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| | Problem requiring Ohm's Law | |
| | Author | Message |
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Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law August 8th 2013, 2:21 pm | |
| You have to have a whale of an appetite for that! Getting back on topic: OK I give up on you guys, My gosh, I thought we would have had some takers. -------> You use ss-diodes. And you can use an AC supply, now remember I only said the requirements was only two single conductor wires between boxes. I mentioned nothing about what could go into either box. So you could use internal batteries, power cords, transformers, switches. If you used AC or DC you would need to connect the diodes to the lamps (this was before the advent of LEDs) and switch the positive or negative pulses to light the lamps. and of course send AC to light both lamps at the same time. Like I said, simple. And if you used a battery you would use a double pole switch to reverse polarity. And maybe a vibrating relay to switch polarity so fast that both lights would appear to be lit at the same time. There are other ways too. _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | FrankB Moderator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law August 7th 2013, 10:36 pm | |
| Cliff- How do you make an elephant float? You start with one elephant and 50,000 gallons of root beer...... |
| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 22nd 2013, 4:15 pm | |
| Well it seems like with little activity I mostly just am looking for appropriate postings vs spam and it takes time to go over everything in detail, of course your right, but I am trying to keep things going and hoping I will get responses. I also monitor TRF as an Admin. and enjoy doing so, but I don't get into most of the non-electronic oriented discussions. -------------------- The two boxes is a very simple design, nothing complex, just a knowledge of electronic theory and common application. I am surprised no one is taking on this challenge. I may have to drop a hint if someone cries uncle. _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | Guest Guest
| Subject: Re: Problem requiring Ohm's Law July 22nd 2013, 2:58 pm | |
| "I skim some times without reading entire context, just to get the gist of things. . . ."
A dangerous habit--and self-contradictory! Without reading the entire context, you cannot get the gist of things.
I assume your math is correct and I grade you A+. A beginner might think to get a voltage answer, he multiplies 0.8 (I) by 1,000,000 (R), then wonders why he gets an impossible answer, 800,000 volts, gets discouraged, doesn't know what he did wrong, and, at worst, drops out of the hobby.
I see no takers, including me, so far for your two boxes problem. |
| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 17th 2013, 12:21 pm | |
| NO just didn't see originated my home town detroit No games intended. I skim some times without reading entire context, just to get the gist of things then get sidetracked.
My original question still remains though. Is the math correct?
_________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | Guest Guest
| Subject: Re: Problem requiring Ohm's Law July 17th 2013, 11:41 am | |
| You said: "Never heard of it but it started in Detroit actually."
I said originally: ". . .Boston Coolers originated in my hometown, Detroit, Mich."
Some kind of verbal game here?
Vernor's Ginger Ale is widely available throughout Michigan and much of elsewhere. It was invented by James Vernor, a prominent Detroit pharmacist of the late 19th century, using chemicals available in his pharmacy. The building still stands (believe it or not) in downtown Detroit--unless it has been destroyed since I quit going to that rotten sewer of a city many years ago).
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| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 3:35 pm | |
| You weren't using me to do your homework now, were you?????? _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 3:33 pm | |
| Never heard of it but it started in Detroit actually. It of course tastes best with Vernors Ginger ale and 3 scoops of vanilla (Tillamook Vanilla Bean Ice-cream being the best of course). Very Hard to find Vernors though! But I imagine Ginger Beer (soda) will run a close second. yum _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | Guest Guest
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 2:56 pm | |
| Root beer floats and Boston Coolers originated in my hometown, Detroit, Mich. Betcha don't know what a Boston Cooler is! |
| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 1:47 pm | |
| is the math Correct? I'm a teetotaler, hence soda, oh my! I feel a root beer float coming in the next two hours. Ever try ice cream in beer? Going to get around 87 here, close to the water Puget Sound, so temperature here moderates and makes a big difference. _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | Guest Guest
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 1:32 pm | |
| Re your physical exhaustion remedy, where do you get the ROOT beer stuff? It's 96 here in central Michigan today, and there is only one kind of beer--the real stuff--that can deal with that heat! |
| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 1:22 pm | |
| looks like its puzzle time. design a circuit that has two separate boxes, the requirements are that between the two boxes you can only have 2 single wire conductors. in one box you shall have 2 light bulbs and be able to light either bulb independently and also be able to light both at the same time. in that box you can make any circuit configuration you wish, in the other box you can use any type of power you wish and circuitry to assist in this requirement, I learned this is the seventh grade and its a very simple circuit, but you can go overboard if you wish as long as you meet the design intent of operation. _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 1:08 pm | |
| Wasn't finished editing, well what's your grade for me teacher????? My gosh that was so taxing I am going to sip some more Root beer. _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
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| | | Guest Guest
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 12:50 pm | |
| Please show your work, as the schoolteachers used to say. (If you multiply 0.8 times 1,000,000., you get an incredible number.) |
| | | Cliff Jones Site Administrator
Join date : 2010-11-22
| Subject: Re: Problem requiring Ohm's Law July 15th 2013, 12:40 pm | |
| .078v No matter what the current draw of a device, it can only drop voltage that is limit by the total current that is available. Since 1 megohm limits the current then you have to use that current for calculations.
250v/1,000,000ohms= .00025a
250v/0.8a = 312ohms
312ohms + 1,000,000 = 1,000,312 ohms
250v/1,000,312 ohms = .00025a
.00025a * 312ohms= .078v
I am assuming this is a series circuit.
250v minus .078v = 249.922 so essentially 250V _________________ I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!
Last edited by Cliff Jones on July 15th 2013, 1:07 pm; edited 1 time in total |
| | | Guest Guest
| Subject: Problem requiring Ohm's Law July 14th 2013, 2:55 pm | |
| In a circuit, at point A the voltage is 250 VDC. What will be the voltage at point B connected to A by a resistor of 1,000,000 ohms and a device that draws a current of 0.8 A.? (Refresher: E = I x R) |
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