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rstofer:
--- Quote from: orolo on April 18, 2016, 02:48:07 pm ---
--- Quote from: rstofer on April 18, 2016, 02:33:20 pm ---So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ). The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V.
Does that seem about right?
--- End quote ---
:-+ Very good reasoning, save for a small error: 9.9/200 of a scale of 2 volts, is about 2*0.05 = 0.1V.
--- End quote ---
Yup! Times 2V...
A little early in the morning, I guess!
Measuring in the region of 100 mV seems practical with many digital meters. When I first saw the problem, the GOhm thing kind of put me off but the problem had to have a solution, it was just a matter of finding it.
Chai:
It would be nice to have an ongoing thread! I just started the lab + aoe text combo and more than a few times you come across lab book sections that say something along the lines of, "you probably already have guessed the answer", without any further explanation! I was pretty bummed about this until I made it to the first Worked Examples section.
And the text box has example problems with no answers! :-/O
Aeternam:
--- Quote from: Chai on April 18, 2016, 06:35:24 pm ---It would be nice to have an ongoing thread!
--- End quote ---
I'd be happy to participate.
Although you'll see more questions from my end than answers, that's for sure :D
rstofer:
--- Quote from: orolo on April 18, 2016, 02:48:07 pm ---
--- Quote from: rstofer on April 18, 2016, 02:33:20 pm ---So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ). The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V.
Does that seem about right?
--- End quote ---
:-+ Very good reasoning, save for a small error: 9.9/200 of a scale of 2 volts, is about 2*0.05 = 0.1V.
--- End quote ---
Of course, you get the same answer if you work out a series voltage divider. The 1 gigaohm resistor connects to the + battery and the top of the 10 megohm resistor (the meter). The bottom of the 10 megohm resistor connects to the - battery. We are interested in the voltage across the 10 megohm resistor. So: 10V / (1 gigaohm + 10 megohm) * 10 megohm = 0.0909V or just about 0.1V
In broad numbers, we would expect the 1 gigaohm resistor to drop 100 times as much as the 10 megohm resistor which would see approximately 0.01V (1/100) if we started with 1V but we start with 10V so 10 * (1/100) = 10 / 100 = 0.1V. Nearly...
Chai:
Was anyone able to find out more information about the mystery boxes used in 1L.5 page 30? The footnote says to see their website for details but I couldn't find something. I guess one would just be a resistor? and the other might contain a resistor in series with diode? :-//
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