 10watts = I x I x 100ohms. I x I = 10/100 = 0,1 I = square root of 0.1 = 0.316amperes. We needed to know this to determine system voltage by the next step. E = I x R -----> E = 0.316 x 100 = 31.6volts. Now we know the voltage drop across the 100ohm resistor. The structure of the question appears to require that this is the supply voltage, ie there are no other resistances in the circuit to confuse the issue. So now we can use E = I x R to determine current in the second case. E = I x R -----> 31.6 = I x 50. I = 31.6/50 = 0.63amperes in the second case, ie double the current in the first case, as is logical since the resistance is halved. Less resistance means more current will flow with the same applied voltage, or voltage drop across the resistor (a better way to say it). So P = I x I x R -----> P = 0.63amps x 0.63amps x 50ohms = 19.85watts which really means 20watts, the discrepancy is due to rounding error of 0.63 which is really 0.6333333333......... . I hope this makes sense. Also I hope it's right. Once the relationships in the two equations at the top are understood, it is quickly obvious that halving the resistance in a circuit, whose voltage is constant, doubles the power in watts that is dissipated Kansasw, did you set the Airservices numeracy test? kansasw22nd May 2007, 18:38Kansasw, did you set the Airservices numeracy test? Nope, I'm a yankee, so suspect that wouldn't be applicable. It took me many years to grok Ohm's Law, and now that I think I do, I'm always pleased with the opportunity to apply it. Now I'm working on Kirchhoff's Law. barit122nd May 2007, 19:00UFO has it: Depens what the other conditions are. Constant voltage supply would double power, constant current supply would halve it. Question meaningless. That's exactly it; I'm 99% sure the question was meant to be "voltage remains constant", but the quiz writer failed to consider the "current remains constant" scenario. You can have one but not both. :rolleyes: FWIW, I judged that the former case was intended, and answered that power would double. But it turned a technical question into a psychology test. :} Farmer 123rd May 2007, 09:49I seem to remember Watts = Volts x Amps. It could be Cole's Law. Kalium Chloride23rd May 2007, 10:46Constant current means ratio of power to resistance stays the same, so the answer is 5W. Constant voltage means the current doubles, so the answer is 20W. I think. :uhoh: Spui1823rd May 2007, 11:20It could be Cole's Law. :confused: :confused: http://www.sofeminine.co.uk/cuisine/boitearecettes/photorecettes/fiche/fiche292.jpg :rolleyes: XXTSGR23rd May 2007, 11:33W = V x I V = I x R So I = V/R and W = V^2/R R x W = V^2 So if you halve the resistance and the voltage remains constant, the Power doubles = 20W barit123rd May 2007, 11:36I don't recall the name "Cole's Law", but KC is correct. The point is that the resistance change implies that the ratio of voltage to current has changed. Re-entry23rd May 2007, 11:40Ohmy. A joule of a question. I amperefectly sure the current answers have it wired. More power to them. R4+Z23rd May 2007, 11:42Kansasw First or second law? XXTSGR23rd May 2007, 11:44I still go for constant voltage. Let's face it, you cannot directly change the current running through a circuit. You can only affect current by changing either the resistance of the circuit or the voltage applied to it. Whilst changing resistance (using, for example, a variable reistor), the only way in which you can keep the current constant is to change voltage. This implies to me that you are altering two variables rather than only one, as the question appears to imply. ZH87523rd May 2007, 18:42How do they pack all that smoke into resistors? Smoke, is the thing which powers all electric circuits, let all the smoke out, and they soon stop working. :p barit123rd May 2007, 22:12XXSTGR has the "common sense" approach, which I'm sure was the quiz's intent. After all, the wall receptacle supplies a constant voltage, as does the accumulator (battery to the colonials) in an auto or airplane. But sparks labs often have a constant-current source at the ready - and the one thing one must never do is to leave it open-circuited, because it will try to pump a constant current through infinite resistance (you do the math!) If the test writer weren't aware of this type of source, then the question posed in post #1 is the likely result. :rolleyes: BlueDiamond24th May 2007, 10:49Ohmy. A joule of a question. I amperefectly sure the current answers have it wired. More power to them.That's a shocking thing to say, Re-entry. Watt were you thinking?? :rolleyes: Here's an easy way to work it out using a "formula wheel." As long as two of the variables are known, you can work out the answer. http://img.photobucket.com/albums/v511/BlueDiamond01/Electronics.jpg If you use that formula, you come up with the answer of 20W ... which is what a few others here have concluded. Binoculars24th May 2007, 15:04Well, I'm so stoopid I don't even know whether the original question was stoopid or not. :uhoh: