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Basic DC theory question

Discussion in 'General Electronics Discussion' started by Ricko, Sep 18, 2012.

  1. Ricko

    Ricko

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    Sep 18, 2012
    If a resistor is connected to a 12 volt DC source in a way that its other end of the resistor is not connected, what will the voltage be at this end of the resistor?
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The question is poorly phrased because voltages are measured with reference to some other point.

    It will have the same voltage as the point it is attached to, however measuring this may cause it to change, and the reference point determines what the 12V source measures.

    If the common point has the voltage source reading 12V, then yes, the voltage on the other end of the resistor is also 12V.
     
  3. wingnut

    wingnut

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    Aug 9, 2012
    What Steve is also saying is that the voltage drop across the resistor is zero.
     
  4. Ricko

    Ricko

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    Sep 18, 2012
    Thanks for your help. To be more concise, if I wanted to measure the voltage on the lead of say a 1k ohm resistor that was attached to a 12 volt battery with respect to the negative terminal of the battery then I would get 12 volts. If it was replaced with a 1M ohm resistor, it would be also 12 volts with respect to the negative terminal, and so on..

    When does this stop? For example, if there was an extremely high resistance attached to the battery , would you still measure 12 volts with respect to the negative terminal? What if the resistance approached an open circuit condition but was not quite open circuit, would you still measure 12 volts?
     
  5. wingnut

    wingnut

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    Aug 9, 2012
    The principle is that in a closed circuit with resistors in series, the voltage drop across any resistor is Vdrop = R/Rt x Vt

    Place resistor upon total resistance times total volts.

    Air is infinite resistance and can be considered as the resistor closing an open circuit.

    So any floating resistor (even billions of ohms) has billions divided by infinite times 12 V which equals zero voltage drop across the billions ohm resistor.

    Thus it would always measure 12V, the same as the battery it is connected to, with no voltage drop.


    Hope this makes sense.
     
    Last edited: Sep 18, 2012
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    If you have a device capable of perfect voltage measurement, then in that case you'll always read 12V.

    However let's assume your real meter has an input impedance of 10Meg.

    The 1k resistor in series with 10M will draw 12/10001000 A, and drop 12*1000/1001000 V = 12/10001 = 0.0012V, so it will read 11.998V -- indistinguishably close to 12V.

    However a 1M resistor in series with 10M will draw 12/11000000 A and drop 12*1000000/11000000 V = 12/11 = 1.091V, so the meter will read 10.909V, significantly different.

    That doesn't mean the voltage is not 12V, it means that as soon as you place a load on it (the meter is a load) the voltage is reduced, so you won't read 12V
     
  7. BillV

    BillV

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    Sep 18, 2012
    The voltage at the resistor will be 12V until its resistance approaches the input impedance of the measuring device, presumably a DVM. This is something that should be listed in the meter's specs and typically is 10 - 100 M ohm. As you approach that the measured voltage will drop as the current drawn by the meter, which also passes through the resistor, causes a voltage drop across the resistor. Old analog meters had much lower impedances.
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The meter will read 50% low if the resistance equals the meter's impedance.

    Depending on your need for accuracy, source impedances (the series resistor) greater than 1% will be too much
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    This circuit is a VOLTAGE DIVIDER.

    Your resistor and the resistance of the voltmeter are connected in SERIES, across the battery, so there is 12V across the series combination. The 12V is divided between the two resistances according to their relative resistances, with the HIGHER resistance having MORE voltage across it.

    Digital multimeters usually have an input resistance of 10 megohms (10,000,000 ohms), which is quite high. If you test with a series resistor of 1 kilohm, the TOTAL resistance is 10,001,000 ohms. The fraction (proportion) of the total applied voltage (12V) that is dropped across each resistor is equal to its resistance as a fraction (proportion) of this total resistance.

    So the voltage across the 1 kilohm resistor will be 12V x (1000 / 10,001,000)
    Which is 0.0012 volts, or 1.2 mV.

    And the voltage across the digital voltmeter will be 12V x (10,000,000 / 10,001,000).
    Which is 11.9988V. That's 1.2 mV shy of 12V, as you would expect, because the 1k resistor is dropping the 1.2 mV to give a total voltage of 12V.

