Recent content by mgibson

Thank you for your response. I am realizing that I probably will not be able to generate a high enough current to use the Lorentz force to accelerate the plasma. I may try and use permanent magnets instead to generate the strong b-field. I plan to use a power supply (similar to a tesla-coil)...

Thank you mgb_phys for your response. The power supply i am using is a 0 - 100kV tesla coil with a current around 1-2mA. I am using it for an application (mini arcjet thruster) were it will be arcing constantly across an inch gap or so, the arc current is used to induce a Lorentz force on the...

Hello, I am trying to find an estimated value (or range) for the current of an electric arc. I imagine this may be a function of the voltage producing the arc, the distance between the electrodes, and other parameters. If this is true, then take the voltage to be around 80kV and the...

Homework Statement The problem asks to find the energy conversion efficiency of a blower-nozzle combination that produces a velocity of 30m/s for air with a density of 1.1kg/mcubed. The nozzle has a diameter of .15m input power is 1 hp Homework Equations given the flow rate =...

Great advice. I agree a bench power supply is a great thing to have when experimenting with circuits. I tried to buy one at the beginning of the year but it was back ordered too long. So I canceled the order. Hopefully sometime this fall I will re-order it. Thanks

For the 3 3W LEDs: I am changing my source to 4 D cells creating 6V. Hopefully this will meet my current requirements and improve battery life. I am also changing my resistors to 3 ohms with up to a 10W power rating. Thanks for your help and patience. I have learned a lot in the process.

Okay, so I think I should re-think the source voltage. I didn't realize so much power was being wasted in heat. Is running the 24 low-power LEDs off of a 12V source (8 AAs) okay? Or is it wasting tons of energy, enough to make it worth re-wiring every resistor/LED pair? For the 3 3W LEDs...

The reason I am using 8 AA's in series is because another part of the light fixture involves 24 basic LED's which came with free resistors for a 12V source. Also, Radioshack had convenient battery holders for 8 AA's. It was either 8 or 4 and 8 seemed most appropriate since I wouldn't have to...

The purpose is for a light fixture I designed in such a way where it would be very inconvenient to redesign the circuit. So I am going to use either: 2 8ohm, 20W resistors in series, giving me 16ohms with up to 20W, which is more than double the 7.84 (.7x.7x16) watt requirement. or 3...

Thanks a lot. I think I have a good understanding of the mistake I made and how to fix it. My only problem now is going to be finding resistors for this at radioshack so I can finish this project on time. Thanks again for all your help :smile:

The three LED circuits are identical so I will describe one of them. The LED is in series with 2 resistors (I thought the 15 Ohm one drops the current while the .47, 5 Watt one should dissipate the heat). The voltage drop across the LED is 3.6-4.2V. When I calculated the resistor needed for...

Hello, I am working on a LED project that involves the use of three 3W LEDS. These LEDs are being driven at about 700mA (I believe). My source is 12V (8 AA's in Series) and I am using 16 Ohm resistors (one for each LED) capable of handling at least 5.4 watts ( ~15 Ohm resistor in series...

The circle is indeed r=2cos(theta) I believe its parameterizing the circle in polar coord. as theta goes from -pi/2 to pi/2. The region is indeed shaped like a moon, and I first approached it with two integrals (the area of circle of r=1 with the shared area of the two circles subtracted from...

I think this integral may work for the Area outside the circle, r=1 and inside the circle, r=2cos(theta) polar coordinates: The double integral (where 1 <= r <= 2cos(theta) and -pi/3 <= theta <= pi/3 ) of differential Area dA=rdrdtheta This seems to make sense when visualizing the double...

I think 2pi/3 would be the answer if the left side of the shape were bounded by straight lines as opposed to curved ones. Since this is an approximation to the area, I'll probably use it if I cannot figure something else out. Thanks so much for your help!