Originally posted by Bad wolfI've got A-level exams in June as well. Maths:mechanics (I hate the m2 mechanics module!), Computing and Geography. I dropped Physics after the first year. 🙂
Well, AS, if you want to get specific. lol 🙂
Maths, that's split into pure and mechanics, physics, accounts and government&politics.
Too much work...
🙁
Maths and physics is a good combination. I'm currently nearly half way though a physics degree. Let me know if you get stuck on any other questions.
PS: even if you cant get the full answer in an exam its always worth staying calm and just working though what you can manage. More often then not you will stumble upon the right course of action whilst working out simpler steps; and even if you dont you can still score heavily without a final answer if you working is correct.
Originally posted by Ari Brenin CymruIndeed. That question was part of my novice ham license test. Simple EIR stuff.
The answer is don't take A-level physics. 😛🙂u
There is an analogous situation (oblique anyway) in AC circuits.
The Buck-boost transformer. It is an autowinding transformer that adds or subtracts a certain voltage, typically 10% to the line voltage, depending on how it is hooked up. The analogous part comes in figuring the power capability of the device. You can use a regular transformer or a giant variac to change the voltage of your house (I have an overvoltage problem in my house, 124 volts coming in, the electric utilty, PP&L here, won't do anything about it because they bribed the state government long time ago to make the legal limit 126 volts. Whats the big deal with that? Do a bit of math here, suppose we are dealing with a resistive element like a stove or electric room heater. Lets make it 100 volts for clarity, then if you draw 10 amps, you have a 1,000 watt heater. But lets bump the voltage to 110. Now you have 110 times not 10 amps as before, but 11 amps,
so now the total draw is 1210 watts. Get it? Increase the voltage 10% and the power drain goes up 21%. So lets do my situation, 110 volts and 11 amps =1210 watts, then I get 124 volts, now it draws 12.4 amps times 124 volts=1537.6 watts drawn, thats the drain they figure out the bill on which just accidently means that heater costs 27% more to run. Not much of a motivation to bring the voltage down, eh.
So there is a solution. Not a humungus transformer which would undoubtedly work and work well, its just that it would weigh in at about 200 pounds and cost something like $2000 bucks or more. So you use the buck-boost transformer. It has a sly little trick, somewhat like the internal resistance problem. It is a tiny thing compared to a 20 KW job you might need for a modern household and weighs about 5 pounds and costs about 150 bucks. The neat thing about it is the power drawn by the transformer itself is only the DIFFERANCE between the two voltages, in and out, times the current so it is passing through 90% of the energy without needing a big massive transformer and the actual energy handled by it is say 1000 watts vs 20000 watts or 10,000 watts you would need in a real transformer. So if the differance is 10 volts and the total current draw is 100 amps, only 1000 watts is actually being handled by the transformer and the rest is bypassed basically. Very advanced device and almost totally unknown. It fixes your voltage, one hook up to boost 10 volts or so and another hookup to reduce the voltage which is what I need in my case. If you do that, you can reduce your electric bill by 20% or more, giving the thing a very quick payback period. Everyone should measure their house voltage and see how much higher it is than 110 volts. If its 125 or so it burns out most electronics that much faster AND you dump wasted money right into the electric utilities pocket. Such a deal, eh.
In my case, those fantastic spiral flourescent bulbs that replace incadescents and are supposed to last, what, 10000 hours or more, shyte, they last about 3 months in my house. Once a Zenith TV caught FIRE! We had to replace a very expensive built in double oven because one day I turned the knob and it EXPLODED!. Electric company is legal however. So I am off to get that little buck boost job and stop them screwing me over. So much for the soap box, just wanted to bring up the idea of the situation where the voltage differance is what you use to calculate the power handled by the device.
Originally posted by sonhouseForgot about that, you're absolutely right.
You can ignore the resistor only for the time the switch is open. That represents the battery with no load, no current flow, so you cannot measure the resistance that way. You have to use the voltage differance between the off and on position.
Well... shows you what I know. 😳
Originally posted by Bad wolfHey again Bad Wolf 🙂 Here are some things that you might find helpful.
http://tinyurl.com/23hmc4
Basically I have to make some notes on this now, for Thursday, when I will write it up in test conditions.
Can you help me makes the notes on it though. 🙂
To vary the wavelength of light you could use a simple prism, which will refract light of different wavelengths by different amounts, but this would require relativly precise placement of the LDR. Perhaps a more practical set up would be to use a Fabry-Perot interferometer to control the wavelength of light. However, from my experience of AS levels, the exam questions are usual based very closely on the lessons so it may be the case that you have been taught about some other interferometer setup.
That leaves you with the problem of measuring the intensity using the LDR. Most likely you will be told how the resitance of the LDR varies with intensity. (e.g: N ohms increase or decrease for a set change in intensity if the relationship is linear), if not then you will need to calibrate it by using known intensities and measureing its resistance.
I would suggest that you use a Wheatstone Bridge set up, with the LDR as the unknown resistance. Balence the bridge for each wavelength to find the resistance of the LDR for each wavelength, then convert this into the intensity.
A Wheatsone bridge is a fairly commen set up for calculating an unknown resistance, if you haven't come across it before it is relativly simple to understand, however feel free to ask...
