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Author Topic: How do you calculate what the voltage will be after a resistor?  (Read 808 times)
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for some reason all of the formulas for this on google involve current. No.

If you have a 12v battery, you put an X Ohm resistor in series, and you measure the voltage after the resistor (at the green terminals in the image below), what will it be? See how that doesn't involve current? Now what is the formula for that? It should only involve initial voltage, resistance, and final voltage. Current should not be a factor.

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If you have a 12v battery, you put an X Ohm resistor in series, and you measure the voltage after the resistor (at the green terminals in the image below), what will it be?

Twelve volts.
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Yes. Given Ohms law: E = IR

Since I is zero because you stated "it does not involve current" then R is irrelevant and the voltage drop will be nil.
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In practice, to measure voltage you have to draw a small amount of current. So if you put a large resistor (say 1 Megohm) as your "R" and measured with a meter there would be a voltage drop, caused by the measuring current drawn by the meter.
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If you have a 12v battery, you put an X Ohm resistor in series, and you measure the voltage after the resistor (at the green terminals in the image below), what will it be?

Twelve volts.
nope. The voltage varies depending on resistance.
Another example is that same battery and a 30ft wire. the voltage measured at the end of that 30ft wire will not be 12v because of the resistance of the wire. if the resistance of the wire is known, what formula will tell me the resultant voltage?
« Last Edit: February 04, 2013, 07:08:12 pm by David82 » Logged

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The voltage varies depending on resistance.

I never contradicted that.

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Another example is that same battery and a 30ft wire. the voltage measured at the end of that 30ft wire will not be 12v because of the resistance of the wire. if the resistance of the wire is known, what formula will tell me the resultant voltage?

Here's a site for calculating voltage loss in a wire...
http://genuinedealz.com/voltage-drop.html

You'll notice there is an input for Enter Load in amps.  Guess what's missing from your description.
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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.
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As I said, measuring the voltage causes current to flow. My meter claims to have a 10 Megohm input impedance. I just measured the current flowing when measuring 5V with the meter, namely 0.45 uA.

Solving for R:

Code:
R = 5 / 0.45e-6 = 11111111

11 Megohms which agrees roughly with the claim by the manufacturer.

Thus to measure 5V I am drawing 0.45 uA.

Measuring 12V I drew 1.2 uA, which agrees exactly with the claimed performance:

Code:
R = 12 / 1.2e-6 = 10000000

So you can't say "with no current flowing". You might say "with 1.2 uA flowing as drawn by my meter."
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if the resistance of the wire is known, what formula will tell me the resultant voltage?
With a voltage divider, the voltage is divided in proportion to the resistance.   

Or, since the current is the same in both resistors, you can calculate the current from the total resistance and total voltage, and then knowing the current through each resistor, you can calculate the voltage across each resistor.

If you connect a voltmeter to your circuit, the resistance of the voltmeter becomes the 2nd series resistor.   The circuit is completed and a tiny amount of current flows.  (The current is small because the resistance of the meter is high).

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nope. The voltage varies depending on resistance.
Are you measuring a voltage drop with a voltmeter/multimeter?    There is no voltage drop until you connect your meter, (or until you connect something to complete the circuit).  because there is no current flow unitl you connect your meter.    And, you will only measure a voltage drop if the resistance is very high.


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Another example is that same battery and a 30ft wire. the voltage measured at the end of that 30ft wire will not be 12v because of the resistance of the wire.
Again, only if there is current.    And in this case, your meter cannot measure the voltage drop, because the resistance is so low compared to the meter resistance.
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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.
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The fact that you can't measure it accurately doesn't change the fact that it's 12 volts. smiley
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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.)
« Last Edit: February 04, 2013, 07:50:08 pm by DVDdoug » Logged

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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
« Last Edit: February 04, 2013, 08:53:12 pm by retrolefty » Logged

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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.
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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.  smiley-wink
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