    So you can see that the high input resistance of a digital multimeter on voltage ranges means that it "has little effect" on circuits when you connect it. It doesn't "load down" the circuit much. This is important, obviously; you don't want to disturb the operation of a circuit in the process of measuring it.

    Another way of looking at a voltage divider is using Ohm's law - probably the most important and fundamental law in electronics, although in its simple form, Ohm's law is normally only applied to resistive or "ohmic" circuit elements - and involving CURRENT.

    Ohm's law states: I = V / R.
    I = current (measured in amps)
    V = voltage (measured in volts)
    R = resistance (measured in ohms).

    When your voltmeter is not connected to your resistor, there is no complete CIRCUIT and no PATH through which CURRENT can flow. When you touch the multimeter probe to the resistor, you complete the circuit, and current flows THROUGH the components as defined by Ohm's law.

    The series circuit has a total resistance of 10,001,000 ohms and the total voltage across that resistance is 12V. From Ohm's law, I = 12 / 10,001,000. This works out to 0.00000119988 amps which is 1.19988 microamps (uA unless you can be bothered typing the mu symbol). This is a very small amount of current.

    This is the current THROUGH the series circuit consisting of the 1k resistor and the voltmeter. In a SERIES circuit, the current through each part of the circuit is the SAME.

    The voltage across each resistance can be calculated by rearranging Ohm's law to state V = I R. In each case, I (current) is the same: 1.19988 uA. But the resistances are different.

    For the 1k resistor, V = 1.19988 uA x 1000 ohms
    = 1.19988e-6 x 1e3
    = 1.19988e-3
    = 1.19988 mV.

    For the multimeter, V = 1.19988 uA x 10 megohms
    = 1.19988e-6 x 10e6
    = 11.9988 volts.

    These results are the same as before; this calculation introduces the idea of CURRENT FLOW. You can also see that the current is very low, and this is mostly because the multimeter's input resistance is so high. Most DMMs have an input resistance of 10 megohms, sometimes higher. With most circuits, this is high enough for us to assume that connecting the meter will not significantly affect the circuit being measured. Some circuits that operate at very low currents can be disturbed when the DMM is connected. In fact every circuit is affected to some extent, but most are not disturbed.

    One analogy for voltage (although this is not the textbook analogy) is distance in one dimension (along a single line). Like distance, voltage is always measured BETWEEN two points. When you think about using a tape measure or a ruler to measure the distance between two parts of your house, you might be surprised if the simple act of measuring the distance actually pulled those points towards each other slightly, but that's what happens when you connect a multimeter to measure voltage in a circuit, because it draws current, however small!

    Back in the day, we used multimeters with mechanical indicators (swinging pointer type). These often drew 50 uA from the circuit, at full scale deflection. Connecting the meter would often affect the circuit noticeably.
     
    Last edited: Sep 18, 2012
  10. Ricko

    Ricko

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    Sep 18, 2012
    Thanks guys so much for some awesome explanations! I understand (I think) pretty much all what has been said.

    The sticking point for me is that as the resistance increases, the voltage remains constant at the unattached end of the resistor - because the circuit is open and so no current flows...

    What if that resistance was so high that it approximated an air-gap? With an actual air-gap as a resistor, obviously there would be no voltage at the end of the resistor as there is no physical attachment to the voltage source - common sense would say that the voltage measured would therefore be zero.

    But with a resistor of high enough resistance to approach the resistance of an air-gap, there would still be 12 volts. This messes with my head.

    Maybe the question should be - When does a resistor stop being a resistor, and becomes an air-gap or open circuit?
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    If you consider the air gap as a resistor, yes, as you increase the resistance, the voltage would fall.

    If both resistors were air, we're talking about a fixed point in space some distance from the battery, then you would have a potential determined by the electric field. In some respects this can be considered as defined by a set of resistors with value determined by length connected to all voltage sources.

    Normally we don't need to consider the electric field, but there are devices that can measure it, and what you'd see would be an estimate of the value (since all measurement affects what is being measured).
     
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