You may be able to get away with simply using a multimeter to measure the resistance of the LDR, depends on where the marks are given for the investigation. However, the Wheatstone bridge has the advantage that if the temperature of the resistors changes (due to prolonged current flow etc) then all resistances change by the same fraction and as you are using ratios the measurement it unaffected. This would mean you did not have to worry as much about maintaining a constant temperature.
To carry out a valid test you must 1st ensure that only a narrow range of wavelengths fall on the LDR at any one time; and that the resistance of the LDR is only effected by changes in the intensity of this light. The first point is sorted by considering the interferometer set up, how you are setting the angle into the interferometer or the plate spacing etc. The second point is adressed by ensuring that the background light does not interfer with your experiment. e.g ensure that the light intensity in the room you are in is constant thoughout the experiment, or (ideally) shield the LDR from all light apart from light from the lamp.
I hope this is helpful - it is not very clear in places so feel free to ask questions. Good luck.
Originally posted by MattPGood point about the Wheatstone bridge, temperature compensation.
Hey again Bad Wolf 🙂 Here are some things that you might find helpful.
To vary the wavelength of light you could use a simple prism, which will refract light of different wavelengths by different amounts, but this would require relativly precise placement of the LDR. Perhaps a more practical set up would be to use a Fabry-Perot interferometer to control t ...[text shortened]... ope this is helpful - it is not very clear in places so feel free to ask questions. Good luck.
One other thing to consider: the LDR, needs to be calibrated for wavelength, that is to say how does the output of this device vary given the same intensity at differant wavelengths. Also, what would be the bandpass of the device, does it give equal results from IR to UV? Does it peak at 550 Nm, (green light), etc. If it is not calibrated and a corresponding table of output v wavelength, the exercise has no quantitative value. Depending on the total bandwidth under consideration, it does not have to be say, +/- 0.1 db in response to all wavelengths, just have a table that relates what the output is vs wavelength, then a reasonable calibration curve can be assembled. These are also called photoresistors or photocells. Here is a link to a brief article about them:
http://optoelectronics.perkinelmer.com/catalog/Category.aspx?CategoryName=Photocells
Another method is called the bolometer, it turns a band of radiation, visible, UV, IR or microwave frequencies into a temperature measurement. Here is another link for a description of one such device:http://www.iop.org/EJ/abstract/0150-536X/27/1/004
- Joined
- 23 Jul 05
- Moves
- 8869
2 edits
Originally posted by MattPThis is just a plan I will be writing up, I won't actually be doing it, also, I would only be able discuss things that are in an AS physics lab.
Hey again Bad Wolf 🙂 Here are some things that you might find helpful.
To vary the wavelength of light you could use a simple prism, which will refract light of different wavelengths by different amounts, but this would require relativly precise placement of the LDR. Perhaps a more practical set up would be to use a Fabry-Perot interferometer to control t ope this is helpful - it is not very clear in places so feel free to ask questions. Good luck.
I discussed what a Fabry-Perot interferometer was with him, but he said that it was unneccessary, in that it would be superfluous at the moment to the experiment. Band-pass filter is only really needed, but he suggested mentioning the Fabry-Perot interferometeras an alternative, but little more.
About the Wheatsone bridge, I'd never heard of that before now, pretty simple when he explained it to me, thanks, I'll use this in the plan. I was able to get a reference about it in a book there, which is necessary so I'll get another mark.
I also need to get another reference, this can be off the internet.
Preferably this one about what you would expect to see on a graph of resistance against light intensity.
I'm only supposed to write about 500 words, so I can't go into mega-detail.
Originally posted by Bad wolfJust remember the photocell has to have a well defined response curve in relation to wavelength or the results of the theoretical study will be of qualitative use only, not quantitative. I assume you know the differance.
This is just a plan I will be writing up, I won't actually be doing it, also, I would only be able discuss things that are in an AS physics lab.
I discussed what a Fabry-Perot interferometer was with him, but he said that it was unneccessary, in that it would be superfluous at the moment to the experiment. Band-pass filter is only really needed, but he su ...[text shortened]... intensity.
I'm only supposed to write about 500 words, so I can't go into mega-detail.
- Joined
- 23 Jul 05
- Moves
- 8869
Originally posted by sonhouseA little confused, how would such curve be useful, how would it be applied?
Just remember the photocell has to have a well defined response curve in relation to wavelength or the results of the theoretical study will be of qualitative use only, not quantitative. I assume you know the differance.
I also believe the answer is 1 ohm for the following reasons.
If a true 1.5v voltage is placed across the resistor you would read 1.5 volts at .375 amps. However the battery cell has internal resistance and at this current can only supply 1.2V. Using the already mentioned formula E (1.2 Volts) Divided by R(4ohms) equals current .3 amps
Now we know that the voltage lost is 1.5 volts minus the 1.2 volts across the resistor. That equals .3 volts.
Using the same formula with the known values we get. E(.3volts) divided by I (.3amps) we get 1 ohm.
To check the answer we use series circuit rules. 1 ohm (cell) plus 4 ohm's (resistor) equal 5 ohms. 1.5 volts divided by 5 ohms equals .3 amps.