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Topic: How do you calculate what the voltage will be after a resistor? (Read 837 times) previous topic - next topic

Papa G

The fact that you can't measure it accurately doesn't change the fact that it's 12 volts. :)

DVDdoug

#11
Feb 05, 2013, 01:48 am Last Edit: Feb 05, 2013, 01:50 am by DVDdoug Reason: 1
Quote
So that would mean that if I have a 100mA load I'm going to get a totally different voltage than I would I I have a 4A load. That's not what I've experienced in practice. The voltage generally remains the same unless something with a massive amp draw is running.
There is something wrong with your measurement....   The ONLY way to get 40 times the current across the same resistance is with 40 times the voltage!   (With a 30-foot wire, you might not be able to measure that accurately.)   Ohm's Law describes the relationship between voltage, current, and resistance.   It's a law of nature* and it's always true.   i.e. If you double the current through a constant resistance, you double the voltage across the resistor.


The units of measure (Volts, Ohms, and Amps) are man-made, but the relationships are determined by God!  (Or nature if you like.)

retrolefty

#12
Feb 05, 2013, 02:42 am Last Edit: Feb 05, 2013, 02:53 am by retrolefty Reason: 1



Theoretically, the voltage at the green points is 12V. Practically - it can't be measured without disrupting the system because we have to use some kind of instrument.


For those still following this thread, there was (and still is) a way to measure the voltage of something without drawing any current from that something (and thereby 'effecting' it's value), and that is with a true Differential Null Voltmeter.

We had these in the Air Force and they were both pretty cool and a pain in the ass to use (time consuming to stepping down to match microvolts or lower differences, and time needed for all equipment to reach stable temperature from first powering on, etc).

  The principal used is it has it's own internal voltage generator that the user adjusts until the meter reads a  zero center difference ('a Null') between the external voltage being measured and the internal generated voltage. At that point no current is flowing to or from either the voltage being tested or from the meter's internal voltage generator, so meter impedance, lead wire resistance (assuming they are of the same length) are all non factors in the measurement results. It is essentially a measurement made requiring no load current being drawn on the voltage source being measured when the external and internal voltages are equal.

Typical look and feel of such an instrument:

http://www.us-instrument.com/objects/catalog/product/image/img3033.jpg

I am a firm believer in Ohm's law, and see no contradictions or mysteries in it's teachings, but sometime some do have misunderstanding and/or misapplication of it's principles.

Lefty

Papa G

We had the Fluke version of the differential voltmeter in my lab at TI in the 60s. I don't think they have changed much in theory since then.

retrolefty


We had the Fluke version of the differential voltmeter in my lab at TI in the 60s. I don't think they have changed much in theory since then.


No, most theories change very slowly and rarely if at all over time. Ohm's law is still hanging in there.  ;)

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