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What does applying 0 Volt to a circuit means?

  1. Jan 31, 2012 #1
    0 V is equivalent to short circuit. But applying 0V between 2 points means not connecting any source to those points (that is open circuit). Is not it?
    Please clear my doubt.

    -Devanand T
     
    Last edited: Jan 31, 2012
  2. jcsd
  3. Jan 31, 2012 #2
    I don't think 0V is equivalent to a short circuit. A short circuit implies 0 resistance, and has nothing to do with how much voltage is across it. It just so happens that ohms law relates that there will be 0 volts across a short circuit when current passes through it. You can see 0V on an open circuit, or 0V on 100 ohms of resistance for example. Keep in mind that this is all the abstract model of ohm's law, and we ignore things like noise and only finite currents and voltages that result from the abstract model.

    Volts says how much work an electric field can transfer to a charged particle as it moves between 2 points in the electric field. It says nothing about the impedance(short circuit, open, etc.) between those 2 points without more information available.

    Applying 0V means that the voltage source applied can do no work on charges in the circuit.
     
    Last edited: Jan 31, 2012
  4. Jan 31, 2012 #3
    Perhaps if you made clear what the context is?

    Are you asking about zero volt release switches for instance?

    Dragonpetter made a good reply interms of Ohm's law
     
  5. Jan 31, 2012 #4
    Thank you sir for your reply. one more doubt. Suppose I have to plot input characteristics of a Common base BJT configuration, applying 0V between collector and base (Vcb=0V) means what? Then will we short collector and base or open it.
     
  6. Jan 31, 2012 #5
    An open or a short will both have 0V in the situation you describe, however, you cannot eliminate noise.

    Charge can be induced on either node through electrostatic, triboelectric noise, piezoelectric noise, thermal noise, electromagnetic radiation, etc. The problem with using an open circuit is that if noise adds a net chargeonto one of the nodes, it will have created a voltage on the node, and there is no way for this voltage to quickly and reliably be removed in your measurement.

    By shorting the two nodes, you allow the charges to move from the voltage noise and go back to 0V.

    You can also rely on a voltage source to actively regulate the voltage to 0V, which just means in the real world, it will be removing the charge from either node when a voltage from noise appears, although this can be impractical for high accuracy and depends on your noise.
     
    Last edited: Jan 31, 2012
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