| Electronics > Beginners |
| Some noob questions |
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| drussell:
--- Quote from: rstofer on July 21, 2018, 06:54:45 pm ---I don't know why I can't link the thread but the title is correct and it's just a few items down this page. --- End quote --- You just missed the last two characters )/ before the /url tag. https://www.eevblog.com/forum/beginners/beginners-that-are-interested-in-learning-electronics-from-the-ground-up-)/ |
| rstofer:
Yup! That works! |
| ArthurDent:
“If you wish to converse with me, define your terms.” ― Voltaire Mr D -“So actually when we say we measure voltage, what we actually mean is we measure current across a known resistance, and infer the voltage.” That isn’t the way to describe it at all. If you have a 9 volt battery and you put a 10 ohm, or a 10,000 ohm, or a 10 megohm resistor in series with the positive terminal, the voltage at the unconnected end of whichever resistor you choose to negative is still exactly 9 volts. This is because there is no current flow and no voltage drop until you connect a load between the unconnected end of the resistor and the negative terminal. Until you have a completed circuit this voltage is only potential energy and is doing absolutely no work. You need a complete circuit or there is no current flow. The ‘work’ could be lighting a lamp, running a motor, or deflecting a meter needle. You don’t “…measure current across a known resistance, and infer the voltage”, you measure the VOLTAGE drop ACROSS a known resistance and convert that to current. You measure the current flowing THROUGH a known resistance and convert that to VOLTAGE. You could measure the voltage across a known resistor and display a voltage but that would NOT be the voltage between the known resistor and the negative terminal, which is what you really want. From the 9V positive terminal if you put a 9K resistor in series with a 1K resistor to the negative battery terminal you have a total resistance across the 9V battery of 10K and because the 1K resistor is 1/10 of the total, the potential between the junction of the 9K/1K and negative is 1/10th of the 9V or 0.9V. If you connect a voltmeter across the 1K resistor with a 10Mohm input resistance almost all of the current will still flow through the 1K resistor and about 1/10,000th will flow through the meter so the reading is basically unaffected. Now if you use a meter with a 1K input resistance, the current from the 9K resistor will split with half going through the 1K resistor and half going through the 1K meter. The equivalent circuit is now 9K in series with 500 ohms (2 parallel 1K) and the meter reading will now be way low. |
| Mr D:
Thanks for the clarification. But is it not indeed the case that when we say we measure voltage or resistance with a DMM, what in fact is happening is that the DMM is measuring current, then telling us (via Ohm's law) what the voltage or resistance is? |
| rstofer:
--- Quote from: Mr D on July 21, 2018, 11:00:07 pm ---Thanks for the clarification. But is it not indeed the case that when we say we measure voltage or resistance with a DMM, what in fact is happening is that the DMM is measuring current, then telling us (via Ohm's law) what the voltage or resistance is? --- End quote --- In the case of the DMM, it is actually measuring the voltage drop across a resistor. It is also true that the voltage drop is proportional to the current flow through the loop. You can look at it either way but the internals of a DMM are fundamentally voltage measuring. The internals of a V-O-M are current measuring. But it doesn't matter because we know Rinternal and we can move back and forth between I & E with ease. Here is a simplified diagram of a V-O-M. Remember that the needle moves in response to current through the winding resistance. https://www.allaboutcircuits.com/textbook/direct-current/chpt-8/multimeters/ I couldn't find a simplified diagram of a DMM but you could just substitute a winding resistor where the meter is shown and then measure the voltage across that resistor. It would be acceptable for discussion if not entirely correct. The DMM measures voltage, the V-O-M measures current. They both get the same answer. Back to the water analogy: I was reminded today that the underlying units for Volts is Joules / Coulomb. How much energy in Joules does it take to move so many Coulombs of charge. An electric potential of 1 Volt will use 1 Joule of energy to move 1 Coulomb of charge. And that's why I did EE and not Physics! http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html#c1 That is the reason we use the water analogy - Volts is just hard to describe when we get down to physical units. We just rename it EMF (Electro Motive Force) or Volts (as a fundamental unit) and move on! Quickly! |
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